How to exit if a command failed? [duplicate] - linux

This question already has answers here:
Aborting a shell script if any command returns a non-zero value
(10 answers)
Closed 1 year ago.
I am a noob in shell-scripting. I want to print a message and exit my script if a command fails. I've tried:
my_command && (echo 'my_command failed; exit)
but it does not work. It keeps executing the instructions following this line in the script. I'm using Ubuntu and bash.

Try:
my_command || { echo 'my_command failed' ; exit 1; }
Four changes:
Change && to ||
Use { } in place of ( )
Introduce ; after exit and
spaces after { and before }
Since you want to print the message and exit only when the command fails ( exits with non-zero value) you need a || not an &&.
cmd1 && cmd2
will run cmd2 when cmd1 succeeds(exit value 0). Where as
cmd1 || cmd2
will run cmd2 when cmd1 fails(exit value non-zero).
Using ( ) makes the command inside them run in a sub-shell and calling a exit from there causes you to exit the sub-shell and not your original shell, hence execution continues in your original shell.
To overcome this use { }
The last two changes are required by bash.

The other answers have covered the direct question well, but you may also be interested in using set -e. With that, any command that fails (outside of specific contexts like if tests) will cause the script to abort. For certain scripts, it's very useful.

If you want that behavior for all commands in your script, just add
set -e
set -o pipefail
at the beginning of the script. This pair of options tell the bash interpreter to exit whenever a command returns with a non-zero exit code. (For more details about why pipefail is needed, see http://petereisentraut.blogspot.com/2010/11/pipefail.html)
This does not allow you to print an exit message, though.

Note also, each command's exit status is stored in the shell variable $?, which you can check immediately after running the command. A non-zero status indicates failure:
my_command
if [ $? -eq 0 ]
then
echo "it worked"
else
echo "it failed"
fi

I've hacked up the following idiom:
echo "Generating from IDL..."
idlj -fclient -td java/src echo.idl
if [ $? -ne 0 ]; then { echo "Failed, aborting." ; exit 1; } fi
echo "Compiling classes..."
javac *java
if [ $? -ne 0 ]; then { echo "Failed, aborting." ; exit 1; } fi
echo "Done."
Precede each command with an informative echo, and follow each command with that same
if [ $? -ne 0 ];... line. (Of course, you can edit that error message if you want to.)

Provided my_command is canonically designed, ie returns 0 when succeeds, then && is exactly the opposite of what you want. You want ||.
Also note that ( does not seem right to me in bash, but I cannot try from where I am. Tell me.
my_command || {
echo 'my_command failed' ;
exit 1;
}

You can also use, if you want to preserve exit error status, and have a readable file with one command per line:
my_command1 || exit
my_command2 || exit
This, however will not print any additional error message. But in some cases, the error will be printed by the failed command anyway.

The trap shell builtin allows catching signals, and other useful conditions, including failed command execution (i.e., a non-zero return status). So if you don't want to explicitly test return status of every single command you can say trap "your shell code" ERR and the shell code will be executed any time a command returns a non-zero status. For example:
trap "echo script failed; exit 1" ERR
Note that as with other cases of catching failed commands, pipelines need special treatment; the above won't catch false | true.

Using exit directly may be tricky as the script may be sourced from other places (e.g. from terminal). I prefer instead using subshell with set -e (plus errors should go into cerr, not cout) :
set -e
ERRCODE=0
my_command || ERRCODE=$?
test $ERRCODE == 0 ||
(>&2 echo "My command failed ($ERRCODE)"; exit $ERRCODE)

Related

Prevent shell script from exiting on certain command when set -e is enabled [duplicate]

