Is there a shell equivalent (in either bash or zsh) of Perl's die function?
I want to set the exit code and print a message in a single line. I know I can make my own simple function but I'm kind of hoping for a built in.
Nope, you need both
echo and exit
Just make a shell function like this :
die() {
[[ $1 ]] || {
printf >&2 -- 'Usage:\n\tdie <message> [return code]\n'
[[ $- == *i* ]] && return 1 || exit 1
}
printf >&2 -- '%s' "$1"
exit ${2:-1}
}
EXAMPLE
die "Oops, there's something wrong!\n" 255
EXPLANATIONS
the first argument is the needed message, the second optional argument is the return code :
${2:-1} is a bash parameter expansion : it exit 1 if the second argument is missing
in shell, 1 is the same as FALSE (1 => 255)
in modern bash, die() { } is preferred as oldish function die {}
redirecting STDERR to STDOUT like Maxwell does, is not the best practice, instead, I redirect to STDERR directly (like perl does)
if you want to use it in interactive shell, put this in ~/.bashrc and then source ~/.bashrc
if you want to use it in scripts, you can source ~/.bashrc in your script or put it manually.
[[ $- == *i* ]] test whether you are in interactive shell or not
Related
I have a function that runs a set of scripts that set variables, functions, and aliases in the current shell.
reloadVariablesFromScript() {
for script in "${scripts[#]}"; do
. "$script"
done
}
If one of the scripts has an error, I want to exit the script and then exit the function, but not to kill the shell.
reloadVariablesFromScript() {
for script in "${scripts[#]}"; do
{(
set -e
. "$script"
)}
if [[ $? -ne 0 ]]; then
>&2 echo $script failed. Skipping remaining scripts.
return 1
fi
done
}
This would do what I want except it doesn't set the variables in the script whether the script succeeds or fails.
Without the subshell, set -e causes the whole shell to exit, which is undesirable.
Is there a way I can either prevent the called script from continuing on an error without killing the shell or else set/export variables, aliases, and functions from within a subshell?
The following script simulates my problem:
test() {
{(
set -e
export foo=bar
false
echo Should not have gotten here!
export bar=baz
)}
local errorCode=$?
echo foo="'$foo'". It should equal 'bar'.
echo bar="'$bar'". It should not be set.
if [[ $errorCode -ne 0 ]]; then
echo Script failed correctly. Exiting function.
return 1
fi
echo Should not have gotten here!
}
test
If worst comes to worse, since these scripts don't actually edit the filesystem, I can run each script in a subshell, check the exit code, and if it succeeds, run it outside of a subshell.
Note that set -e has a number of surprising behaviors -- relying on it is not universally considered a good idea. That caveat being give, though: We can shuffle environment variables, aliases, and shell functions out as text:
envTest() {
local errorCode newVars
newVars=$(
set -e
{
export foo=bar
false
echo Should not have gotten here!
export bar=baz
} >&2
# print generate code which, when eval'd, recreates our functions and variables
declare -p | egrep -v '^declare -[^[:space:]]*r'
declare -f
alias -p
); errorCode=$?
if (( errorCode == 0 )); then
eval "$newVars"
fi
printf 'foo=%q. It should equal %q\n' "$foo" "bar"
printf 'bar=%q. It should not be set.\n' "$bar"
if [[ $errorCode -ne 0 ]]; then
echo 'Script failed correctly. Exiting function.'
return 1
fi
echo 'Should not have gotten here!'
}
envTest
Note that this code only evaluates either export should the entire script segment succeed; the question text and comments appear to indicate that this is acceptable if not desired.
I want to write code like this:
command="some command"
safeRunCommand $command
safeRunCommand() {
cmnd=$1
$($cmnd)
if [ $? != 0 ]; then
printf "Error when executing command: '$command'"
exit $ERROR_CODE
fi
}
But this code does not work the way I want. Where did I make the mistake?
