Bash script to pass variable across functions - bash

I have removed parentheses, but still I was not able to fetch ENV_NODE value in second function scpTAR. Please let me know what is wrong.
set -x
MASTER_HOSTNAME=`hostname | cut -d . -f1`
TARGET_ENVIRONMENT = it
evaluateEnvProp(){
if [ ${TARGET_ENVIRONMENT} = it ]; then
ENV_NAME=it && ENV_NODE=1cf62108e084
fi
}
scpTAR() {
echo ENV_NODE
echo ${ENV_NODE}
if [ ENV_NODE = ${MASTER_HOSTNAME} ] ; then
echo "scpTAR ENV_NODE = ${MASTER_HOSTNAME} "
else
"echo 'scpTAR ssh other node than jenkins server ENV_NODE=${MASTER_HOSTNAME}'"
fi
}
main(){
scpTAR
}
main

As #cyrus said, variables are global by default. What you did is setting the variables in a subshell:
( ENV_NAME=it && ENV_NODE=xyx && ENV_WLS_DOMAIN=user1 && ENV_NODE_PATH=path )
Because of that, these are gone (not propagated to calling script's environment) once the subshell exists. This is why you do not see their values set in scpTAR function. Remove the parentheses and your code should start working.
Update
Updated version of your code (based on answer by itChi) has another major error. You put spaces around the assignment operator when setting TARGET_ENVIRONMENT = it. This syntax is invalid and as a result TARGET_ENVIRONMENT is not assigned the specified value, thus the condition inside evaluateEnvProp function evaluates to false and ENV_NODE variable is not being set. Removing the spaces should solve the problem. You also did not call evaluateEnvProp as pointed out in update to #itChi's answer.
I'd highly recommend that you start using ShellCheck to verify correctness of your scripts.

As mentioned, variables are global by default, so if you reference them in your scpTAR function, you will get a return value.
However, as a second method, should you wish, you can reference it like so:
scpTAR $ENV_NAME $ENV_NODE $ENV_WLS_DOMAIN $ENV_NODE_PATH
Then in your scpTAR function reference them as:
echo "$1 $2 $3 $4"
it xyx user1 path
Particularly useful should you wish to run code on another machine, run a remote script, or set your script up to pass variables as arguments from bash.
EDIT:
Sorry if I tread on someone's toes, but here is your answer without subshell:
evalEnvProp(){
if [ ${TARGET_ENVIRONMENT} = "it" ]; then
ENV_NAME=it
ENV_NODE=xyx
ENV_WLS_DOMAIN=user1
ENV_NODE_PATH=path
fi
}
scpTAR() {
echo $ENV_NODE_PATH
}
main(){
evalEnvProp
scpTAR
}
main
Update2:
#!/bin/bash
set -x
MASTER_HOSTNAME=`hostname | cut -d . -f1`
TARGET_ENVIRONMENT=it
evaluateEnvProp(){
if [ ${TARGET_ENVIRONMENT} = "it" ]; then
ENV_NAME=it && ENV_NODE=1cf62108e084
fi
}
scpTAR() {
echo ENV_NODE
echo ${ENV_NODE}
if [ ENV_NODE == ${MASTER_HOSTNAME} ] ; then
echo "scpTAR ENV_NODE = ${MASTER_HOSTNAME} " else
"echo 'scpTAR ssh other node than jenkins server >ENV_NODE=${MASTER_HOSTNAME}'"
fi
}
main(){
evaluateEnvProp
scpTAR
}
main
MASTER_HOSTNAME needs `` for the command. you are calling hostname. You can also accomplish this with $()
Spacing in the IF statement either side of the = sign. otherwise you are not evaluating, you are setting a variable.
In your edit you missed out a function call to evaluateEnvProp which failed because you didn't set the variable.