I am running my scripts with:
#!/bin/bash -eu
Which aborts the script whenever a problem occurs, as wanted. But sometimes I expect one of the commands to eventually fail, and I would like to tell bash to ignore the fail condition. In make you can ignore the status of one command with the handy:
-command
Is there something similar in bash? The only thing that comes to mind is the ugly:
set +e
command
set -e
You could just do a no-op on the command failure or set an explicit true condition as
command || true
or for no-op as
command || :
Doing so forces the command-list (even a pipeline) to return an exit status of 0 on failure. See
true | false
echo $?
1
true | false || true
echo $?
0
true | false || :
echo $?
0
Just prepend a ! to the command so that its exit status does not make the script exit when running it with e:
! command
As seen in What's the meaning of a ! before a command in the shell?, having ! command negates the exit status of the given command and, used with set -e, prevents the shell to exit whatever the exit result is on that line.
From Bash Reference Manual → Pipelines:
Each command in a pipeline is executed in its own subshell. The exit status of a pipeline is the exit status of the last command in the pipeline (...). If the reserved word ‘!’ precedes the pipeline, the exit status is the logical negation of the exit status as described above. The shell waits for all commands in the pipeline to terminate before returning a value.
Then we have the info about 4.3.1 The set Builtin:
-e
Exit immediately if a pipeline (see Pipelines), which may consist of a single simple command (see Simple Commands), a list (see Lists), or a compound command (see Compound Commands) returns a non-zero status. The shell does not exit if the command that fails is part of the command list immediately following a while or until keyword, part of the test in an if statement, part of any command executed in a && or || list except the command following the final && or ||, any command in a pipeline but the last, or if the command’s return status is being inverted with !.
All together, and quoting my own answer:
When you have:
set -e
! command1
command2
What you are doing is to by-pass the set -e flag in the command1.
Why?
if command1 runs properly, it will return a zero status. ! will negate it, but set -e won't trigger an exit by the because it comes
from a return status inverted with !, as described above.
if command1 fails, it will return a non-zero status. ! will negate it, so the line will end up returning a zero status and the
script will continue normally.
Don't think there is
Could just write your own function though
#!/bin/bash -eu
-(){
set +e
"$#"
set -e
}
- command
echo got here
May want to use a function name since - is already used in bash.
As chepner pointed out it only works for simple commands, though, not pipelines or lists.