Below is the fixed code:
#!/bin/ksh
safeRunCommand() {
typeset cmnd="$*"
typeset ret_code
echo cmnd=$cmnd
eval $cmnd
ret_code=$?
if [ $ret_code != 0 ]; then
printf "Error: [%d] when executing command: '$cmnd'" $ret_code
exit $ret_code
fi
}
command="ls -l | grep p"
safeRunCommand "$command"
Now if you look into this code, the few things that I changed are:
use of typeset is not necessary, but it is a good practice. It makes cmnd and ret_code local to safeRunCommand
use of ret_code is not necessary, but it is a good practice to store the return code in some variable (and store it ASAP), so that you can use it later like I did in printf "Error: [%d] when executing command: '$command'" $ret_code
pass the command with quotes surrounding the command like safeRunCommand "$command". If you don’t then cmnd will get only the value ls and not ls -l. And it is even more important if your command contains pipes.
you can use typeset cmnd="$*" instead of typeset cmnd="$1" if you want to keep the spaces. You can try with both depending upon how complex is your command argument.
'eval' is used to evaluate so that a command containing pipes can work fine
Note: Do remember some commands give 1 as the return code even though there isn't any error like grep. If grep found something it will return 0, else 1.
I had tested with KornShell and Bash. And it worked fine. Let me know if you face issues running this.
Try
safeRunCommand() {
"$#"
if [ $? != 0 ]; then
printf "Error when executing command: '$1'"
exit $ERROR_CODE
fi
}
It should be $cmd instead of $($cmd). It works fine with that on my box.
Your script works only for one-word commands, like ls. It will not work for "ls cpp". For this to work, replace cmd="$1"; $cmd with "$#". And, do not run your script as command="some cmd"; safeRun command. Run it as safeRun some cmd.
Also, when you have to debug your Bash scripts, execute with '-x' flag. [bash -x s.sh].
There are several things wrong with your script.
Functions (subroutines) should be declared before attempting to call them. You probably want to return() but not exit() from your subroutine to allow the calling block to test the success or failure of a particular command. That aside, you don't capture 'ERROR_CODE' so that is always zero (undefined).
It's good practice to surround your variable references with curly braces, too. Your code might look like:
#!/bin/sh
command="/bin/date -u" #...Example Only
safeRunCommand() {
cmnd="$#" #...insure whitespace passed and preserved
$cmnd
ERROR_CODE=$? #...so we have it for the command we want
if [ ${ERROR_CODE} != 0 ]; then
printf "Error when executing command: '${command}'\n"
exit ${ERROR_CODE} #...consider 'return()' here
fi
}
safeRunCommand $command
command="cp"
safeRunCommand $command
The normal idea would be to run the command and then use $? to get the exit code. However, sometimes you have multiple cases in which you need to get the exit code. For example, you might need to hide its output, but still return the exit code, or print both the exit code and the output.
ec() { [[ "$1" == "-h" ]] && { shift && eval $* > /dev/null 2>&1; ec=$?; echo $ec; } || eval $*; ec=$?; }
This will give you the option to suppress the output of the command you want the exit code for. When the output is suppressed for the command, the exit code will directly be returned by the function.
I personally like to put this function in my .bashrc file.
Below I demonstrate a few ways in which you can use this:
# In this example, the output for the command will be
# normally displayed, and the exit code will be stored
# in the variable $ec.
$ ec echo test
test
$ echo $ec
0
# In this example, the exit code is output
# and the output of the command passed
# to the `ec` function is suppressed.
$ echo "Exit Code: $(ec -h echo test)"
Exit Code: 0
# In this example, the output of the command
# passed to the `ec` function is suppressed
# and the exit code is stored in `$ec`
$ ec -h echo test
$ echo $ec
0
Solution to your code using this function
#!/bin/bash
if [[ "$(ec -h 'ls -l | grep p')" != "0" ]]; then
echo "Error when executing command: 'grep p' [$ec]"
exit $ec;
fi
You should also note that the exit code you will be seeing will be for the grep command that's being run, as it is the last command being executed. Not the ls.
