I'm seeing some behavior that doesn't make sense to me when I run a bash script with the -e option that has multiple commands strung together with &&s and one of them fails. I would expect the script to stop on the failed command and return the exit status, but instead it just executes the rest of the script happily.
Here are examples that make sense to me:
$ false && true; echo $?
1
$ bash -xe -c "false && true"; echo $?
+ false
1
$ bash -xe -c "false; true"; echo $?
+ false
1
And here is the one that does not make sense to me:
$ bash -xe -c "false && true; true"; echo $?
+ false
+ true
0
This is where I don't understand what is going on. false && true returns status 1 so shouldn't the script stop executing and return status 1, like it does when the script is false; true?
While experimenting, I found that it works the way I would expect if I surround the chain of commands with parentheses:
$ bash -xe -c "(false && true); true"; echo $?
+ false
1
Can anybody give an explanation for this?
The logic here is that your use of && already is error-checking. The same way bash doesn't treat a failure within an if condition as worth aborting, even with set -e.
When you wrap the commands in a parenthesis, you are actually running those commands within a subshell, so the script itself only sees the return of that subshell, ie: it isn't aware that && is involved at all, so it aborts as you expect.
Quoth the reference manual:
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 !
To avoid exiting bash scripts (that has set -e) when evaluating false conditions, it is safer to use If-else statements, rather than short-circuiting.
But if you prefer the shorter syntax (like I do ;-), replace:
[[ condition ]] && command
With:
[[ ! condition ]] || command
To illustrate:
$ false && do_whatever ; echo rc=$?
rc=1 # i.e. Will exit bash at this point
$ ! false || do_whatever ; echo rc=$?
rc=0 # i.e. Bash will continue...
Related
What is the purpose of a shell command (part of a shell script) starting with an exclamation mark?
Concrete example:
In foo.sh:
#!/usr/bin/env bash
set -e
! docker stop foo
! docker rm -f foo
# ... other stuff
I know that without the space the exclamation mark is used for history replacements and ! <expression> according to the man page can be used to evaluate "True if expr is false". But in the example context that does not make sense to me.
TL;DR: This is just by-passing the set -e flag in the specific line where you are using it.
Adding add to hek2mgl's correct and useful answer.
You have:
set -e
! command
Bash Reference Manual → Pipelines describes:
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.
This means that ! preceding a command is negating the exit status of it:
$ echo 23
23
$ echo $?
0
# But
$ ! echo 23
23
$ echo $?
1
Or:
$ echo 23 && echo "true" || echo "fail"
23
true
$ ! echo 23 && echo "true" || echo "fail"
23
fail
The exit status is useful in many ways. In your script, used together with set -e makes the script exit whenever a command returns a non-zero status.
Thus, when you have:
set -e
command1
command2
If command1 returns a non-zero status, the script will finish and won't proceed to command2.
However, there is also an interesting point to mention, described in 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 !. If a compound command other than a subshell returns a non-zero status because a command failed while -e was being ignored, the shell does not exit. A trap on ERR, if set, is executed before the shell exits.
Taking all of these into consideration, 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.
If you don't want the script to fail in both cases, error or success of the command, you can also use this alternative:
set -e
docker stop foo || true
The boolean or true makes the pipeline always have 0 as the return value.
I am using following options
set -o pipefail
set -e
In bash script to stop execution on error. I have ~100 lines of script executing and I don't want to check return code of every line in the script.
But for one particular command, I want to ignore the error. How can I do that?
The solution:
particular_script || true
Example:
$ cat /tmp/1.sh
particular_script()
{
false
}
set -e
echo one
particular_script || true
echo two
particular_script
echo three
$ bash /tmp/1.sh
one
two
three will be never printed.
Also, I want to add that when pipefail is on,
it is enough for shell to think that the entire pipe has non-zero exit code
when one of commands in the pipe has non-zero exit code (with pipefail off it must the last one).
$ set -o pipefail
$ false | true ; echo $?
1
$ set +o pipefail
$ false | true ; echo $?
0
Just add || true after the command where you want to ignore the error.