Raise error in a Bash script

I want to raise an error in a Bash script with message "Test cases Failed !!!". How to do this in Bash?
For example:
if [ condition ]; then
raise error "Test cases failed !!!"
fi
This depends on where you want the error message be stored.
You can do the following:
echo "Error!" > logfile.log
exit 125
Or the following:
echo "Error!" 1>&2
exit 64
When you raise an exception you stop the program's execution.
You can also use something like exit xxx where xxx is the error code you may want to return to the operating system (from 0 to 255). Here 125 and 64 are just random codes you can exit with. When you need to indicate to the OS that the program stopped abnormally (eg. an error occurred), you need to pass a non-zero exit code to exit.
As #chepner pointed out, you can do exit 1, which will mean an unspecified error.
Basic error handling
If your test case runner returns a non-zero code for failed tests, you can simply write:
test_handler test_case_x; test_result=$?
if ((test_result != 0)); then
printf '%s\n' "Test case x failed" >&2 # write error message to stderr
exit 1 # or exit $test_result
fi
Or even shorter:
if ! test_handler test_case_x; then
printf '%s\n' "Test case x failed" >&2
exit 1
fi
Or the shortest:
test_handler test_case_x || { printf '%s\n' "Test case x failed" >&2; exit 1; }
To exit with test_handler's exit code:
test_handler test_case_x || { ec=$?; printf '%s\n' "Test case x failed" >&2; exit $ec; }
Advanced error handling
If you want to take a more comprehensive approach, you can have an error handler:
exit_if_error() {
local exit_code=$1
shift
[[ $exit_code ]] && # do nothing if no error code passed
((exit_code != 0)) && { # do nothing if error code is 0
printf 'ERROR: %s\n' "$#" >&2 # we can use better logging here
exit "$exit_code" # we could also check to make sure
# error code is numeric when passed
}
}
then invoke it after running your test case:
run_test_case test_case_x
exit_if_error $? "Test case x failed"
or
run_test_case test_case_x || exit_if_error $? "Test case x failed"
The advantages of having an error handler like exit_if_error are:
we can standardize all the error handling logic such as logging, printing a stack trace, notification, doing cleanup etc., in one place
by making the error handler get the error code as an argument, we can spare the caller from the clutter of if blocks that test exit codes for errors
if we have a signal handler (using trap), we can invoke the error handler from there
Error handling and logging library
Here is a complete implementation of error handling and logging:
https://github.com/codeforester/base/blob/master/lib/stdlib.sh
Related posts
Error handling in Bash
The 'caller' builtin command on Bash Hackers Wiki
Are there any standard exit status codes in Linux?
BashFAQ/105 - Why doesn't set -e (or set -o errexit, or trap ERR) do what I expected?
Equivalent of __FILE__, __LINE__ in Bash
Is there a TRY CATCH command in Bash
To add a stack trace to the error handler, you may want to look at this post: Trace of executed programs called by a Bash script
Ignoring specific errors in a shell script
Catching error codes in a shell pipe
How do I manage log verbosity inside a shell script?
How to log function name and line number in Bash?
Is double square brackets [[ ]] preferable over single square brackets [ ] in Bash?
There are a couple more ways with which you can approach this problem. Assuming one of your requirement is to run a shell script/function containing a few shell commands and check if the script ran successfully and throw errors in case of failures.
The shell commands in generally rely on exit-codes returned to let the shell know if it was successful or failed due to some unexpected events.
So what you want to do falls upon these two categories
exit on error
exit and clean-up on error
Depending on which one you want to do, there are shell options available to use. For the first case, the shell provides an option with set -e and for the second you could do a trap on EXIT
Should I use exit in my script/function?
Using exit generally enhances readability In certain routines, once you know the answer, you want to exit to the calling routine immediately. If the routine is defined in such a way that it doesn’t require any further cleanup once it detects an error, not exiting immediately means that you have to write more code.
So in cases if you need to do clean-up actions on script to make the termination of the script clean, it is preferred to not to use exit.
Should I use set -e for error on exit?
No!
set -e was an attempt to add "automatic error detection" to the shell. Its goal was to cause the shell to abort any time an error occurred, but it comes with a lot of potential pitfalls for example,
The commands that are part of an if test are immune. In the example, if you expect it to break on the test check on the non-existing directory, it wouldn't, it goes through to the else condition
set -e
f() { test -d nosuchdir && echo no dir; }
f
echo survived
Commands in a pipeline other than the last one, are immune. In the example below, because the most recently executed (rightmost) command's exit code is considered ( cat) and it was successful. This could be avoided by setting by the set -o pipefail option but its still a caveat.
set -e
somecommand that fails | cat -
echo survived
Recommended for use - trap on exit
The verdict is if you want to be able to handle an error instead of blindly exiting, instead of using set -e, use a trap on the ERR pseudo signal.
The ERR trap is not to run code when the shell itself exits with a non-zero error code, but when any command run by that shell that is not part of a condition (like in if cmd, or cmd ||) exits with a non-zero exit status.
The general practice is we define an trap handler to provide additional debug information on which line and what cause the exit. Remember the exit code of the last command that caused the ERR signal would still be available at this point.
cleanup() {
exitcode=$?
printf 'error condition hit\n' 1>&2
printf 'exit code returned: %s\n' "$exitcode"
printf 'the command executing at the time of the error was: %s\n' "$BASH_COMMAND"
