I have two scripts main script and sub script and I called the subscript using source script, if the specified package is not installed then it should return exit code 1.
If I run the main script by using bash main.sh I am unable to get the subScriptExitCode from the main script
main script
source "sub.sh"
subScriptExitCode=$?
log "subScript ExitCode: $subScriptExitCode"
if [ $subScriptExitCode -ne 0 ]; then
exit $subScriptExitCode
fi
sub script
type -p <package>
subScriptExitCode=$?
if [ $subScriptExitCode -ne 0 ]; then
exit 1
fi
When a file is sourced, don't use exit, as this will terminate the whole execution. Instead, use return in sub.sh :
return [n]
Causes a function to exit with the return value specified
by n. If used outside a function, but during execution of a script
by the . (source) command, it causes the shell to stop executing that
script and return either n or the exit status of the last command
executed within the script as the exit status of the script. If used
outside a function and not during execution of a script by ., the
return status is false.
sub.sh
type -p <package>
subScriptExitCode="$?"
if [ "$subScriptExitCode" -ne 0 ]; then
return 1
fi
Instead of sourcing the sub script run it as below and check the return code
Main.sh
sh sub.sh
subScriptExitCode=$?
log "subScript ExitCode: $subScriptExitCode"
if [ $subScriptExitCode -ne 0 ]; then
exit $subScriptExitCode
fi
If you have a look at the manual of Bash, then you read
source filename [arguments]:
Read and execute commands from filename in the current shell environment and return the exit status of the last command executed from filename. If filename does not contain ...
source: man bash
These are two very important properties which are related to your problem:
If your sub_script encounters a subScriptExitCode which is different from zero. It will terminate the main_script instantaneously due to the exit statement.
The main_script will set subScriptExitCode to the exit state of the if-statement. This is zero in case subScriuptExitCode of sub_script equals 0.
if list; then list; [ elif list; then list; ] ... [ else list; ] fi: ... The exit status is the exit status of the last command executed, or zero if no condition tested true.
source: man bash
A possible way to solve your problem, making use only of the properties of source would be:
sub_script:
type -p <package>
[ $? -eq 0 ]
Here, the test command will exit with the state 0 if type p <package> terminated with zero, otherwise the test-command will exit with state 1. This state is then picked up in your main_source as $?. However, since type -p can only return 0 or 1, you can just get rid of the test and reduce sub_script to:
type -p <package>
type [-aftpP] name [name ...]: ... type returns true if all of the arguments are found, false if any are not found.
[source: man bash]
Related
I wrote a simple script which will prompt pppd to work. I have read in the pppd documentation that it can return an exit code in the range 1-20. I would like my script to return the value of pppd if it is not 0. I tried this way but unfortunately it failed.
myscript.sh:
#!/bin/sh
exec pppd call ${#:-provider}
if [ $? !-eq 0 ]
then
exit $?
fi
How could i get the exit code from pppd?
Just do:
#!/bin/sh
pppd call "${#:-provider}" || exit
...
or
if pppd call "${#:-provider}"; then : ; else exit; fi
If pppd fails, the script will exit and the value it returns will be that returned by pppd. (You could explicitly write exit $?, but that is the default value returned by exit when no argument is given and is not necessary.) If pppd succeeds, the script will continue.
The problem with your script (other than the ill-advised usage of exec, which I will mostly ignore) is that calling [ $? -ne 0 ] resets $?. You could re-write your script as:
#!/bin/sh
pppd call "${#:-provider}"
save_val=$?
if ! [ $save_val -eq 0 ]
then
exit $save_val
fi
but that seems excessively verbose.
Note that in your original script, you would get an error on the line if [ $? !-eq 0 ], since !-eq is not a valid operator. However, you never see that error because your early invocation of exec makes it so that line is never executed.
I want to know whether any commands in a bash script exited with a non-zero status.
I want something similar to set -e functionality, except that I don't want it to exit when a command exits with a non-zero status. I want it to run the whole script, and then I want to know that either:
a) all commands exited with exit status 0
-or-
b) one or more commands exited with a non-zero status
e.g., given the following:
#!/bin/bash
command1 # exits with status 1
command2 # exits with status 0
command3 # exits with status 0
I want all three commands to run. After running the script, I want an indication that at least one of the commands exited with a non-zero status.
Set a trap on ERR:
#!/bin/bash
err=0
trap 'err=1' ERR
command1
command2
command3
test $err = 0 # Return non-zero if any command failed
You might even throw in a little introspection to get data about where the error occurred:
#!/bin/bash
for i in 1 2 3; do
eval "command$i() { echo command$i; test $i != 2; }"
done
err=0
report() {
err=1
printf '%s' "error at line ${BASH_LINENO[0]}, in call to "
sed -n ${BASH_LINENO[0]}p $0
} >&2
trap report ERR
command1
command2
command3
exit $err
You could try to do something with a trap for the DEBUG pseudosignal, such as
trap '(( $? && ++errcount ))' DEBUG
The DEBUG trap is executed
before every simple command, for command, case command, select command, every arithmetic for command, and before the first command executes in a shell function
(quote from manual).