Don't stop and also save exit status
Just in case if you want your script not to stop if a particular command fails and you also want to save error code of failed command:
set -e
EXIT_CODE=0
command || EXIT_CODE=$?
echo $EXIT_CODE
More concisely:
! particular_script
From the POSIX specification regarding set -e (emphasis mine):
When this option is on, if a simple command fails for any of the reasons listed in Consequences of Shell Errors or returns an exit status value >0, and is not part of the compound list following a while, until, or if keyword, and is not a part of an AND or OR list, and is not a pipeline preceded by the ! reserved word, then the shell shall immediately exit.
Instead of "returning true", you can also use the "noop" or null utility (as referred in the POSIX specs) : and just "do nothing". You'll save a few letters. :)
#!/usr/bin/env bash
set -e
man nonexistentghing || :
echo "It's ok.."
Thanks for the simple solution here from above:
<particular_script/command> || true
The following construction could be used for additional actions/troubleshooting of script steps and additional flow control options:
if <particular_script/command>
then
echo "<particular_script/command> is fine!"
else
echo "<particular_script/command> failed!"
#exit 1
fi
We can brake the further actions and exit 1 if required.
I found another way to solve this:
set +e
find "./csharp/Platform.$REPOSITORY_NAME/obj" -type f -iname "*.cs" -delete
find "./csharp/Platform.$REPOSITORY_NAME.Tests/obj" -type f -iname "*.cs" -delete
set -e
You can turn off failing on errors by set +e this will now ignore all errors after that line. Once you are done, and you want the script to fail again on any error, you can use set -e.
After applying set +e the find does not fail the whole script anymore, when files are not found. At the same time, error messages
from find are still printed, but the whole script continues to execute. So it is easy to debug if that causes the problem.
This is useful for CI & CD (for example in GitHub Actions).
If you want to prevent your script failing and collect the return code:
command () {
return 1 # or 0 for success
}
set -e
command && returncode=$? || returncode=$?
echo $returncode
returncode is collected no matter whether command succeeds or fails.
output=$(*command* 2>&1) && exit_status=$? || exit_status=$?
echo $output
echo $exit_status
Example of using this to create a log file
log_event(){
timestamp=$(date '+%D %T') #mm/dd/yy HH:MM:SS
echo -e "($timestamp) $event" >> "$log_file"
}
output=$(*command* 2>&1) && exit_status=$? || exit_status=$?
if [ "$exit_status" = 0 ]
then
event="$output"
log_event
else
event="ERROR $output"
log_event
fi
I have been using the snippet below when working with CLI tools and I want to know if some resource exist or not, but I don't care about the output.
if [ -z "$(cat no_exist 2>&1 >/dev/null)" ]; then
echo "none exist actually exist!"
fi
while || true is preferred one, but you can also do
var=$(echo $(exit 1)) # it shouldn't fail
I kind of like this solution :
: `particular_script`
The command/script between the back ticks is executed and its output is fed to the command ":" (which is the equivalent of "true")
$ false
$ echo $?
1
$ : `false`
$ echo $?
0
edit: Fixed ugly typo
If i copy and paste all the commands into the terminal..
some do not even go through.
so the solution is perhaps to turn the file into an executable file
and then execute it.
but what if some commands fail.
the script keeps on executing the other commands.
obviously there is no solution to this right ?
The easiest way to do this is to use the -e option in your shell. For example:
#!/bin/sh -e
command1
command2
In this script, if command1 fails, then the script as a whole will fail at that point without running any further commands.
You can check the error code from commands you run
#!/bin/bash
function test {
"$#"
status=$?
if [ $status -ne 0 ]; then
echo "error with $1"
exit 255
fi
return $status
}
test ls
test ps -ef
test not_a_command
taken from here for more information Checking Bash exit status of several commands efficiently
#Terminal, you were almost there.
If you just stick && on the end of each command, then execution will stop with the first failure (ie. the first command that returns a non-zero exit code).
Example:
#!/bin/sh
true &&
echo 'got here' &&
echo 'got here too' &&
false &&
echo 'also got here'
produces the output
got here
got here too
(Actually, I thought it would also require line-continuation markers too: && \, but a quick test showed otherwise.)
