I have a shell script which updates different firmwares with different executables.
I need to know if one of the executable has hung and not returning back to shell.
Can I introduce any kind of timeout ?
Sample shell script below. How to handle if updatefw command hangs and does not return.
#!/bin/sh
updatefw -c config.cfg
if [ $? != 0 ]; then
echo "exec1 failed"
exit 1
fi
exit 0
I suggest with timeout from GNU core utilities:
#!/bin/bash
timeout 30 updatefw -c config.cfg
if [[ $? == 124 ]]; then
echo "update failed"
exit 1
fi
When timeout quits updatefw, the return code is 124.
I assume here that the update will never take longer than 30 seconds.
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
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 $?
1
#!/bin/bash -e
should suffice.
I am just throwing in another one for reference since there was an additional question to Mark Edgars input and here is an additional example and touches on the topic overall:
[[ `cmd` ]] && echo success_else_silence
Which is the same as cmd || exit errcode as someone showed.
For example, I want to make sure a partition is unmounted if mounted:
[[ `mount | grep /dev/sda1` ]] && umount /dev/sda1
I'm not very good on bash I've been modifying a code to create a lock file so a cron don't execute a second time if the first process hasn't finish.
LOCK_FILE=./$(hostname)-lock
(set -C; : > $LOCK_FILE) 2> /dev/null
if [ $? != "0" ]; then
echo "already running (lock file exists); exiting..."
exit 1
fi
trap 'rm $LOCK_FILE' INT TERM EXIT
when I run it for the first time I get the message already running as if the file already existed.
Perhaps I'm missing something
#!/bin/sh
(
# Wait for lock on /tmp/lock
flock -x -w 10 200 || exit 127 # you can use or not use -w
#your stuff here
) 200> /tmp/lock
check man page flock.
This is the tool for you.
And it comes with example in man page :)
How can I have my shell script echo to me that the script that it calls has failed?
#!/bin/sh
test="/Applications/test.sh"
sh $test
exit 0
exit 1
#!/bin/sh
if sh /Applications/test.sh; then
echo "Well done $USER"
exit 0
else
echo "script failed with code [$?]" >&2
exit 1
fi
The /Applications/test.sh script should be well coded to exit with conventional status. 0 if it's ok and > 0 if it fails.
Like you can see, no need to test the special variable $?, we use boolean expression directly.
I usually take the following approach:
#!/usr/bin/env bash
test="/Applications/test.sh"
sh "${test}"
exit_status=$?
if [[ ${exit_status} ]] ; then
echo "Error: ${test} failed with status ${exit_status}." >&2
else
echo "Success!"
fi
In terms of best practice, you should not. If a script fails, it should emit an error message before it terminates so that its parent doesn't have to. The main reason for this is that the process knows why it is failing, while the parent can only guess. In other words, you should just write:
#!/bin/sh
test="/Applications/test.sh"
sh $test
Although really, it would be more typical to just write:
#!/bin/sh
/Applications/test.sh
test.sh will emit the necessary error message, and your script will return the same value as did test.sh. Also, in its current form your script will always be successful, even if test.sh actually failed because exit 0; exit 1 is pretty pointless: the exit 1 will never be called.
Greetings all. I'm setting up a cron job to execute a bash script, and I'm worried that the next one may start before the previous one ends. A little googling reveals that a popular way to address this is the flock command, used in the following manner:
flock -n lockfile myscript.sh
if [ $? -eq 1 ]; then
echo "Previous script is still running! Can't execute!"
fi
This works great. However, what do I do if I want to check the exit code of myscript.sh? Whatever exit code it returns will be overwritten by flock's, so I have no way of knowing if it executed successfully or not.
It looks like you can use the alternate form of flock, flock <fd>, where <fd> is a file descriptor. If you put this into a subshell, and redirect that file descriptor to your lock file, then flock will wait until it can write to that file (or error out if it can't open it immediately and you've passed -n). You can then do everything in your subshell, including testing the return value of scripts you run:
(
if flock -n 200
then
myscript.sh
echo $?
fi
) 200>lockfile
According to the flock man page, flock has a -E or --exit-conflict-code flag you can use to set what the exit code of flock should be in the case a conflict occurs:
-E, --conflict-exit-code number
The exit status used when the -n option is in use, and the conflicting lock exists, or the -w option is in use, and the timeout is reached. The default value is 1. The number has to be in the range of 0 to 255.
The man page also states:
EXIT STATUS
The command uses sysexits.h exit status values for everything, except when using either of the options -n or -w which report a failure to acquire the lock with a exit status given by the -E option, or 1 by default. The exit status given by -E has to be in the range of 0 to 255.
When using the command variant, and executing the child worked, then the exit status is that of the child command.
So, in the case of the -n or -w flags while using the "command" variant, you can see both exit statuses.
Example:
$ flock --exclusive /tmp/flock.lock bash -c 'exit 42'; echo $?
42
$ flock --exclusive /tmp/flock.lock flock --exclusive --nonblock --conflict-exit-code 100 /tmp/flock.lock bash -c 'exit 42'; echo $?
100
In the first example, we see that we get back the exit status of the process we're running with flock. In the second example, we are creating contention for the lock. In that case, flock itself returns the status code we tell it (100). If you do not specify a value with the --conflict-exit-code flag, it will return 1 instead. However, I prefer setting less common values to prevent confusion from other processess/scripts which also might return a value of 1.
#!/bin/bash
if ! pgrep myscript.sh; then
flock -n lockfile myscript.sh
fi
If I understand you right, you want to make sure 'myscript.sh' is not running before cron attempts to run your command again. Assuming that's right, we check to see if pgrep failed to find myscript.sh in the processes list and if so we run the flock command again.
Perhaps something like this would work for you.
#!/bin/bash
RETVAL=0
lockfailed()
{
echo "cannot flock"
exit 1
}
(
flock -w 2 42 || lockfailed
false
RETVAL=$?
echo "original retval $RETVAL"
exit $RETVAL
) 42>|/tmp/flocker
RETVAL=$?
echo "returned $RETVAL"
exit $RETVAL