printf 'command present on line: %d' "${BASH_LINENO[0]}"
# Some more clean up code can be added here before exiting
exit $exitcode
}
and we just use this handler as below on top of the script that is failing
trap cleanup ERR
Putting this together on a simple script that contained false on line 15, the information you would be getting as
error condition hit
exit code returned: 1
the command executing at the time of the error was: false
command present on line: 15
The trap also provides options irrespective of the error to just run the cleanup on shell completion (e.g. your shell script exits), on signal EXIT. You could also trap on multiple signals at the same time. The list of supported signals to trap on can be found on the trap.1p - Linux manual page
Another thing to notice would be to understand that none of the provided methods work if you are dealing with sub-shells are involved in which case, you might need to add your own error handling.
On a sub-shell with set -e wouldn't work. The false is restricted to the sub-shell and never gets propagated to the parent shell. To do the error handling here, add your own logic to do (false) || false
set -e
(false)
echo survived
The same happens with trap also. The logic below wouldn't work for the reasons mentioned above.
trap 'echo error' ERR
(false)
Here's a simple trap that prints the last argument of whatever failed to STDERR, reports the line it failed on, and exits the script with the line number as the exit code. Note these are not always great ideas, but this demonstrates some creative application you could build on.
trap 'echo >&2 "$_ at $LINENO"; exit $LINENO;' ERR
I put that in a script with a loop to test it. I just check for a hit on some random numbers; you might use actual tests. If I need to bail, I call false (which triggers the trap) with the message I want to throw.
For elaborated functionality, have the trap call a processing function. You can always use a case statement on your arg ($_) if you need to do more cleanup, etc. Assign to a var for a little syntactic sugar -
trap 'echo >&2 "$_ at $LINENO"; exit $LINENO;' ERR
throw=false
raise=false
while :
do x=$(( $RANDOM % 10 ))
case "$x" in
0) $throw "DIVISION BY ZERO" ;;
3) $raise "MAGIC NUMBER" ;;
*) echo got $x ;;
esac
done
Sample output:
# bash tst
got 2
got 8
DIVISION BY ZERO at 6
# echo $?
6
Obviously, you could
runTest1 "Test1 fails" # message not used if it succeeds
Lots of room for design improvement.
The draw backs include the fact that false isn't pretty (thus the sugar), and other things tripping the trap might look a little stupid. Still, I like this method.
You have 2 options: Redirect the output of the script to a file, Introduce a log file in the script and
Redirecting output to a file:
Here you assume that the script outputs all necessary info, including warning and error messages. You can then redirect the output to a file of your choice.
./runTests &> output.log
The above command redirects both the standard output and the error output to your log file.
Using this approach you don't have to introduce a log file in the script, and so the logic is a tiny bit easier.
Introduce a log file to the script:
In your script add a log file either by hard coding it:
logFile='./path/to/log/file.log'
or passing it by a parameter:
logFile="${1}" # This assumes the first parameter to the script is the log file
It's a good idea to add the timestamp at the time of execution to the log file at the top of the script:
date '+%Y%-m%d-%H%M%S' >> "${logFile}"
You can then redirect your error messages to the log file
if [ condition ]; then
echo "Test cases failed!!" >> "${logFile}";
fi
This will append the error to the log file and continue execution. If you want to stop execution when critical errors occur, you can exit the script:
if [ condition ]; then
echo "Test cases failed!!" >> "${logFile}";
# Clean up if needed
exit 1;
fi
Note that exit 1 indicates that the program stop execution due to an unspecified error. You can customize this if you like.
Using this approach you can customize your logs and have a different log file for each component of your script.
If you have a relatively small script or want to execute somebody else's script without modifying it to the first approach is more suitable.
If you always want the log file to be at the same location, this is the better option of the 2. Also if you have created a big script with multiple components then you may want to log each part differently and the second approach is your only option.
I often find it useful to write a function to handle error messages so the code is cleaner overall.
# Usage: die [exit_code] [error message]
die() {
local code=$? now=$(date +%T.%N)
if [ "$1" -ge 0 ] 2>/dev/null; then # assume $1 is an error code if numeric
code="$1"
shift
fi
echo "$0: ERROR at ${now%???}${1:+: $*}" >&2
exit $code
}
This takes the error code from the previous command and uses it as the default error code when exiting the whole script. It also notes the time, with microseconds where supported (GNU date's %N is nanoseconds, which we truncate to microseconds later).
If the first option is zero or a positive integer, it becomes the exit code and we remove it from the list of options. We then report the message to standard error, with the name of the script, the word "ERROR", and the time (we use parameter expansion to truncate nanoseconds to microseconds, or for non-GNU times, to truncate e.g. 12:34:56.%N to 12:34:56). A colon and space are added after the word ERROR, but only when there is a provided error message. Finally, we exit the script using the previously determined exit code, triggering any traps as normal.
Some examples (assume the code lives in script.sh):
if [ condition ]; then die 123 "condition not met"; fi
# exit code 123, message "script.sh: ERROR at 14:58:01.234564: condition not met"
$command |grep -q condition || die 1 "'$command' lacked 'condition'"
# exit code 1, "script.sh: ERROR at 14:58:55.825626: 'foo' lacked 'condition'"
$command || die
# exit code comes from command's, message "script.sh: ERROR at 14:59:15.575089"