Note: All of the above assumes that your shell is bash; I can't speak for other shells.
I am using following options
set -o pipefail
set -e
In bash script to stop execution on error. I have ~100 lines of script executing and I don't want to check return code of every line in the script.
But for one particular command, I want to ignore the error. How can I do that?
The solution:
particular_script || true
Example:
$ cat /tmp/1.sh
particular_script()
{
false
}
set -e
echo one
particular_script || true
echo two
particular_script
echo three
$ bash /tmp/1.sh
one
two
three will be never printed.
Also, I want to add that when pipefail is on,
it is enough for shell to think that the entire pipe has non-zero exit code
when one of commands in the pipe has non-zero exit code (with pipefail off it must the last one).
$ set -o pipefail
$ false | true ; echo $?
1
$ set +o pipefail
$ false | true ; echo $?
0
Just add || true after the command where you want to ignore the error.
Don't stop and also save exit status
Just in case if you want your script not to stop if a particular command fails and you also want to save error code of failed command:
set -e
EXIT_CODE=0
command || EXIT_CODE=$?
echo $EXIT_CODE
More concisely:
! particular_script
From the POSIX specification regarding set -e (emphasis mine):
When this option is on, if a simple command fails for any of the reasons listed in Consequences of Shell Errors or returns an exit status value >0, and is not part of the compound list following a while, until, or if keyword, and is not a part of an AND or OR list, and is not a pipeline preceded by the ! reserved word, then the shell shall immediately exit.
Instead of "returning true", you can also use the "noop" or null utility (as referred in the POSIX specs) : and just "do nothing". You'll save a few letters. :)
#!/usr/bin/env bash
set -e
man nonexistentghing || :
echo "It's ok.."
Thanks for the simple solution here from above:
<particular_script/command> || true
The following construction could be used for additional actions/troubleshooting of script steps and additional flow control options:
if <particular_script/command>
then
echo "<particular_script/command> is fine!"
else
echo "<particular_script/command> failed!"
#exit 1
fi
We can brake the further actions and exit 1 if required.
I found another way to solve this:
set +e
find "./csharp/Platform.$REPOSITORY_NAME/obj" -type f -iname "*.cs" -delete
find "./csharp/Platform.$REPOSITORY_NAME.Tests/obj" -type f -iname "*.cs" -delete
set -e
You can turn off failing on errors by set +e this will now ignore all errors after that line. Once you are done, and you want the script to fail again on any error, you can use set -e.
After applying set +e the find does not fail the whole script anymore, when files are not found. At the same time, error messages
from find are still printed, but the whole script continues to execute. So it is easy to debug if that causes the problem.
This is useful for CI & CD (for example in GitHub Actions).
If you want to prevent your script failing and collect the return code:
command () {
return 1 # or 0 for success
}
set -e
command && returncode=$? || returncode=$?
echo $returncode
returncode is collected no matter whether command succeeds or fails.
output=$(*command* 2>&1) && exit_status=$? || exit_status=$?
echo $output
echo $exit_status
Example of using this to create a log file
log_event(){
timestamp=$(date '+%D %T') #mm/dd/yy HH:MM:SS
echo -e "($timestamp) $event" >> "$log_file"
}
output=$(*command* 2>&1) && exit_status=$? || exit_status=$?
if [ "$exit_status" = 0 ]
then
event="$output"
log_event
else
event="ERROR $output"
log_event
fi
I have been using the snippet below when working with CLI tools and I want to know if some resource exist or not, but I don't care about the output.
if [ -z "$(cat no_exist 2>&1 >/dev/null)" ]; then
echo "none exist actually exist!"
fi
while || true is preferred one, but you can also do
var=$(echo $(exit 1)) # it shouldn't fail
I kind of like this solution :
: `particular_script`
The command/script between the back ticks is executed and its output is fed to the command ":" (which is the equivalent of "true")
$ false
$ echo $?
1
$ : `false`
$ echo $?
0
edit: Fixed ugly typo
I have a Bash shell script that invokes a number of commands.
I would like to have the shell script automatically exit with a return value of 1 if any of the commands return a non-zero value.