Related

Perl one liner in Bash script

I have a bash script that runs, and I'm trying to use a Perl one-liner to replace some text in a file variables.php
However, I would like to check if the Perl one-liner runs successfully and that's where I get hung up. I could just output the one-liner and it would work fine, but I would like to know for sure that it ran.
Basically, the function replace_variables() is the function that does the update, and it's the if statement there that I would like to check if my one-liner worked properly.
I've tried using the run_command function in that if statement, but that did not work, and I've tried putting the one-liner directly there, which also didn't work.
If I don't wrap it in an if statement, and just call the one-liner directly, everything works as intended.
here's the full file
#!/bin/bash
export CLI_CWD="$PWD"
site_variables() {
if [ -f "$CLI_CWD/variables.php" ]; then
return true
else
return false
fi
}
replace_variables() {
# perl -pi -e 's/(dbuser)(\s+)=\s.*;$/\1 = Config::get("db")["user"];/; s/(dbpass)(\s+)=\s.*;$/\1 = Config::get("db")["pass"];/; s/(dbname)(\s+)=\s.*;$/\1 = Config::get("db")["database"];/' "$CLI_CWD/variables.php"
if [run_command ]; then
echo "Updated variables.php successfully"
else
echo "Did not update variables.php"
fi
}
run_command() {
perl -pi -e 's/(dbuser)(\s+)=\s.*;$/\1 = Config::get("db")["user"];/; s/(dbpass)(\s+)=\s.*;$/\1 = Config::get("db")["pass"];/; s/(dbname)(\s+)=\s.*;$/\1 = Config::get("db")["database"];/' "$CLI_CWD/variables.php"
}
if [ site_variables ]; then
replace_variables
else
>&2 echo "Current directory ($(pwd)) is not a project root directory"
exit 4
fi
here's the function where the if statement fails
replace_variables() {
# perl -pi -e 's/(dbuser)(\s+)=\s.*;$/\1 = Config::get("db")["user"];/; s/(dbpass)(\s+)=\s.*;$/\1 = Config::get("db")["pass"];/; s/(dbname)(\s+)=\s.*;$/\1 = Config::get("db")["database"];/' "$CLI_CWD/variables.php"
if [run_command ]; then
echo "Updated variables.php successfully"
else
echo "Did not update variables.php"
fi
}
You can see that I commented out the one-liner just before the if statement, it works if I let that run and remove the if/else check.
here is the original file snippet before the update
//Load from Settings DB
$dbuser = 'username';
$dbpass = 'password';
$dbname = 'database_name';
here is the file snippet after the update would run
//Load from Settings DB
$dbuser = Config::get("db")["user"];
$dbpass = Config::get("db")["pass"];
$dbname = Config::get("db")["database"];
tl;dr and Solution
This usage of if with [ ] will not give you the result you expect.
What you're looking for
...
if run_command; then
...
Longer explanation
Basics of if
if is a shell feature
based on the condition, it executes the body contained in between then and fi
the "condition" that if checks is a command
commands usually have a return/exit code. typically
0 for success
1 (common) and everything else for some error
e.g. 127 for command not found
when the return/exit code is 0, the body is executed
otherwise it is skipped; or control is passed to elif or else
the syntax is if <command>; then...
Where does that [ ] come from?
test is a command that can check file types and compare values
refer man test and help test (bash only)
[ ... ] is a synonym for test
NB the brackets should be surrounded by spaces on both sides
if [ -f "/path/to/$filename" ]; then
exception: when terminated by new line or ; space not required
test (or [ ]) evaluates expressions and cannot execute other commands or functions
if [ expr ]; then is alternate syntax for if test expr; then
PS: good practice to "quote" your "$variables" when used with test or [ ]
PPS: [[ ... ]] is a different thing altogether. not POSIX; available only in some shells. take a look at this thread on the UNIX Stack Exchange

returning values from functions in bash [duplicate]

