Motivation
I'm in a situation where I have to run multiple bash commands with a single bash invocation without the possibility to write a full script file (use case: Passing multiple commands to a container in Kubernetes). A common solution is to combine commands with ; or &&, for instance:
bash -c " \
echo \"Hello World\" ; \
ls -la ; \
run_some_command "
In practice writing bash scripts like that turns out to be error prone, because I often forget the semicolon leading to subtle bugs.
Inspired by this question, I was experiment with writing scripts in a more standard style by using a heredoc:
bash <<EOF
echo "Hello World"
ls -la
run_some_command
EOF
Unfortunately, I noticed that there is a difference in exit code error handling when using a heredoc. For instance:
bash -c " \
run_non_existing_command ; \
echo $? "
outputs (note that $? properly captures the exit code):
bash: run_non_existing_command: command not found
127
whereas
bash <<EOF
run_non_existing_command
echo $?
EOF
outputs (note that $? fails to capture the exit code compared to standard script execution):
bash: line 1: run_non_existing_command: command not found
0
Why is the heredoc version behaving differently? Is it possible to write the script in the heredoc style and maintaining normal exit code handling?
Why is the heredoc version behaving differently?
Because $? is expanded before running the command.
The following will output 1, that is the exit status of false command:
false
bash <<EOF
run_non_existing_command
echo $?
EOF
It's the same in principle as the following, which will print 5:
variable=5
bash <<EOF
variable="This is ignored"
echo $variable
EOF
Is it possible to write the script in the heredoc style and maintaining normal exit code handling?
If you want to have the $? expanded inside the subshell, then:
bash <<EOF
run_non_existing_command
echo \$?
EOF
or
bash <<'EOF'
run_non_existing_command
echo $?
EOF
Also note that:
bash -c \
run_non_existing_command ;
echo $? ;
is just equal to:
bash -c run_non_existing_command
echo $?
The echo $? is not executed inside bash -c.
I want to run the bash script, StartSomething.sh, as a specific user. I use runuser command for that. Also I want to know an exit code from this bash script. So I write an exit code to the file when the command is finished or interrupted. Here is the code:
runuser myuser -s /bin/bash -c "./StartSomething.sh --pidfile=${pidfile}; \
echo $? > ${statusfile};" &
sleep 5
pid=$(cat ${pidfile})
while ps -p ${pid} > /dev/null; do sleep 1; done
end=$(cat ${statusfile})
echo "End code: ${end}"
exit ${end}
Problem is that exit code is still 0, though bash script is interrupted. What can be wrong?
If I have separate file, start.sh, with this code:
./StartSomething.sh --pidfile=${pidfile}
echo $? > ${statusfile}
and runuser command look like this:
runuser myuser -s /bin/bash -c "./start.sh" &
everything is working fine. I want to use first example without separate file. Can someone tell me what can be wrong? Is there better solution for this problem?
If all you want to do is to run the program in the background, and wait for it to finish, I think you could also use wait to get the return value (runuser passes it through, unless something exceptional happens):
runuser myuser ./StartSomething.sh --pidfile=${pidfile} &
pid=$!
# do something else
wait $!
echo "it returned $?"
or
runuser myuser ./StartSomething.sh --pidfile=${pidfile} &
pid=$!
echo -n "waiting"
while kill -0 $pid 2>/dev/null; do
echo -n "."
sleep 1
done
echo
wait $!
echo "it returned $?"
There is problem with escaping special character $. Correct command:
runuser myuser -s /bin/bash -c "./StartSomething.sh --pidfile=${pidfile}; \
echo \$? > ${statusfile};" &
Replace $? with \$?.
I'm trying to figure out what this means/how $? gets populated in linux, I tried doing a search but if someone could clarify that would be great:
exitstat=$?
if [ $exitstat -ne 0 ]
then
echo -e "Could Not Extract"
echo -e "Aborting Script `date`"
exit $exitstat
fi
The code above that is:
_xfile << %% 2> /files/thefile-7000.log | _afile -x -r 10 2> /files/thefile-7000.log > /files/thefile.7000
OperatorHigh = $finalnumber
%%
$? expands to the exit status of the most recent foreground command.
Since your prior command is a pipeline, the exit status is that of the last command in the pipeline -- in this case, _afile -- unless the pipefail shell option is set, in which case failures elsewhere in the pipeline can also make exit status nonzero.
I'm trying to call a custom shell script through sh:
/bin/sh -c 'myscript.sh` >log.txt 2>&1 & echo $!
Output of this command is a PID of a created background process. I want to instruct /bin/sh to save return code of myscript.sh to some file. Is it possible?
echo $? >> /path/to/return_code
$? has the return code of the last statement in bash.
(/bin/sh -c "myscript.sh" >log.txt 2>&1 ; echo $? >somefile) & echo $!
(
/bin/sh -c 'myscript.sh` >log.txt 2>&1
echo $? > some_file
) &
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