Is this possible without explicitly checking the result of each command?
For example,
dosomething1
if [[ $? -ne 0 ]]; then
exit 1
fi
dosomething2
if [[ $? -ne 0 ]]; then
exit 1
fi
Add this to the beginning of the script:
set -e
This will cause the shell to exit immediately if a simple command exits with a nonzero exit value. A simple command is any command not part of an if, while, or until test, or part of an && or || list.
See the bash manual on the "set" internal command for more details.
It's really annoying to have a script stubbornly continue when something fails in the middle and breaks assumptions for the rest of the script. I personally start almost all portable shell scripts with set -e.
If I'm working with bash specifically, I'll start with
set -Eeuo pipefail
This covers more error handling in a similar fashion. I consider these as sane defaults for new bash programs. Refer to the bash manual for more information on what these options do.
To add to the accepted answer:
Bear in mind that set -e sometimes is not enough, specially if you have pipes.
For example, suppose you have this script
#!/bin/bash
set -e
./configure > configure.log
make
... which works as expected: an error in configure aborts the execution.
Tomorrow you make a seemingly trivial change:
#!/bin/bash
set -e
./configure | tee configure.log
make
... and now it does not work. This is explained here, and a workaround (Bash only) is provided:
#!/bin/bash
set -e
set -o pipefail
./configure | tee configure.log
make
The if statements in your example are unnecessary. Just do it like this:
dosomething1 || exit 1
If you take Ville Laurikari's advice and use set -e then for some commands you may need to use this:
dosomething || true
The || true will make the command pipeline have a true return value even if the command fails so the the -e option will not kill the script.
If you have cleanup you need to do on exit, you can also use 'trap' with the pseudo-signal ERR. This works the same way as trapping INT or any other signal; bash throws ERR if any command exits with a nonzero value:
# Create the trap with
# trap COMMAND SIGNAME [SIGNAME2 SIGNAME3...]
trap "rm -f /tmp/$MYTMPFILE; exit 1" ERR INT TERM
command1
command2
command3
# Partially turn off the trap.
trap - ERR
# Now a control-C will still cause cleanup, but
# a nonzero exit code won't:
ps aux | grep blahblahblah
Or, especially if you're using "set -e", you could trap EXIT; your trap will then be executed when the script exits for any reason, including a normal end, interrupts, an exit caused by the -e option, etc.
The $? variable is rarely needed. The pseudo-idiom command; if [ $? -eq 0 ]; then X; fi should always be written as if command; then X; fi.
The cases where $? is required is when it needs to be checked against multiple values:
command
case $? in
(0) X;;
(1) Y;;
(2) Z;;
esac
or when $? needs to be reused or otherwise manipulated:
if command; then
echo "command successful" >&2
else
ret=$?
echo "command failed with exit code $ret" >&2
exit $ret
fi
Run it with -e or set -e at the top.
Also look at set -u.
On error, the below script will print a RED error message and exit.
Put this at the top of your bash script:
# BASH error handling:
# exit on command failure
set -e
# keep track of the last executed command
trap 'LAST_COMMAND=$CURRENT_COMMAND; CURRENT_COMMAND=$BASH_COMMAND' DEBUG
# on error: print the failed command
trap 'ERROR_CODE=$?; FAILED_COMMAND=$LAST_COMMAND; tput setaf 1; echo "ERROR: command \"$FAILED_COMMAND\" failed with exit code $ERROR_CODE"; put sgr0;' ERR INT TERM
An expression like
dosomething1 && dosomething2 && dosomething3
will stop processing when one of the commands returns with a non-zero value. For example, the following command will never print "done":
cat nosuchfile && echo "done"
echo $?
1
#!/bin/bash -e
should suffice.
I am just throwing in another one for reference since there was an additional question to Mark Edgars input and here is an additional example and touches on the topic overall:
[[ `cmd` ]] && echo success_else_silence
Which is the same as cmd || exit errcode as someone showed.
For example, I want to make sure a partition is unmounted if mounted:
[[ `mount | grep /dev/sda1` ]] && umount /dev/sda1