So if you add this trap and as the last command something to print the error count, you get the proper value:
#!/usr/bin/env bash
trap '(( $? && ++errcount ))' DEBUG
true
false
true
echo "Errors: $errcount"
returns Errors: 1 and
#!/usr/bin/env bash
trap '(( $? && ++errcount ))' DEBUG
true
false
true
false
echo "Errors: $errcount"
prints Errors: 2. Beware that that last statement is actually required to account for the second false because the trap is executed before the commands, so the exit status for the second false is only checked when the trap for the echo line is executed.
I am not sure if there is a ready-made solution for your requirement. I would write a function like this:
function run_cmd_with_check() {
"$#"
[[ $? -ne 0 ]] && ((non_zero++))
}
Then, use the function to run all the commands that need tracking:
run_cmd_with_check command1
run_cmd_with_check command2
run_cmd_with_check command3
printf "$non_zero commands exited with non-zero exit code\n"
If required, the function can be enhanced to store all failed commands in an array which can be printed out at the end.
You may want to take a look at this post for more info: Error handling in Bash
You have the magic variable $? available in bash which tells the exit code of last command:
#!/bin/bash
command1 # exits with status 1
C1_output=$? # will be 1
command2 # exits with status 0
C2_output=$? # will be 0
command3 # exits with status 0
C3_output=$? # will be 0
For each command you could do this:
if ! Command1 ; then an_error=1; fi
And repeat this for all commands
At the end an_error will be 1 if any of them failed.
If you want a count of failures set an_error to 0 at the beginning and do $((an_error++)). Instead of an_error=1
You could place your list of commands into an array and then loop over the commands. Any that return an error code your keep the results for later viewing.
declare -A results
commands=("your" "commands")
for cmd in "${commands[#]}"; do
out=$($cmd 2>&1)
[[ $? -eq 0 ]] || results[$cmd]="$out"
done
Then to see any non zero exit codes:
for cmd in "${!results[#]}"; do echo "$cmd = ${results[$cmd]}"; done
If the length of results is 0, there were no errors on your list of commands.
This requires Bash 4+ (for the associative array)
You can use the DEBUG trap like:
trap 'code+=$?' DEBUG
code=0
# run commands here normally
exit $code
I have a bash script that I use to execute multiple commands in sequence and I need to return non-zero exit code if at least one command in the sequence returns non-zero exit code. I know there is a wait command for that but I'm not sure I understand how to use it.
UPD The script looks like this:
#!/bin/bash
command1
command2
command3
All the commands run in foreground. All the commands need to run regardless of which exit status the previous command returned (so it must not behave as "exit on first error"). Basically I need to gather all the exit statuses and return global exit status accordingly.
Just do it:
EXIT_STATUS=0
command1 || EXIT_STATUS=$?
command2 || EXIT_STATUS=$?
command3 || EXIT_STATUS=$?
exit $EXIT_STATUS
Not sure which of the statuses it should return if several of commands have failed.
If by sequence you mean pipe then you need to set pipefail in your script like set -o pipefail. From man bash:
The return status of a pipeline is the exit status of the last command,
unless the pipefail option is
enabled. If pipefail is enabled, the pipeline's return status is the value of the last (rightmost)
command to exit with a non-zero status, or zero if all commands exit successfully. If the reserved
word ! precedes a pipeline, the exit status of that pipeline is the logical negation of the exit
status as described above. The shell waits for all commands in the pipeline to terminate before
returning a value.
If you just mean sequential commands then just check the exit status of each command and set a flag if the exit status is none zero. Have your script return the value of the flag like:
#!/bin/bash
EXIT=0
grep -q A <<< 'ABC' || EXIT=$? # Will exit with 0
grep -q a <<< 'ABC' || EXIT=$? # Will exit with 1
grep -q A <<< 'ABC' || EXIT=$? # Will exit with 0
echo $EXIT # Will print 1
exit $EXIT # Exit status of script will be 1
This uses the logical operator OR || to only set EXIT if the command fails. If multiple commands fail the exit status from the last failed command will be return by the script.
If these commands are not running in the background then wait isn't relevant here.
If you wish to know which command failed, but not neccessarily its return code you could use:
#!/bin/bash
rc=0;
counter=0;
command1 || let "rc += 1 << $counter"; let counter+=1;
command2 || let "rc += 1 << $counter"; let counter+=1;
command3 || let "rc += 1 << $counter"; let counter+=1;
exit $rc
This uses bit shifting in bash in order to set the bit corresponding to which command failed.
Hence if the first command failed you'll get an return code of 1 (=2^0), if the third failed you would get a return code of 8 (=2^3), and if both the first and the third command failed you would get 9 as the return code.