I'd like to return a string from a Bash function.
I'll write the example in java to show what I'd like to do:
public String getSomeString() {
return "tadaa";
}
String variable = getSomeString();
The example below works in bash, but is there a better way to do this?
function getSomeString {
echo "tadaa"
}
VARIABLE=$(getSomeString)
There is no better way I know of. Bash knows only status codes (integers) and strings written to the stdout.
You could have the function take a variable as the first arg and modify the variable with the string you want to return.
#!/bin/bash
set -x
function pass_back_a_string() {
eval "$1='foo bar rab oof'"
}
return_var=''
pass_back_a_string return_var
echo $return_var
Prints "foo bar rab oof".
Edit: added quoting in the appropriate place to allow whitespace in string to address #Luca Borrione's comment.
Edit: As a demonstration, see the following program. This is a general-purpose solution: it even allows you to receive a string into a local variable.
#!/bin/bash
set -x
function pass_back_a_string() {
eval "$1='foo bar rab oof'"
}
return_var=''
pass_back_a_string return_var
echo $return_var
function call_a_string_func() {
local lvar=''
pass_back_a_string lvar
echo "lvar='$lvar' locally"
}
call_a_string_func
echo "lvar='$lvar' globally"
This prints:
+ return_var=
+ pass_back_a_string return_var
+ eval 'return_var='\''foo bar rab oof'\'''
++ return_var='foo bar rab oof'
+ echo foo bar rab oof
foo bar rab oof
+ call_a_string_func
+ local lvar=
+ pass_back_a_string lvar
+ eval 'lvar='\''foo bar rab oof'\'''
++ lvar='foo bar rab oof'
+ echo 'lvar='\''foo bar rab oof'\'' locally'
lvar='foo bar rab oof' locally
+ echo 'lvar='\'''\'' globally'
lvar='' globally
Edit: demonstrating that the original variable's value is available in the function, as was incorrectly criticized by #Xichen Li in a comment.
#!/bin/bash
set -x
function pass_back_a_string() {
eval "echo in pass_back_a_string, original $1 is \$$1"
eval "$1='foo bar rab oof'"
}
return_var='original return_var'
pass_back_a_string return_var
echo $return_var
function call_a_string_func() {
local lvar='original lvar'
pass_back_a_string lvar
echo "lvar='$lvar' locally"
}
call_a_string_func
echo "lvar='$lvar' globally"
This gives output:
+ return_var='original return_var'
+ pass_back_a_string return_var
+ eval 'echo in pass_back_a_string, original return_var is $return_var'
++ echo in pass_back_a_string, original return_var is original return_var
in pass_back_a_string, original return_var is original return_var
+ eval 'return_var='\''foo bar rab oof'\'''
++ return_var='foo bar rab oof'
+ echo foo bar rab oof
foo bar rab oof
+ call_a_string_func
+ local 'lvar=original lvar'
+ pass_back_a_string lvar
+ eval 'echo in pass_back_a_string, original lvar is $lvar'
++ echo in pass_back_a_string, original lvar is original lvar
in pass_back_a_string, original lvar is original lvar
+ eval 'lvar='\''foo bar rab oof'\'''
++ lvar='foo bar rab oof'
+ echo 'lvar='\''foo bar rab oof'\'' locally'
lvar='foo bar rab oof' locally
+ echo 'lvar='\'''\'' globally'
lvar='' globally
All answers above ignore what has been stated in the man page of bash.
All variables declared inside a function will be shared with the calling environment.
All variables declared local will not be shared.
Example code
#!/bin/bash
f()
{
echo function starts
local WillNotExists="It still does!"
DoesNotExists="It still does!"
echo function ends
}
echo $DoesNotExists #Should print empty line
echo $WillNotExists #Should print empty line
f #Call the function
echo $DoesNotExists #Should print It still does!
echo $WillNotExists #Should print empty line
And output
$ sh -x ./x.sh
+ echo
+ echo
+ f
+ echo function starts
function starts
+ local 'WillNotExists=It still does!'
+ DoesNotExists='It still does!'
+ echo function ends
function ends
+ echo It still 'does!'
It still does!
+ echo
Also under pdksh and ksh this script does the same!
Bash, since version 4.3, feb 2014(?), has explicit support for reference variables or name references (namerefs), beyond "eval", with the same beneficial performance and indirection effect, and which may be clearer in your scripts and also harder to "forget to 'eval' and have to fix this error":
declare [-aAfFgilnrtux] [-p] [name[=value] ...]
typeset [-aAfFgilnrtux] [-p] [name[=value] ...]
Declare variables and/or give them attributes
...
-n Give each name the nameref attribute, making it a name reference
to another variable. That other variable is defined by the value
of name. All references and assignments to name, except for⋅
changing the -n attribute itself, are performed on the variable
referenced by name's value. The -n attribute cannot be applied to
array variables.
...
When used in a function, declare and typeset make each name local,
as with the local command, unless the -g option is supplied...
and also:
PARAMETERS
A variable can be assigned the nameref attribute using the -n option to the
declare or local builtin commands (see the descriptions of declare and local
below) to create a nameref, or a reference to another variable. This allows
variables to be manipulated indirectly. Whenever the nameref variable is⋅
referenced or assigned to, the operation is actually performed on the variable
specified by the nameref variable's value. A nameref is commonly used within
shell functions to refer to a variable whose name is passed as an argument to⋅
the function. For instance, if a variable name is passed to a shell function
as its first argument, running
declare -n ref=$1
inside the function creates a nameref variable ref whose value is the variable
name passed as the first argument. References and assignments to ref are
treated as references and assignments to the variable whose name was passed as⋅
$1. If the control variable in a for loop has the nameref attribute, the list
of words can be a list of shell variables, and a name reference will be⋅
established for each word in the list, in turn, when the loop is executed.
Array variables cannot be given the -n attribute. However, nameref variables
can reference array variables and subscripted array variables. Namerefs can be⋅
unset using the -n option to the unset builtin. Otherwise, if unset is executed
with the name of a nameref variable as an argument, the variable referenced by⋅
the nameref variable will be unset.
For example (EDIT 2: (thank you Ron) namespaced (prefixed) the function-internal variable name, to minimize external variable clashes, which should finally answer properly, the issue raised in the comments by Karsten):
# $1 : string; your variable to contain the return value
function return_a_string () {
declare -n ret=$1
local MYLIB_return_a_string_message="The date is "
MYLIB_return_a_string_message+=$(date)
ret=$MYLIB_return_a_string_message
}
and testing this example:
$ return_a_string result; echo $result
The date is 20160817
Note that the bash "declare" builtin, when used in a function, makes the declared variable "local" by default, and "-n" can also be used with "local".
I prefer to distinguish "important declare" variables from "boring local" variables, so using "declare" and "local" in this way acts as documentation.
EDIT 1 - (Response to comment below by Karsten) - I cannot add comments below any more, but Karsten's comment got me thinking, so I did the following test which WORKS FINE, AFAICT - Karsten if you read this, please provide an exact set of test steps from the command line, showing the problem you assume exists, because these following steps work just fine:
$ return_a_string ret; echo $ret
The date is 20170104
(I ran this just now, after pasting the above function into a bash term - as you can see, the result works just fine.)
Like bstpierre above, I use and recommend the use of explicitly naming output variables:
