if statement not working in cron - linux

I have the following code: (record.sh)
cd $(dirname $0)
dt=$(date '+%d/%m/%Y %H:%M:%S');
echo $dt;
read action < /home/nfs/sauger/web/pi/action.txt
echo $action;
if [[ $action == *"start"* ]]
then
echo "start recording"
./gone.sh
exit 1
elif [[ $action == *"stop"* ]]
then
echo "stop recording"
./gone.sh
exit 1
else
#More stuff done here
fi
When I run this script manually the output is the following:
19/01/2016 19:07:11
start
start recording
If the same script is run via a (root) cronjob, the output is the following:
19/01/2016 19:07:01
start
As you can see, the file "action.txt" has been read without a problem ("start" is logged both times) so this should not be an issue of permissions or wrong paths. But when run as a cronjob, the if-statement is not called. No "start recording" appears.
So my question is: Why does the if-statement work when I call the script manually, but not when this is done via cron?

Your script is written for bash; these errors are almost certainly indicative of it being run with /bin/sh instead.
Either add an appropriate shebang and ensure that it's being called in a way that honors it (/path/to/script rather than sh /path/to/script), or fix it to be compatible. For instance:
case $action in
*start*)
echo "start recording"
./gone.sh
exit 1
;;
*stop*)
echo "stop recording"
./gone.sh
exit 1
;;
esac

Related

can shell script make itself run in background after running some steps?

I have BBB based custom Embedded Linux based board with busybox shell(ash)
I have a situation where my script must run in background with following condition
There must only one instance of the script.
wrapper script need to know if script started successfully in background or not.
There is another wrapper script which starts and stops my script, wrapper script is as mentioned below.
#!/bin/sh
export PATH=/bin:/sbin:/usr/bin:/usr/sbin
readonly TEST_SCRIPT_PATH="/home/testscript.sh"
readonly TEST_SCRIPT_LOCK_PATH="/var/run/${TEST_SCRIPT_PATH##*/}.lock"
start_test_script()
{
local pid_of_testscript=0
local status=0
#Run test script in background
"${TEST_SCRIPT_PATH}" &
#---------Now When this point is hit, lock file must be created.-----
if [ -f "${TEST_SCRIPT_LOCK_PATH}" ];then
pid_of_testscript=$(head -n1 ${TEST_SCRIPT_LOCK_PATH})
if [ -n "${pid_of_testscript}" ];then
kill -0 ${pid_of_testscript} &> /dev/null || status="${?}"
if [ ${status} -ne 0 ];then
echo "Error starting testscript"
else
echo "testscript start successfully"
fi
else
echo "Error starting testscript.sh"
fi
fi
}
stop_test_script()
{
local pid_of_testscript=0
local status=0
if [ -f "${TEST_SCRIPT_LOCK_PATH}" ];then
pid_of_testscript=$(head -n1 ${TEST_SCRIPT_LOCK_PATH})
if [ -n "${pid_of_testscript}" ];then
kill -0 ${pid_of_testscript} &> /dev/null || status="${?}"
if [ ${status} -ne 0 ];then
echo "testscript not running"
rm "${TEST_SCRIPT_LOCK_PATH}"
else
#send SIGTERM signal
kill -SIGTERM "${pid_of_testscript}"
fi
fi
fi
}
#Script starts from here.
case ${1} in
'start')
start_test_script
;;
'stop')
stop_test_script
;;
*)
echo "Usage: ${0} [start|stop]"
exit 1
;;
esac
Now actual script "testscript.sh" looks something like this,
#!/bin/sh
#Filename : testscript.sh
export PATH=/bin:/sbin:/usr/bin:/usr/sbin
set -eu
LOCK_FILE="/var/run/${0##*/}.lock"
FLOCK_CMD="/bin/flock"
FLOCK_ID=200
eval "exec ${FLOCK_ID}>>${LOCK_FILE}"
"${FLOCK_CMD}" -n "${FLOCK_ID}" || exit 0
echo "${$}" > "${LOCK_FILE}"
# >>>>>>>>>>-----Now run the code in background---<<<<<<
handle_sigterm()
{
# cleanup
"${FLOCK_CMD}" -u "${FLOCK_ID}"
if [ -f "${LOCK_FILE}" ];then
rm "${LOCK_FILE}"
fi
}
trap handle_sigterm SIGTERM
while true
do
echo "do something"
sleep 10
done
Now in above script you can see "---Now run the code in background--" at that point I am sure that either lock file is successfully created or instance of this script is already running. So Then I can safely run other code in background and wrapper script can check for lockfile and find out if the process mentioned in the lock file is running or not.
can shellscript itself make it to run in background ?
if not is there a better way to meet all the conditions ?
I think you can look into job control built-in, specifically bg.
Job Control Commands
When processes say they background themselves, what they actually do is fork and exit the parent. You can do the same by running whichever commands, functions or statements you want with & and then exiting.
#!/bin/sh
echo "This runs in the foreground"
sleep 3
while true
do
sleep 10
echo "doing background things"
done &