In a bash script that starts with set -e, can I set the exit code to a different value than the first failed command's?

I'm currently working on some init scripts that should both use set -e and confirm to the Linux Standard Base core specification. Now those two don't really work together:
Since due to the set -e the first command that fails causes the scripts to exit with the return value of the failed command I cannot set the exit status of the script to something LSB conformant for commands that fail with LSB incompatible return values. I could unset -e before each such command, but that's quite a hassle and in that case I'd actually rather not use set -e at all?
I guess another way to achieve the same result as with unset -e would be to do something like
returns_1() { return 1; }
...
cmd_that_fails_with_non_LSB_return_values || returns_1
but that seems quite bulky as well and I'd again have to check each and every command's possible return values.
Is there a way to set the error code returned by the script when it is terminated due to set -e to a fixed value so it would return 1 (i.e. LSB general/unspecified error) no mather what return value the failed command had? Should I not bother with LSB conformant return codes and/or set -e? (this will probably turn into a discussion about the merrits of set -e anyway judging from the amount of search results you get for that)
Small code snippet to illustrate the problem:
#!/bin/bash
# init script for service foo
set -e
start() {
echo "bar"
cmd_fails_with_return_code_3 # script exits with return code 3, not LSB conformant
echo "baz"
}
...
case "$1" in
start)
start
;;
...
esac
From the man page, it appears you can set a trap on ERR:
A trap on ERR, if set, is executed before the shell exits.
I haven't tried this but this would mean something like this might help you:
trap "exit 1" ERR
Be sure to read the man page for other useful options such as -E to inherit the ERR trap in subshells.
You basically have it nailed, although the returns function is superfluous.
cmd_that_fails_with_non_LSB_return_values || exit 1
More typically, you would actually somehow handle the error, if only just to report what happened.
die () {
echo "$0: $#" >&2
exit 1
}
:
cmd_that_fails_with_non_LSB_return_values ||
die "Long command with underscores failed; aborting"
Philosophically, I suppose the purpose of set -e is mainly to "gently remind" (that is, force) you to handle all possible error conditions.