function some_func() # OUTVAR ARG1
{
local _outvar=$1
local _result # Use some naming convention to avoid OUTVARs to clash
... some processing ....
eval $_outvar=\$_result # Instead of just =$_result
}
Note the use of quoting the $. This will avoid interpreting content in $result as shell special characters. I have found that this is an order of magnitude faster than the result=$(some_func "arg1") idiom of capturing an echo. The speed difference seems even more notable using bash on MSYS where stdout capturing from function calls is almost catastrophic.
It's ok to send in a local variables since locals are dynamically scoped in bash:
function another_func() # ARG
{
local result
some_func result "$1"
echo result is $result
}
You could also capture the function output:
#!/bin/bash
function getSomeString() {
echo "tadaa!"
}
return_var=$(getSomeString)
echo $return_var
# Alternative syntax:
return_var=`getSomeString`
echo $return_var
Looks weird, but is better than using global variables IMHO. Passing parameters works as usual, just put them inside the braces or backticks.
The most straightforward and robust solution is to use command substitution, as other people wrote:
assign()
{
local x
x="Test"
echo "$x"
}
x=$(assign) # This assigns string "Test" to x
The downside is performance as this requires a separate process.
The other technique suggested in this topic, namely passing the name of a variable to assign to as an argument, has side effects, and I wouldn't recommend it in its basic form. The problem is that you will probably need some variables in the function to calculate the return value, and it may happen that the name of the variable intended to store the return value will interfere with one of them:
assign()
{
local x
x="Test"
eval "$1=\$x"
}
assign y # This assigns string "Test" to y, as expected
assign x # This will NOT assign anything to x in this scope
# because the name "x" is declared as local inside the function
You might, of course, not declare internal variables of the function as local, but you really should always do it as otherwise you may, on the other hand, accidentally overwrite an unrelated variable from the parent scope if there is one with the same name.
One possible workaround is an explicit declaration of the passed variable as global:
assign()
{
local x
eval declare -g $1
x="Test"
eval "$1=\$x"
}
If name "x" is passed as an argument, the second row of the function body will overwrite the previous local declaration. But the names themselves might still interfere, so if you intend to use the value previously stored in the passed variable prior to write the return value there, be aware that you must copy it into another local variable at the very beginning; otherwise the result will be unpredictable!
Besides, this will only work in the most recent version of BASH, namely 4.2. More portable code might utilize explicit conditional constructs with the same effect:
assign()
{
if [[ $1 != x ]]; then
local x
fi
x="Test"
eval "$1=\$x"
}
Perhaps the most elegant solution is just to reserve one global name for function return values and
use it consistently in every function you write.
As previously mentioned, the "correct" way to return a string from a function is with command substitution. In the event that the function also needs to output to console (as #Mani mentions above), create a temporary fd in the beginning of the function and redirect to console. Close the temporary fd before returning your string.
#!/bin/bash
# file: func_return_test.sh
returnString() {
exec 3>&1 >/dev/tty
local s=$1
s=${s:="some default string"}
echo "writing directly to console"
exec 3>&-
echo "$s"
}
my_string=$(returnString "$*")
echo "my_string: [$my_string]"
executing script with no params produces...
# ./func_return_test.sh
writing directly to console
my_string: [some default string]
hope this helps people
-Andy
You could use a global variable:
declare globalvar='some string'
string ()
{
eval "$1='some other string'"
} # ---------- end of function string ----------
string globalvar
echo "'${globalvar}'"
This gives
'some other string'
To illustrate my comment on Andy's answer, with additional file descriptor manipulation to avoid use of /dev/tty:
#!/bin/bash
exec 3>&1
returnString() {
exec 4>&1 >&3
local s=$1
s=${s:="some default string"}
echo "writing to stdout"
echo "writing to stderr" >&2
exec >&4-
echo "$s"
}
my_string=$(returnString "$*")
echo "my_string: [$my_string]"
Still nasty, though.
The way you have it is the only way to do this without breaking scope. Bash doesn't have a concept of return types, just exit codes and file descriptors (stdin/out/err, etc)
Addressing Vicky Ronnen's head up, considering the following code:
function use_global
{
eval "$1='changed using a global var'"
}
function capture_output
{
echo "always changed"
}
function test_inside_a_func
{
local _myvar='local starting value'
echo "3. $_myvar"
use_global '_myvar'
echo "4. $_myvar"
_myvar=$( capture_output )
echo "5. $_myvar"
}
function only_difference
{
local _myvar='local starting value'
echo "7. $_myvar"
local use_global '_myvar'
echo "8. $_myvar"
local _myvar=$( capture_output )
echo "9. $_myvar"
}
declare myvar='global starting value'
echo "0. $myvar"
use_global 'myvar'
echo "1. $myvar"
myvar=$( capture_output )
echo "2. $myvar"
test_inside_a_func
echo "6. $_myvar" # this was local inside the above function
only_difference
will give
0. global starting value
1. changed using a global var
2. always changed
3. local starting value
4. changed using a global var
5. always changed
6.
7. local starting value
8. local starting value
9. always changed
Maybe the normal scenario is to use the syntax used in the test_inside_a_func function, thus you can use both methods in the majority of cases, although capturing the output is the safer method always working in any situation, mimicking the returning value from a function that you can find in other languages, as Vicky Ronnen correctly pointed out.
The options have been all enumerated, I think. Choosing one may come down to a matter of the best style for your particular application, and in that vein, I want to offer one particular style I've found useful. In bash, variables and functions are not in the same namespace. So, treating the variable of the same name as the value of the function is a convention that I find minimizes name clashes and enhances readability, if I apply it rigorously. An example from real life:
UnGetChar=
function GetChar() {
# assume failure
GetChar=
# if someone previously "ungot" a char
if ! [ -z "$UnGetChar" ]; then
GetChar="$UnGetChar"
UnGetChar=
return 0 # success
# else, if not at EOF
elif IFS= read -N1 GetChar ; then
return 0 # success
else
return 1 # EOF
fi
}
function UnGetChar(){
UnGetChar="$1"
}
And, an example of using such functions:
function GetToken() {
# assume failure
GetToken=
# if at end of file
if ! GetChar; then
return 1 # EOF
# if start of comment
elif [[ "$GetChar" == "#" ]]; then
while [[ "$GetChar" != $'\n' ]]; do
GetToken+="$GetChar"
GetChar
done
UnGetChar "$GetChar"
# if start of quoted string
elif [ "$GetChar" == '"' ]; then
# ... et cetera
As you can see, the return status is there for you to use when you need it, or ignore if you don't. The "returned" variable can likewise be used or ignored, but of course only after the function is invoked.
Of course, this is only a convention. You are free to fail to set the associated value before returning (hence my convention of always nulling it at the start of the function) or to trample its value by calling the function again (possibly indirectly). Still, it's a convention I find very useful if I find myself making heavy use of bash functions.
As opposed to the sentiment that this is a sign one should e.g. "move to perl", my philosophy is that conventions are always important for managing the complexity of any language whatsoever.
