I am new to Bash scripting, having a lot more experience with C-type languages. I have written a few scripts with a conditional that checks the value of a non-instantiated variable and if it doesn't exist or match a value sets the variable. On top of that the whole thing is in a for loop. Something like this:
for i in ${!my_array[#]}; do
if [ "${my_array[i]}" = true ]
then
#do something
else
my_array[i]=true;
fi
done
This would fail through a null pointer in Java since my_array[i] is not instantiated until after it is checked. Is this good practice in Bash? My script is working the way I designed, but I have learned that just because a kluge works now doesn't mean it will work in the future.
Thanks!
You will find this page on parameter expansion helpful, as well as this one on conditionals.
An easy way to test a variable is to check it for nonzero length.
if [[ -n "$var" ]]
then : do stuff ...
I also like to make it fatal to access a nonexisting variable; this means extra work, but better safety.
set -u # unset vars are fatal to access without exception handling
if [[ -n "${var:-}" ]] # handles unset during check
then : do stuff ...
By default, referencing undefined (or "unset") variable names in shell scripts just gives the empty string. But is an exception: if the shell is run with the -u option or set -u has been run in it, expansions of unset variables are treated as errors and (if the shell is not interactive) cause the shell to exit. Bash applies this principle to array elements as well:
$ array=(zero one two)
$ echo "${array[3]}"
$ echo "array[3] = '${array[3]}'"
array[3] = ''
$ set -u
$ echo "array[3] = '${array[3]}'"
-bash: array[3]: unbound variable
There are also modifiers you can use to control what expansions do if a variable (or array element) is undefined and/or empty (defined as the empty string):
$ array=(zero one '')
$ echo "array[2] is ${array[2]-unset}, array[3] is ${array[3]-unset}"
array[2] is , array[3] is unset
$ echo "array[2] is ${array[2]:-unset or empty}, array[3] is ${array[3]:-unset or empty}"
array[2] is unset or empty, array[3] is unset or empty
There are a bunch of other variants, see the POSIX shell syntax standard, section 2.6.2 (Parameter Expansion).
BTW, you do need to use curly braces (as I did above) around anything other than a plain variable reference. $name[2] is a reference to the plain variable name (or element 0 if it's an array), followed by the string "[2]"; ${name[2]}, on the other hand, is a reference to element 2 of the array name. Also, you pretty much always want to wrap variable references in double-quotes (or include them in double-quoted strings), to prevent the shell from "helpfully" splitting them into words and/or expanding them into lists of matching files. For example, this test:
if [ $my_array[i] = true ]
is (mostly) equivalent to:
if [ ${my_array[0]}[i] = true ]
...which isn't what you want at all. But this one:
if [ ${my_array[i]} = true ]
still doesn't work, because if my_array[i] is unset (or empty) it'll expand to the equivalent of:
if [ = true ]
...which is bad test expression syntax. You want this:
if [ "${my_array[i]}" = true ]
I have these two functions:
function two() {
local -a -x var=( ${var[#]} )
echo "${var[#]}"
}
function one() {
local -a -x var=(11 22 33)
two
}
If I call one, then nothing is printed. Why is that?
nothing is print. Why is that?
Here you're having the same identifier name var in both the functions
The var you defined in one could accessed by two because two is called from one. However,
when declaring and setting a local variable in a single command,
apparently the order of operations is to first set the variable, and
only afterwards restrict it to local scope.
So in
local -a -x var=( "${var[#]}" )
the ${var[#]} part will be empty as the variable var is set local first.
To verify this you could change the variable name in one to var1 and and in two do
local -a -x var=( "${var1[#]}" ) # var1 though local to one should be accessible here.
You could use #inian's answer as a work-around to pass variables easily and yet not bother about such dark corners in bash.
Your code is not reflecting what are you trying to do! The locals you've defined are only within the scope of the function Yes! but if you are passing it to the other function, pass it as positional arguments "$#". In the function below when you do two "${var[#]}", you are passing the local array as a positional argument array to be used in the other function.
two() {
local -a -x var=( "$#" )
echo "${var[#]}"
}
The argument list "$#" represents the argument list passed to the function two, now from the function one pass it as
one() {
local -a -x var=(11 22 33)
two "${var[#]}"
}
Also the use of function keyword is non-standard. POSIX does not recommend using it. If you are planning to re-use script for multiple shells, drop the keyword. Also quote the variables/array to avoid them being string-splited and glob-expanded. It could result in unexpected values in the final array.
Also worth noting that variables/arrays are global by default unless you override with local keyword inside a function.
$ x=2
$ test_local(){ local x=1; }
$ test_local; echo "$x"
2
But the same without local would print the value as 1 which proves the point explained above.
When you declare var in two that declaration hides the one in one. Curiously, local variables are visible in called functions. The easiest way to make this work is to do nothing: simply access $var in two.
two() {
echo "${var[#]}"
}
one() {
local var=(11 22 33)
two "${var[#]}"
}
I don't necessarily recommend doing this, though. It makes it hard to understand what two does just by reading it. It's better to explicitly pass the values as arguments.
two() {
local var=("$#")
echo "${var[#]}"
}
one() {
local var=(11 22 33)
two "${var[#]}"
}
By the way, you should always quote your variable expansions to prevent them from being subjected to word splitting and globbing. In your original code you should quote ${var[#]}:
local -a -x var=( "${var[#]}" )
Also, for portability you should either write one() or function one, but not both. I prefer the former.