Any way to exit bash script, but not quitting the terminal

When I use exit command in a shell script, the script will terminate the terminal (the prompt). Is there any way to terminate a script and then staying in the terminal?
My script run.sh is expected to execute by directly being sourced, or sourced from another script.
EDIT:
To be more specific, there are two scripts run2.sh as
...
. run.sh
echo "place A"
...
and run.sh as
...
exit
...
when I run it by . run2.sh, and if it hit exit codeline in run.sh, I want it to stop to the terminal and stay there. But using exit, the whole terminal gets closed.
PS: I have tried to use return, but echo codeline will still gets executed....
The "problem" really is that you're sourcing and not executing the script. When you source a file, its contents will be executed in the current shell, instead of spawning a subshell. So everything, including exit, will affect the current shell.
Instead of using exit, you will want to use return.
Yes; you can use return instead of exit. Its main purpose is to return from a shell function, but if you use it within a source-d script, it returns from that script.
As §4.1 "Bourne Shell Builtins" of the Bash Reference Manual puts it:
return [n]
Cause a shell function to exit with the return value n.
If n is not supplied, the return value is the exit status of the
last command executed in the function.
This may also be used to terminate execution of a script being executed
with the . (or source) builtin, returning either n or
the exit status of the last command executed within the script as the exit
status of the script.
Any command associated with the RETURN trap is executed
before execution resumes after the function or script.
The return status is non-zero if return is used outside a function
and not during the execution of a script by . or source.
You can add an extra exit command after the return statement/command so that it works for both, executing the script from the command line and sourcing from the terminal.
Example exit code in the script:
if [ $# -lt 2 ]; then
echo "Needs at least two arguments"
return 1 2>/dev/null
exit 1
fi
The line with the exit command will not be called when you source the script after the return command.
When you execute the script, return command gives an error. So, we suppress the error message by forwarding it to /dev/null.
Instead of running the script using . run2.sh, you can run it using sh run2.sh or bash run2.sh
A new sub-shell will be started, to run the script then, it will be closed at the end of the script leaving the other shell opened.
Actually, I think you might be confused by how you should run a script.
If you use sh to run a script, say, sh ./run2.sh, even if the embedded script ends with exit, your terminal window will still remain.
However if you use . or source, your terminal window will exit/close as well when subscript ends.
for more detail, please refer to What is the difference between using sh and source?
This is just like you put a run function inside your script run2.sh.
You use exit code inside run while source your run2.sh file in the bash tty.
If the give the run function its power to exit your script and give the run2.sh
its power to exit the terminator.
Then of cuz the run function has power to exit your teminator.
#! /bin/sh
# use . run2.sh
run()
{
echo "this is run"
#return 0
exit 0
}
echo "this is begin"
run
echo "this is end"
Anyway, I approve with Kaz it's a design problem.
I had the same problem and from the answers above and from what I understood what worked for me ultimately was:
Have a shebang line that invokes the intended script, for example,
#!/bin/bash uses bash to execute the script
I have scripts with both kinds of shebang's. Because of this, using sh or . was not reliable, as it lead to a mis-execution (like when the script bails out having run incompletely)
The answer therefore, was
Make sure the script has a shebang, so that there is no doubt about its intended handler.
chmod the .sh file so that it can be executed. (chmod +x file.sh)
Invoke it directly without any sh or .
(./myscript.sh)
Hope this helps someone with similar question or problem.
To write a script that is secure to be run as either a shell script or sourced as an rc file, the script can check and compare $0 and $BASH_SOURCE and determine if exit can be safely used.
Here is a short code snippet for that
[ "X$(basename $0)" = "X$(basename $BASH_SOURCE)" ] && \
echo "***** executing $name_src as a shell script *****" || \
echo "..... sourcing $name_src ....."
I think that this happens because you are running it on source mode
with the dot
. myscript.sh
You should run that in a subshell:
/full/path/to/script/myscript.sh
'source' http://ss64.com/bash/source.html
It's correct that sourced vs. executed scripts use return vs. exit to keep the same session open, as others have noted.
Here's a related tip, if you ever want a script that should keep the session open, regardless of whether or not it's sourced.
The following example can be run directly like foo.sh or sourced like . foo.sh/source foo.sh. Either way it will keep the session open after "exiting". The $# string is passed so that the function has access to the outer script's arguments.
#!/bin/sh
foo(){
read -p "Would you like to XYZ? (Y/N): " response;
[ $response != 'y' ] && return 1;
echo "XYZ complete (args $#).";
return 0;
echo "This line will never execute.";
}
foo "$#";
Terminal result:
$ foo.sh
$ Would you like to XYZ? (Y/N): n
$ . foo.sh
$ Would you like to XYZ? (Y/N): n
$ |
(terminal window stays open and accepts additional input)
This can be useful for quickly testing script changes in a single terminal while keeping a bunch of scrap code underneath the main exit/return while you work. It could also make code more portable in a sense (if you have tons of scripts that may or may not be called in different ways), though it's much less clunky to just use return and exit where appropriate.
Also make sure to return with expected return value. Else if you use exit when you will encounter an exit it will exit from your base shell since source does not create another process (instance).
Improved the answer of Tzunghsing, with more clear results and error re-direction, for silent usage:
#!/usr/bin/env bash
echo -e "Testing..."
if [ "X$(basename $0 2>/dev/null)" = "X$(basename $BASH_SOURCE)" ]; then
echo "***** You are Executing $0 in a sub-shell."
exit 0
else
echo "..... You are Sourcing $BASH_SOURCE in this terminal shell."
return 0
fi
echo "This should never be seen!"
Or if you want to put this into a silent function:
function sExit() {
# Safe Exit from script, not closing shell.
[ "X$(basename $0 2>/dev/null)" = "X$(basename $BASH_SOURCE)" ] && exit 0 || return 0
}
...
# ..it have to be called with an error check, like this:
sExit && return 0
echo "This should never be seen!"
Please note that:
if you have enabled errexit in your script (set -e) and you return N with N != 0, your entire script will exit instantly. To see all your shell settings, use, set -o.
when used in a function, the 1st return 0 is exiting the function, and the 2nd return 0 is exiting the script.
if your terminal emulator doesn't have -hold you can sanitize a sourced script and hold the terminal with:
#!/bin/sh
sed "s/exit/return/g" script >/tmp/script
. /tmp/script
read
otherwise you can use $TERM -hold -e script
If a command succeeded successfully, the return value will be 0. We can check its return value afterwards.
Is there a “goto” statement in bash?
Here is some dirty workaround using trap which jumps only backwards.
#!/bin/bash
set -eu
trap 'echo "E: failed with exitcode $?" 1>&2' ERR
my_function () {
if git rev-parse --is-inside-work-tree > /dev/null 2>&1; then
echo "this is run"
return 0
else
echo "fatal: not a git repository (or any of the parent directories): .git"
goto trap 2> /dev/null
fi
}
my_function
echo "Command succeeded" # If my_function failed this line is not printed
Related:
https://stackoverflow.com/a/19091823/2402577
How to use $? and test to check function?
I couldn't find solution so for those who want to leave the nested script without leaving terminal window:
# this is just script which goes to directory if path satisfies regex
wpr(){
leave=false
pwd=$(pwd)
if [[ "$pwd" =~ ddev.*web ]]; then
# echo "your in wordpress instalation"
wpDir=$(echo "$pwd" | grep -o '.*\/web')
cd $wpDir
return
fi
echo 'please be in wordpress directory'
# to leave from outside the scope
leave=true
return
}
wpt(){
# nested function which returns $leave variable
wpr
# interupts the script if $leave is true
if $leave; then
return;
fi
echo 'here is the rest of the script, executes if leave is not defined'
}
I have no idea whether this is useful for you or not, but in zsh, you can exit a script, but only to the prompt if there is one, by using parameter expansion on a variable that does not exist, as follows.
${missing_variable_ejector:?}
Though this does create an error message in your script, you can prevent it with something like the following.
{ ${missing_variable_ejector:?} } 2>/dev/null
1) exit 0 will come out of the script if it is successful.
2) exit 1 will come out of the script if it is a failure.
You can try these above two based on ur req.