In my programs, by convention, this is what the pre-existing $REPLY variable is for, which read uses for that exact purpose.
function getSomeString {
REPLY="tadaa"
}
getSomeString
echo $REPLY
This echoes
tadaa
But to avoid conflicts, any other global variable will do.
declare result
function getSomeString {
result="tadaa"
}
getSomeString
echo $result
If that isn’t enough, I recommend Markarian451’s solution.
They key problem of any 'named output variable' scheme where the caller can pass in the variable name (whether using eval or declare -n) is inadvertent aliasing, i.e. name clashes: From an encapsulation point of view, it's awful to not be able to add or rename a local variable in a function without checking ALL the function's callers first to make sure they're not wanting to pass that same name as the output parameter. (Or in the other direction, I don't want to have to read the source of the function I'm calling just to make sure the output parameter I intend to use is not a local in that function.)
The only way around that is to use a single dedicated output variable like REPLY (as suggested by Evi1M4chine) or a convention like the one suggested by Ron Burk.
However, it's possible to have functions use a fixed output variable internally, and then add some sugar over the top to hide this fact from the caller, as I've done with the call function in the following example. Consider this a proof of concept, but the key points are
The function always assigns the return value to REPLY, and can also return an exit code as usual
From the perspective of the caller, the return value can be assigned to any variable (local or global) including REPLY (see the wrapper example). The exit code of the function is passed through, so using them in e.g. an if or while or similar constructs works as expected.
Syntactically the function call is still a single simple statement.
The reason this works is because the call function itself has no locals and uses no variables other than REPLY, avoiding any potential for name clashes. At the point where the caller-defined output variable name is assigned, we're effectively in the caller's scope (technically in the identical scope of the call function), rather than in the scope of the function being called.
#!/bin/bash
function call() { # var=func [args ...]
REPLY=; "${1#*=}" "${#:2}"; eval "${1%%=*}=\$REPLY; return $?"
}
function greet() {
case "$1" in
us) REPLY="hello";;
nz) REPLY="kia ora";;
*) return 123;;
esac
}
function wrapper() {
call REPLY=greet "$#"
}
function main() {
local a b c d
call a=greet us
echo "a='$a' ($?)"
call b=greet nz
echo "b='$b' ($?)"
call c=greet de
echo "c='$c' ($?)"
call d=wrapper us
echo "d='$d' ($?)"
}
main
Output:
a='hello' (0)
b='kia ora' (0)
c='' (123)
d='hello' (0)
You can echo a string, but catch it by piping (|) the function to something else.
You can do it with expr, though ShellCheck reports this usage as deprecated.
bash pattern to return both scalar and array value objects:
definition
url_parse() { # parse 'url' into: 'url_host', 'url_port', ...
local "$#" # inject caller 'url' argument in local scope
local url_host="..." url_path="..." # calculate 'url_*' components
declare -p ${!url_*} # return only 'url_*' object fields to the caller
}
invocation
main() { # invoke url parser and inject 'url_*' results in local scope
eval "$(url_parse url=http://host/path)" # parse 'url'
echo "host=$url_host path=$url_path" # use 'url_*' components
}
Although there were a lot of good answers, they all did not work the way I wanted them to. So here is my solution with these key points:
Helping the forgetful programmer
Atleast I would struggle to always remember error checking after something like this: var=$(myFunction)
Allows assigning values with newline chars \n
Some solutions do not allow for that as some forgot about the single quotes around the value to assign. Right way: eval "${returnVariable}='${value}'" or even better: see the next point below.
Using printf instead of eval
Just try using something like this myFunction "date && var2" to some of the supposed solutions here. eval will execute whatever is given to it. I only want to assign values so I use printf -v "${returnVariable}" "%s" "${value}" instead.
Encapsulation and protection against variable name collision
If a different user or at least someone with less knowledge about the function (this is likely me in some months time) is using myFunction I do not want them to know that he must use a global return value name or some variable names are forbidden to use. That is why I added a name check at the top of myFunction:
if [[ "${1}" = "returnVariable" ]]; then
echo "Cannot give the ouput to \"returnVariable\" as a variable with the same name is used in myFunction()!"
echo "If that is still what you want to do please do that outside of myFunction()!"
return 1
fi
Note this could also be put into a function itself if you have to check a lot of variables.
If I still want to use the same name (here: returnVariable) I just create a buffer variable, give that to myFunction and then copy the value returnVariable.
So here it is:
myFunction():
myFunction() {
if [[ "${1}" = "returnVariable" ]]; then
echo "Cannot give the ouput to \"returnVariable\" as a variable with the same name is used in myFunction()!"
echo "If that is still what you want to do please do that outside of myFunction()!"
return 1
fi
if [[ "${1}" = "value" ]]; then
echo "Cannot give the ouput to \"value\" as a variable with the same name is used in myFunction()!"
echo "If that is still what you want to do please do that outside of myFunction()!"
return 1
fi
local returnVariable="${1}"
local value=$'===========\nHello World\n==========='
echo "setting the returnVariable now..."
printf -v "${returnVariable}" "%s" "${value}"
}
Test cases:
var1="I'm not greeting!"
myFunction var1
[[ $? -eq 0 ]] && echo "myFunction(): SUCCESS" || echo "myFunction(): FAILURE"
printf "var1:\n%s\n" "${var1}"
# Output:
# setting the returnVariable now...
# myFunction(): SUCCESS
# var1:
# ===========
# Hello World
# ===========
returnVariable="I'm not greeting!"
myFunction returnVariable
[[ $? -eq 0 ]] && echo "myFunction(): SUCCESS" || echo "myFunction(): FAILURE"
printf "returnVariable:\n%s\n" "${returnVariable}"
# Output
# Cannot give the ouput to "returnVariable" as a variable with the same name is used in myFunction()!
# If that is still what you want to do please do that outside of myFunction()!
# myFunction(): FAILURE
# returnVariable:
# I'm not greeting!
var2="I'm not greeting!"
myFunction "date && var2"
[[ $? -eq 0 ]] && echo "myFunction(): SUCCESS" || echo "myFunction(): FAILURE"
printf "var2:\n%s\n" "${var2}"
# Output
# setting the returnVariable now...
# ...myFunction: line ..: printf: `date && var2': not a valid identifier
# myFunction(): FAILURE
# var2:
# I'm not greeting!
myFunction var3
[[ $? -eq 0 ]] && echo "myFunction(): SUCCESS" || echo "myFunction(): FAILURE"
printf "var3:\n%s\n" "${var3}"
# Output
# setting the returnVariable now...
# myFunction(): SUCCESS
# var3:
# ===========
# Hello World
# ===========
#Implement a generic return stack for functions:
STACK=()
push() {
STACK+=( "${1}" )
}
pop() {
export $1="${STACK[${#STACK[#]}-1]}"
unset 'STACK[${#STACK[#]}-1]';
}
#Usage:
my_func() {
push "Hello world!"
push "Hello world2!"
}
my_func ; pop MESSAGE2 ; pop MESSAGE1
echo ${MESSAGE1} ${MESSAGE2}
agt#agtsoft:~/temp$ cat ./fc
#!/bin/sh
fcall='function fcall { local res p=$1; shift; fname $*; eval "$p=$res"; }; fcall'
function f1 {
res=$[($1+$2)*2];
}
function f2 {
local a;
eval ${fcall//fname/f1} a 2 3;
echo f2:$a;
}
a=3;
f2;
echo after:a=$a, res=$res
agt#agtsoft:~/temp$ ./fc
f2:10
after:a=3, res=