In perl, you can exit with an error msg with die "some msg". Is there an equivalent single command in bash? Right now, I'm achieving this using commands: echo "some msg" && exit 1
You can roll your own easily enough:
die() { echo "$*" 1>&2 ; exit 1; }
...
die "Kaboom"
Here's what I'm using. It's too small to put in a library so I must have typed it hundreds of times ...
warn () {
echo "$0:" "$#" >&2
}
die () {
rc=$1
shift
warn "$#"
exit $rc
}
Usage: die 127 "Syntax error"
This is a very close function to perl's "die" (but with function name):
function die
{
local message=$1
[ -z "$message" ] && message="Died"
echo "$message at ${BASH_SOURCE[1]}:${FUNCNAME[1]} line ${BASH_LINENO[0]}." >&2
exit 1
}
And bash way of dying if built-in function is failed (with function name)
function die
{
local message=$1
[ -z "$message" ] && message="Died"
echo "${BASH_SOURCE[1]}: line ${BASH_LINENO[0]}: ${FUNCNAME[1]}: $message." >&2
exit 1
}
So, Bash is keeping all needed info in several environment variables:
LINENO - current executed line number
FUNCNAME - call stack of functions, first element (index 0) is current function, second (index 1) is function that called current function
BASH_LINENO - call stack of line numbers, where corresponding FUNCNAME was called
BASH_SOURCE - array of source file, where corresponfing FUNCNAME is stored
Yep, that's pretty much how you do it.
You might use a semicolon or newline instead of &&, since you want to exit whether or not echo succeeds (though I'm not sure what would make it fail).
Programming in a shell means using lots of little commands (some built-in commands, some tiny programs) that do one thing well and connecting them with file redirection, exit code logic and other glue.
It may seem weird if you're used to languages where everything is done using functions or methods, but you get used to it.
# echo pass params and print them to a log file
wlog(){
# check terminal if exists echo
test -t 1 && echo "`date +%Y.%m.%d-%H:%M:%S` [$$] $*"
# check LogFile and
test -z $LogFile || {
echo "`date +%Y.%m.%d-%H:%M:%S` [$$] $*" >> $LogFile
} #eof test
}
# eof function wlog
# exit with passed status and message
Exit(){
ExitStatus=0
case $1 in
[0-9]) ExitStatus="$1"; shift 1;;
esac
Msg="$*"
test "$ExitStatus" = "0" || Msg=" ERROR: $Msg : $#"
wlog " $Msg"
exit $ExitStatus
}
#eof function Exit
I'm looking for exception handling mechanism in shell script. Is there any try,catch equivalent mechanism in shell script ?
There is not really a try/catch in bash (i assume you're using bash), but you can achieve a quite similar behaviour using && or ||.
In this example, you want to run fallback_command if a_command fails (returns a non-zero value):
a_command || fallback_command
And in this example, you want to execute second_command if a_command is successful (returns 0):
a_command && second_command
They can easily be mixed together by using a subshell, for example, the following command will execute a_command, if it succeeds it will then run other_command, but if a_command or other_command fails, fallback_command will be executed:
(a_command && other_command) || fallback_command
The if/else structure and exit codes can help you fake some of it. This should work in Bash or Bourne (sh).
if foo ; then
else
e=$? # return code from if
if [ "${e}" -eq "1"]; then
echo "Foo returned exit code 1"
elif [ "${e}" -gt "1"]; then
echo "Foo returned BAD exit code ${e}"
fi
fi
{
# command which may fail and give an error
} || {
# command which should be run instead of the above failing command
}
Here are two simple bashfunctions which enable eventhandling in bash:
You could use it for basic exceptionhandling like this:
onFoo(){
echo "onFoo() called width arg $1!"
}
foo(){
[[ -f /tmp/somefile ]] || throw EXCEPTION_FOO_OCCURED "some arg"
}
addListener EXCEPTION_FOO_OCCURED onFoo
Exceptionhandling using try/catch blocks is not supported in bash, however, you might wanna try looking at the BANGSH framework which supports it (its a bit like jquery for bash).
However, exceptionhandling without cascading try/catch-blocks is similar to eventhandling, which is possible in almost any language with array-support.
If you want to keep your code nice and tidy (without if/else verbosity), I would recommend to use events.