Does a bash script have return code itself?

I know the return code will be contained in $? after a command was executed, but what does $? mean after a script was executed? The return code of the last command in that script?
Can I tell if a script has been excuted from head to tail and not interrupted by some unexpected system halt or something?
If I have a script like below excuted,
Command A;
if [ $? -eq 0]
then
echo "OK" >> log
else
echo "failed" >> log
fi
and the system halted while A was running, what will I find in that log file? "OK", "failed" or nothing?
Yes, or the value passed after exit, e.g. exit 31.
Not without taking measures within the other script to make it explicit.
$? reads the exit status of the last command executed. After a function returns, $? gives the exit status of the last command executed in the function. This is Bash's way of giving functions a "return value.
Example
#!/bin/bash
echo hello
echo $? # Exit status 0 returned because command executed successfully.
lskdf # Unrecognized command.
echo $? # Non-zero exit status returned because command failed to execute.
echo
exit 113 # Will return 113 to shell.
# To verify this, type "echo $?" after script terminates.
# By convention, an 'exit 0' indicates success,
#+ while a non-zero exit value means an error or anomalous condition
the return code of the script is indeed the return code of the last command executed, some commands allow you to finish execution at any point and arbitrarily set the return code; those are exit for scripts and return for functions but in both cases if you omit the argument they'll just use the return code of the previous command.

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