If you wish to know which command failed:
#!/bin/bash
EXITCODE_RESULT=0
command1
EXIT_CODE_1=$?
command2
EXIT_CODE_2=$?
command3
EXIT_CODE_3=$?
for i in ${!EXIT_CODE_*}
do
# check if the values of the EXIT_CODE vars contain 1
EXITCODE_RESULT=$(($EXITCODE_RESULT || ${!i}))
if [ ${!i} -ne 0 ]
then
var_fail+="'$i' "
else
var_succ+="'$i' "
fi
done
In $var_fail you get a list of the failed EXIT_CODE vars and in $var_succ a list of the successful ones
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
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 have a shell script that executes a number of commands. How do I make the shell script exit if any of the commands exit with a non-zero exit code?
After each command, the exit code can be found in the $? variable so you would have something like:
ls -al file.ext
rc=$?; if [[ $rc != 0 ]]; then exit $rc; fi
You need to be careful of piped commands since the $? only gives you the return code of the last element in the pipe so, in the code:
ls -al file.ext | sed 's/^/xx: /"
will not return an error code if the file doesn't exist (since the sed part of the pipeline actually works, returning 0).
The bash shell actually provides an array which can assist in that case, that being PIPESTATUS. This array has one element for each of the pipeline components, that you can access individually like ${PIPESTATUS[0]}:
pax> false | true ; echo ${PIPESTATUS[0]}
1
Note that this is getting you the result of the false command, not the entire pipeline. You can also get the entire list to process as you see fit:
pax> false | true | false; echo ${PIPESTATUS[*]}
1 0 1
If you wanted to get the largest error code from a pipeline, you could use something like:
true | true | false | true | false
rcs=${PIPESTATUS[*]}; rc=0; for i in ${rcs}; do rc=$(($i > $rc ? $i : $rc)); done
echo $rc
This goes through each of the PIPESTATUS elements in turn, storing it in rc if it was greater than the previous rc value.
If you want to work with $?, you'll need to check it after each command, since $? is updated after each command exits. This means that if you execute a pipeline, you'll only get the exit code of the last process in the pipeline.
Another approach is to do this:
set -e
set -o pipefail
If you put this at the top of the shell script, it looks like Bash will take care of this for you. As a previous poster noted, "set -e" will cause Bash to exit with an error on any simple command. "set -o pipefail" will cause Bash to exit with an error on any command in a pipeline as well.
See here or here for a little more discussion on this problem. Here is the Bash manual section on the set builtin.
"set -e" is probably the easiest way to do this. Just put that before any commands in your program.
If you just call exit in Bash without any parameters, it will return the exit code of the last command. Combined with OR, Bash should only invoke exit, if the previous command fails. But I haven't tested this.
command1 || exit;
command2 || exit;
Bash will also store the exit code of the last command in the variable $?.
[ $? -eq 0 ] || exit $?; # Exit for nonzero return code
http://cfaj.freeshell.org/shell/cus-faq-2.html#11
How do I get the exit code of cmd1 in cmd1|cmd2
First, note that cmd1 exit code could be non-zero and still don't mean an error. This happens for instance in
cmd | head -1
You might observe a 141 (or 269 with ksh93) exit status of cmd1, but it's because cmd was interrupted by a SIGPIPE signal when head -1 terminated after having read one line.
To know the exit status of the elements of a pipeline
cmd1 | cmd2 | cmd3
a. with Z shell (zsh):
The exit codes are provided in the pipestatus special array.
cmd1 exit code is in $pipestatus[1], cmd3 exit code in
$pipestatus[3], so that $? is always the same as
$pipestatus[-1].
b. with Bash:
The exit codes are provided in the PIPESTATUS special array.
cmd1 exit code is in ${PIPESTATUS[0]}, cmd3 exit code in
${PIPESTATUS[2]}, so that $? is always the same as
${PIPESTATUS: -1}.
...
For more details see Z shell.
For Bash:
# This will trap any errors or commands with non-zero exit status
# by calling function catch_errors()
trap catch_errors ERR;
#
# ... the rest of the script goes here
#
function catch_errors() {
# Do whatever on errors
#
#
echo "script aborted, because of errors";
exit 0;
}
In Bash this is easy. Just tie them together with &&:
command1 && command2 && command3
You can also use the nested if construct:
if command1
then
if command2
then
do_something
else
exit
fi
else
exit
fi
#
#------------------------------------------------------------------------------
# purpose: to run a command, log cmd output, exit on error
# usage:
# set -e; do_run_cmd_or_exit "$cmd" ; set +e
#------------------------------------------------------------------------------
do_run_cmd_or_exit(){
cmd="$#" ;
do_log "DEBUG running cmd or exit: \"$cmd\""
msg=$($cmd 2>&1)
export exit_code=$?
# If occurred during the execution, exit with error
error_msg="Failed to run the command:
\"$cmd\" with the output:
\"$msg\" !!!"
if [ $exit_code -ne 0 ] ; then
do_log "ERROR $msg"
do_log "FATAL $msg"
do_exit "$exit_code" "$error_msg"
else
# If no errors occurred, just log the message
do_log "DEBUG : cmdoutput : \"$msg\""
fi
}