Is there an elegant way to store and evaluate return values in bash scripts?

I have a rather complex series of commands in bash that ends up returning a meaningful exit code. Various places later in the script need to branch conditionally on whether the command set succeed or not.
Currently I am storing the exit code and testing it numerically, something like this:
long_running_command | grep -q trigger_word
status=$?
if [ $status -eq 0 ]; then
: stuff
else
: more code
if [ $status -eq 0 ]; then
: stuff
else
For some reason it feels like this should be simpler. We have a simple exit code stored and now we are repeatedly typing out numerical test operations to run on it. For example I can cheat use the string output instead of the return code which is simpler to test for:
status=$(long_running_command | grep trigger_word)
if [ $status ]; then
: stuff
else
: more code
if [ $status ]; then
: stuff
else
On the surface this looks more straight forward, but I realize it's dirty.
If the other logic wasn't so complex and I was only running this once, I realize I could embed it in place of the test operator, but this is not ideal when you need to reuse the results in other locations without re-running the test:
if long_running_command | grep -q trigger_word; then
: stuff
else
The only thing I've found so far is assigning the code as part of command substitution:
status=$(long_running_command | grep -q trigger_word; echo $?)
if [ $status -eq 0 ]; then
: stuff
else
Even this is not technically a one shot assignment (although some may argue the readability is better) but the necessary numerical test syntax still seems cumbersome to me. Maybe I'm just being OCD.
Am I missing a more elegant way to assign an exit code to a variable then branch on it later?
The simple solution:
output=$(complex_command)
status=$?
if (( status == 0 )); then
: stuff with "$output"
fi
: more code
if (( status == 0 )); then
: stuff with "$output"
fi
Or more eleganter-ish
do_complex_command () {
# side effects: global variables
# store the output in $g_output and the status in $g_status
g_output=$(
command -args | commands | grep -q trigger_word
)
g_status=$?
}
complex_command_succeeded () {
test $g_status -eq 0
}
complex_command_output () {
echo "$g_output"
}
do_complex_command
if complex_command_succeeded; then
: stuff with "$(complex_command_output)"
fi
: more code
if complex_command_succeeded; then
: stuff with "$(complex_command_output)"
fi
Or
do_complex_command () {
# side effects: global variables
# store the output in $g_output and the status in $g_status
g_output=$(
command -args | commands
)
g_status=$?
}
complex_command_output () {
echo "$g_output"
}
complex_command_contains_keyword () {
complex_command_output | grep -q "$1"
}
if complex_command_contains_keyword "trigger_word"; then
: stuff with "$(complex_command_output)"
fi
If you don't need to store the specific exit status, just whether the command succeeded or failed (e.g. whether grep found a match), I's use a fake boolean variable to store the result:
if long_running_command | grep trigger_word; then
found_trigger=true
else
found_trigger=false
fi
# ...later...
if ! $found_trigger; then
# stuff to do if the trigger word WASN'T found
fi
#...
if $found_trigger; then
# stuff to do if the trigger WAS found
fi
Notes:
The shell doesn't really have boolean (true/false) variables. What's actually happening here is that "true" and "false" are stored as strings in the found_trigger variable; when if $found_trigger; then executes, it runs the value of $found_trigger as a command, and it just happens that the true command always succeeds and the false command always fails, thus causing "the right thing" to happen. In if ! $found_trigger; then, the "!" toggles the success/failure status, effectively acting as a boolean "not".
if long_running_command | grep trigger_word; then is equivalent to running the command, then using if [ $? -ne 0 ]; then to check its exit status. I find it a little cleaner, but you have to get used to thinking of if as checking the success/failure of a command, not just testing boolean conditions. If "active" if commands aren't intuitive to you, use a separate test instead.
As Charles Duffy pointed out in a comment, this trick executes data as a command, and if you don't have full control over that data... you don't have control over what your script is going to do. So never set a fake-boolean variable to anything other than the fixed strings "true" and "false", and be sure to set the variable before using it. If you have any nontrivial execution flow in the script, set all fake-boolean variables to sane default values (i.e. "true" or "false") before the execution flow gets complicated.
Failure to follow these rules can lead to security holes large enough to drive a freight train through.
Why don't you set flags for the stuff that needs to happen later?
cheeseballs=false
nachos=false
guppies=false
command
case $? in
42) cheeseballs=true ;;
17 | 31) cheeseballs=true; nachos=true; guppies=true;;
66) guppies=true; echo "Bingo!";;
esac
$cheeseballs && java -crash -burn
$nachos && python ./tex.py --mex
if $guppies; then
aquarium --light=blue --door=hidden --decor=squid
else
echo SRY
fi
As pointed out by #CharlesDuffy in the comments, storing an actual command in a variable is slightly dubious, and vaguely triggers Bash FAQ #50 warnings; the code reads (slightly & IMHO) more naturally like this, but you have to be really careful that you have total control over the variables at all times. If you have the slightest doubt, perhaps just use string values and compare against the expected value at each junction.
[ "$cheeseballs" = "true" ] && java -crash -burn
etc etc; or you could refactor to some other implementation structure for the booleans (an associative array of options would make sense, but isn't portable to POSIX sh; a PATH-like string is flexible, but perhaps too unstructured).
Based on the OP's clarification that it's only about success v. failure (as opposed to the specific exit codes):
long_running_command | grep -q trigger_word || failed=1
if ((!failed)); then
: stuff
else
: more code
if ((!failed)); then
: stuff
else
Sets the success-indicator variable only on failure (via ||, i.e, if a non-zero exit code is returned).
Relies on the fact that variables that aren't defined evaluate to false in an arithmetic conditional (( ... )).
Care must be taken that the variable ($failed, in this example) hasn't accidentally been initialized elsewhere.
(On a side note, as #nos has already mentioned in a comment, you need to be careful with commands involving a pipeline; from man bash (emphasis mine):
The return status of a pipeline is the exit status of the last command,
unless the pipefail option is enabled. If pipefail is enabled, the
pipeline's return status is the value of the last (rightmost) command
to exit with a non-zero status, or zero if all commands exit successfully.
To set pipefail (which is OFF by default), use set -o pipefail; to turn it back off, use set +o pipefail.)
If you don't care about the exact error code, you could do:
if long_running_command | grep -q trigger_word; then
success=1
: success
else
success=0
: failure
fi
if ((success)); then
: success
else
: failure
fi
Using 0 for false and 1 for true is my preferred way of storing booleans in scripts. if ((flag)) mimics C nicely.
If you do care about the exit code, then you could do:
if long_running_command | grep -q trigger_word; then
status=0
: success
else
status=$?
: failure
fi
if ((status == 0)); then
: success
else
: failure
fi
I prefer an explicit test against 0 rather than using !, which doesn't read right.
(And yes, $? does yield the correct value here.)
Hmm, the problem is a bit vague - if possible, I suggest considering refactoring/simplify, i.e.
function check_your_codes {
# ... run all 'checks' and store the results in an array
}
###
function process_results {
# do your 'stuff' based on array values
}
###
create_My_array
check_your_codes
process_results
Also, unless you really need to save the exit code then there is no need to store_and_test - just test_and_do, i.e. use a case statement as suggested above or something like:
run_some_commands_and_return_EXIT_CODE_FROM_THE_LAST_ONE
if [[ $? -eq 0 ]] ; then do_stuff else do_other_stuff ; fi
:)
Dale