The suggestion which MatToufoutu recommends (using || and &&) is not recommended for functions, but ok for simple commands. (see BashPitfalls about the risks)
Use following to handle error properly where error_exit is function that accepts one argument. In case if argument is not passed then it will throw unknown error with LineNo where actually error is happening. Please experiment before actually uses for production -
#!/bin/bash
PROGNAME=$(basename $0)
error_exit()
{
echo "${PROGNAME}: ${1:-"Unknown Error"}" 1>&2
exit 1
}
echo "Example of error with line number and message"
error_exit "$LINENO: An error has occurred."
I'd like to return an exit code from a BASH script that is called within another script, but could also be called directly. It roughly looks like this:
#!/bin/bash
dq2-get $1
if [ $? -ne 0 ]; then
echo "ERROR: ..."
# EXIT HERE
fi
# extract, do some stuff
# ...
Now in the line EXIT HERE the script should exit and return exit code 1. The problem is that
I cannot use return, because when I forget to source the script instead of calling it, return will not exit, and the rest of the script will be executed and mess things up.
I cannot use exit, because this closes the shell.
I cannot use the nice trick kill -SIGINT $$, because this doesn't allow to return an exit code.
Is there any viable alternative that I have overlooked?
The answer to the question title (not in the body as other answers have addressed) is:
Return an exit code without closing shell
(exit 33)
If you need to have -e active and still avoid exiting the shell with a non-zero exit code, then do:
(exit 33) && true
The true command is never executed but is used to build a compound command that is not exited by the -e shell flag.
That sets the exit code without exiting the shell (nor a sourced script).
For the more complex question of exiting (with an specific exit code) either if executed or sourced:
#!/bin/bash
[ "$BASH_SOURCE" == "$0" ] &&
echo "This file is meant to be sourced, not executed" &&
exit 30
return 88
Will set an exit code of 30 (with an error message) if executed.
And an exit code of 88 if sourced.
Will exit both the execution or the sourcing without affecting the calling shell.
Use this instead of exit or return:
[ $PS1 ] && return || exit;
Works whether sourced or not.
You can use x"${BASH_SOURCE[0]}" == x"$0" to test if the script was sourced or called (false if sourced, true if called) and return or exit accordingly.
Another option is to use a function and put the return values in that and then simply either source the script (source processStatus.sh) or call the script (./processStatus.sh) . For example consider the processStatus.sh script that needs to return a value to the stopProcess.sh script but also needs to be called separately from say the command line without using source (only relevant parts included)
Eg:
check_process ()
{
if [ $1 -eq "50" ]
then
return 1
else
return 0
fi
}
and
source processStatus.sh $1
RET_VALUE=$?
if [ $RET_VALUE -ne "0" ]
then
exit 0
fi
You can use return if you use set -e in the beginning of the script.
If you just want to check if the function returned no errors, I'd rather suggest rewriting your code like this:
#!/bin/bash
set -e # exit program if encountered errors
dq2-get ()
{
# define the function here
# ...
if [ $1 -eq 0 ]
then
return 0
else
return 255
# Note that nothing will execute from this point on,
# because `return` terminates the function.
}
# ...
# lots of code ...
# ...
# Now, the test:
# This won't exit the program.
if $(dq2-get $1); then
echo "No errors, everything's fine"
else
echo "ERROR: ..."
fi
# These commands execute anyway, no matter what
# `dq2-get $1` returns (i.e. {0..255}).
# extract, do some stuff
# ...
Now, the code above won't leave the program if the function dq2-get $1 returns errors. But, implementing the function all by itself will exit the program because of the set -e. The code below describes this situation:
# The function below will stop the program and exit
# if it returns anything other than `0`
# since `set -e` means stop if encountered any errors.
$(dq2-get $1)
# These commands execute ONLY if `dq2-get $1` returns `0`
# extract, do some stuff
# ...
Thanks for the question, my case was to source a file for some setup, but end the script and skip the setup actions if certain conditions were not met.
I had hit the issue of an attempt to use exit() actually causing the closing of my terminal, and found myself here :D
After reviewing the options for the specific solution i just went with something like the below, I also think Deepaks answer is worth reviewing if this approach works in your case.
if [ -z "$REQUIRED_VAR" ]; then
echo "please check/set \$REQUIRED_VAR ..."
echo "skipping logic"
else
echo "starting logic"
doStuff()
echo "completed logic"
fi