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
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
This question already has answers here:
Why does set -e; true && false && true not exit?
(3 answers)
Closed 1 year ago.
set -eo pipefail
commandThatFails || exitFunction
exitFunction
So this script is running the exitMethod twice... I thought set -e exited immediately on any non zero exit code and set -o pipefail made sure that during the pipeline any failure is the final exit status code not the most recent command?
Therefore I thought :
commandThatFails
Executes exitFunction
set -o pipefail returns non zero exit code as the first command fails
Non zero exit code is detected by set -e and exits immediately
In the docs it states:
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.
I thought that the exitfunction is the command following the final || so therefore would be counted and picked up and exited immediately.
I can resolve the issue with:
commandThatFails || { exitFunction; exit 1; }
but it doesn't seem like the more elegant way to handle this, any thoughts appreciated!
|| is a flow control operator, not a pipeline component. pipefail has no effect on it.
If set -e caused flow control operators to exit, then you could never have an else branch of your script run with it active; it would be completely useless.
For that reason, && and || suppress set -e behavior for their left-hand sides, just like if condition; then success; else fail; fi suppresses that behavior for condition.
General practice is to make your exitFunction actually call exit, so you can write:
die() {
rc=$?
(( $# )) && printf '%s\n' "$*" >&2
exit "$(( rc == 0 ? 1 : rc ))"
}
commandThatFails || die "commandThatFails failed"
...but if you want to call something else first, yes, you need to use a grouping, just as you did in the question.
I am new to bash scripting and I am trying to understand the following line:
# Verify pre-req environment
command -v kubectl > /dev/null 2>&1 || { echo "kubectl pre-req is missing."; exit 1; }
I unerstand this part: command -v kubectl > /dev/null 2>&1 so if it fails or not we redirect the output to /dev/null in which if I understand it correctly means ignore it in all cases. So in what cases the following line will be executed then?
{ echo "kubectl pre-req is missing."; exit 1; }
and how || behaves in this case?
|| is used for OR condition in bash.
For a simplified example lets look at
cmd1 || cmd2
What happens is that if cmd1 succeeds (i.e. exit code 0), then cmd2 will NOT be performed. cmd2 will only be performed if cmd1 fails.
This is one reason why returning exit codes is so important.
For more details, please look at this post: Boolean operators ( &&, -a, ||, -o ) in Bash
I unerstand this part: command -v kubectl > /dev/null 2>&1 so if it fails or not we redirect the output to /dev/null in which if I understand it correctly means ignore it in all cases.
You've misunderstood that; no part of command -v kubectl > /dev/null 2>&1 has any connection to "if it fails" vs. "not".
Rather, > and 2> redirect two different types of output; > (or 1>) redirects whatever the command writes to standard output, and 2> redirects whatever the command writes to standard error. Whether the command writes anything to standard output is independent of whether it ends up succeeding or failing; whether it writes anything to standard error is independent of whether it ends up succeeding or failing; and whether it writes anything to standard output is independent of whether it writes anything to standard error.
For example, here is a command that prints to both standard output and standard error, and then returns successfully:
( echo "this is on standard output" ; echo "this is on standard error" >&2 ; exit 0 )
and here is one that prints to both standard output and standard error, and then fails:
( echo "this is on standard output" ; echo "this is on standard error" >&2 ; exit 1 )
By contrast, || really does have to do with whether a command succeeds (exiting with zero) or fails (exiting with something other than zero).
If the kubectl command exits with a non-zero exit status, the code after the || will be executed.
$ sh -c 'exit 42' && echo "exit status zero" || echo "exit status non-zero"
exit status non-zero
$ sh -c 'exit 0' && echo "exit status zero" || echo "exit status non-zero"
exit status zero
|| is a logic or and acts based on previous command exit code (0 means true and other value is false).
In this case, it will only execute the echo part when the first command (kubectl..) exit code is not 0 (i.e: fails). The exit 1 is to make the whole command fail, otherwise the exit code would be 0 from echo command.
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
We are writing shell scripts with set -e as policy which means that it will exit if any unhandled non-zero exit status appears.
#!/bin/sh -e
if some_command; then
experience_happyness
else
print error status of some_command to log or standard error
experience_sadness
exit 1
fi
The $? expression evaluates to 0 at the beginning of the else branch. If I don't run the some_command inside if then an eventual error will terminate the shell script immediately.
How can I know the exit status of a program when set -e is effective without terminating the script?
I'm interested in bash specific solutions too if pure sh solutions are not available.
EDIT: My bad. as #"that other guy" answered I was mistaken when I told that "$?" evaluates to 0 at the beginning of the else branch. I tried it, but I made some mistake when tried it. Sorry.
I think we may keep this question because of the pro quality answer. Should we?
The $? expression evaluates to 0 at the beginning of the else branch.
No it doesn't.
#!/bin/sh -e
some_command() { return 42; }
if some_command; then
echo "Worked"
else
echo "Command failed with $?"
exit 1
fi
will print Command failed with 42.