Why do I get different bash script results when invoked with 'set -x', and how do I fix it?

I've found that the results of my bash script will change depending upon if I execute it with debugging or not (i.e. invoking set -x). I don't mean that I get more output, but that the result of the program itself differs.
I'm assuming this isn't the desired behavior, and I'm hoping that you can teach me how to correc this.
The bash script below is a contrived example, I tried reducing the logic from the script I'm investigating so that the problem can be easily reproducible and obvious.
#!/bin/bash
# Base function executes command (print working directory) stores the value in
# the destination and returns the status.
function get_cur_dir {
local dest=$1
local result
result=$((pwd) 2>&1)
status=$?
eval $dest="\"$result\""
return $status
}
# 2nd level function uses the base function to execute the command and store
# the result in the desired location. However if the base function fails it
# terminates the script. Yes, I know 'pwd' won't fail -- this is a contrived
# example to illustrate the types of problems I am seeing.
function get_cur_dir_nofail {
local dest=$1
local gcdnf_result
local status
get_cur_dir gcdnf_result
status=$?
if [ $status -ne 0 ]; then
echo "ERROR: Command failure"
exit 1
fi
eval dest="\"$gcdnf_result\""
}
# Cause blarg to be loaded with the current directory, use the results to
# create a flag_file name, and do logic with the flag_file name.
function main {
get_cur_dir blarg
echo "Current diregtory is:$blarg"
local flag_file=/tmp/$blarg.flag
echo -e ">>>>>>>> $flag_file"
if [ "/tmp//root.flag" = "$flag_file" ]; then
echo "Match"
else
echo "No Match"
fi
}
main
.
.
When I execute without the set -x it works as I expect as illustrated below:
Current diregtory is:/root
>>>>>>>> /tmp//root.flag
Match
.
.
However, when I add the debugging output with -x it doesn't work, as illustrated below:
root#psbu-jrr-lnx:# bash -x /tmp/example.sh
+ main
+ get_cur_dir blarg
+ local dest=blarg
+ local result
+ result='++ pwd
/root'
+ status=0
+ eval 'blarg="++ pwd
/root"'
++ blarg='++ pwd
/root'
+ return 0
+ echo 'Current diregtory is:++ pwd
/root'
Current diregtory is:++ pwd
/root
+ local 'flag_file=/tmp/++ pwd
/root.flag'
+ echo -e '>>>>>>>> /tmp/++ pwd
/root.flag'
>>>>>>>> /tmp/++ pwd
/root.flag
+ '[' /tmp//root.flag = '/tmp/++ pwd
/root.flag' ']'
+ echo 'No Match'
No Match
root#psbu-jrr-lnx:#
I think what happens is you capture the debugging logging output produced by the shell when you run it with set -x, this line, for example, does it:
result=$((pwd) 2>&1)
In the above line you shouldn't really need to redirect standard error to standard output, so remove 2>&1.
Changing...
result=$((pwd) 2>&1)
...into...
result=$(pwd 2>&1)
...will allow you to capture the output of pwd without capturing the debug info generated by set -x.
The reason the the $PWD variable exists is to free your script from having to run a separate process or interpret its output (which in this case has been modified by -x). Use $PWD instead.