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=
So I have an array like:
al_ap_version=('ap_version' '[[ $data -ne $version ]]')
And the condition gets evaluated inside a loop like:
for alert in alert_list; do
data=$(tail -1 somefile)
condition=$(eval echo \${$alert[1]})
if eval "$condition" ; then
echo SomeAlert
fi
done
Whilst this generally works with many scenarios, if $data returns something like "-/-" or "4.2.9", I get errors as it doesn't seem to like complex strings in the variable.
Obviously I can't enclose the variable in single quotes as it won't expand so I'm after any ideas to expand the $data variable (or indeed the $version var which suffers the same possible fate) in a way that the evaluation can handle?
Ignoring the fact that eval is probably super dangerous to use here (unless the data in somefile is controlled by you and only you), there are a few issues to fix in your example code.
In your for loop, alert_list needs to be $alert_list.
Also, as pointed out by #choroba, you should be using != instead of -ne since your input isn't always an integer.
Finally, while debugging, you can add set -x to the top of your script, or add -x to the end of your shebang line to enable verbose output (helps to determine how bash is expanding your variables).
This works for me:
#!/bin/bash -x
data=2.2
version=1
al_ap_version=('ap_version' '[[ $data != $version ]]')
alert_list='al_ap_version'
for alert in $alert_list; do
condition=$(eval echo \${$alert[1]})
if eval "$condition"; then
echo "alert"
fi
done
You could try a more functional approach, even though bash is only just barely capable of such things. On the whole, it is usually a lot easier to pack an action to be executed into a bash function and refer to it with the name of the function, than to try to maintain the action as a string to be evaluated.
But first, the use of an array of names of arrays is awkward. Let's get rid of it.
It's not clear to me the point of element 0, ap_version, in the array al_ap_version but I suppose it has something to do with error messages. If the order of alert processing isn't important, you could replace the list of names of arrays with a single associative array:
declare -A alert_list
alert_list[ap_version]=... # see below
alert_list[os_dsk]=...
and then process them with:
for alert_name in ${!alert_list[#]}; do
alert=${alert_list[$alert_name]}
...
done
Having done that, we can get rid of the eval, with its consequent ugly necessity for juggling quotes, by creating a bash function for each alert:
check_ap_version() {
(($version != $1))
}
Edit: It seems that $1 is not necessarily numeric, so it would be better to use a non-numeric comparison, although exact version match might not be what you're after either. So perhaps it would be better to use:
check_ap_version() {
[[ $version != $1 ]]
}
Note the convention that the first argument of the function is the data value.
Now we can insert the name of the function into the alert array, and call it indirectly in the loop:
declare -A alert_list
alert_list[ap_version]=check_ap_version
alert_list[os_dsk]=check_op_dsk
check_alerts() {
local alert_name alert
local data=$(tail -1 somefile)
for alert_name in ${!alert_list[#]}; do
alert=${alert_list[$alert_name]}
if $alert "$data"; then
signal_alert $alert_name
fi
done
}
If you're prepared to be more disciplined about the function names, you can avoid the associative array, and thereby process the alerts in order. Suppose, for example, that every function has the name check_<alert_name>. Then the above could be:
alert_list=(ap_version os_dsk)
check_alerts() {
local alert_name
local data=$(tail -1 somefile)
for alert_name in $alert_list[#]; do
if check_$alert_name "$data"; then
signal_alert $alert_name
fi
done
}
I want to save the variable name and its contents easily from my script.
Currently :-
LOGFILE=/root/log.txt
TEST=/file/path
echo "TEST : ${TEST}" >> ${LOGFILE}
Desired :-
LOGFILE=/root/log.txt
function save()
{
echo "$1 : $1" >> ${LOGFILE}
}
TEST=/file/path
save TEST
Obviously the above save function just saves TEST : TEST
Want I want it to save is TEST : /file/path
Can this be done? How? Many thanks in advance!
You want to use Variable Indirection. Also, don't use the function keyword, it is not POSIX and also not necessary as long as you have () at the end of your function name.
LOGFILE=/root/log.txt
save()
{
echo "$1 : ${!1}" >> ${LOGFILE}
}
TEST=/file/path
save TEST
Proof of Concept
$ TEST=foo; save(){ echo "$1 : ${!1}"; }; save TEST
TEST : foo
Yes, using indirect expansion:
echo "$1 : ${!1}"
Quoting from Bash reference manual:
The basic form of parameter expansion is ${parameter} [...] If the first character of parameter is an exclamation point (!), a level of variable indirection is introduced. Bash uses the value of the variable formed from the rest of parameter as the name of the variable; this variable is then expanded and that value is used in the rest of the substitution, rather than the value of parameter itself. This is known as indirect expansion
Consider using the printenv function. It does exactly what it says on the tin, prints your environment. It can also take parameters
$ printenv
SSH_AGENT_PID=2068
TERM=xterm
SHELL=/bin/bash
LANG=en_US.UTF-8
HISTCONTROL=ignoreboth
...etc
You could do printenv and then grep for any vars you know you have defined and be done in two lines, such as:
$printenv | grep "VARNAME1\|VARNAME2"
VARNAME1=foo
VARNAME2=bar