Set in a bash script vs function parameter

I am changing a bash script which has a structure as such:
#somewhere in the code
sim_counts=#... some value
function_name()
{
set $sim_counts
for hostname in $linux_hostnames; do
if [ $1 -eq 0 ]; then # if sim_counts equal 0
shift # jump forward in sim_counts
continue
fi
# ... more code
shift
done
}
Then it is called in the script:
function_name
I want to introduce a parameter to this function:
#somewhere in the code
sim_counts=#... some value
function_name()
{
ip=$1
set $sim_counts
for hostname in $linux_hostnames; do
if [ $1 -eq 0 ]; then # if sim_counts equal 0
shift # jump forward in sim_counts
continue
fi
# ... more code
shift
done
}
And call the function in following way:
function_name 10.255.192.123
What should I do to avoid $1 conflict of function parameter and the other value from set command ?
If I am correctly reading the set builtin page in the Bash Reference Manual, I believe the code as you have written it will just work. Quoting from that page:
The remaining N arguments are positional parameters and are assigned, in order, to $1, $2, … $N. The special parameter # is set to N.
In essence, any pre-existing values for the positional variables will be blown away. The first sentence on that manual page is also interesting:
This builtin is so complicated that it deserves its own section.
In short, I think your code should simply work as expected. You've saved the initial value of $1 (from the function call) into a temporary variable; as long as you refer to $ip for that specific value, you should be good. In my own test script, it seems that $1 gets blown away as I expect it should.

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