How to run bash script while it returns code 0?

I have bash script with many lines of code and I need run it while it returns $? == 0, but in case if it has error I need stop it and exit with code 1?
The question is how to do it?
I tried to use set -e command, but Jenkins does not marks build as failed, for him it looks like Success
I also need to get the Error message to show it in my Jenkins log
I managed to get error code(in my case it will be 126), but how to get error message?
main file
fileWithError.sh
rc=$?; if [[ $rc != 0 ]]; then
echo "exit {$rc} ";
fi
fileWithError.sh
#!/bin/sh
set -e
echo "Test"
agjfsjgfshgd
echo "Test2"
echo "Test3"
Just add the command set -e to the beginning of the file
This should look something similar to this
#!/bin/sh
set -e
#...Your code...
I think you just want:
#!/bin/sh
while fileWithError.sh; do
sleep 1;
done
echo fileWithError.sh failed!! >&2
Note that if the script is written well, then the echo is
redundant as fileWithError.sh should have written a decent
error message already. Also, the sleep may not be needed, but is useful to prevent a fast loop if the script succeeds quickly.
You can get the explicit return value, but it requires a bit of refactoring.
#!/bin/sh
true
while test $? = 0; do fileWithError.sh; done
echo fileWithError.sh failed with status $?!! >&2
since the return value of the while script will be the
return value of sleep in the first construction.
Its not quite easy to get an error code only.
How about this ...
#!/bin/bash
Msg=$(fileWithError.sh 2>&1) # redirect all error messages to stdout
if [ "$?" -ne 0 ] # Not Equal
then
echo "$Msg"
exit 1
fi
exit 0
You catch all messages created by fileWithError.sh and if the programm returned an error code then you have the error message already saved in a variable.
But this will make a disadvantage, because you will temporary store all messages created by fileWithError.sh till the error appears.
You can filter the error message with echo "$Msg" |tail -n 1, but its not 100% save.
You should also do some changes in fileWithError.sh...
Switch set -e with trap "exit 1" ERR. this will close the script on errors.
Hope this will help.

Check all commands exit code within a bash script

Consider the case where I have a very long bash script with several commands. Is there a simple way to check the exit status for ALL of them easily. So if there's some failure I can show which command has failed and its return code.
I mean, I don't want to use the test for each one of them checks like the following:
my_command
if [ $status -ne 0 ]; then
#error case
echo "error while executing " my_command " ret code:" $?
exit 1
fi
You can do trap "cmd" ERR, which invokes cmd when a command fails. However this solution has a couple of drawbacks. For example, it does not catch a failure inside a pipe.
In a nutshell, you are better off doing your error management properly, on a case by case basis.
You can write a function that launches:
function test {
"$#"
local status=$?
if [ $status -ne 0 ]; then
echo "error with $1" >&2
fi
return $status
}
test command1
test command2
One can test the value of $? or simply put set -e at the top of the script which will cause it to exit upon any command which errors.
#!/bin/sh
set -xe
my_command1
# never makes it here if my_command1 fails
my_command2

shell to find a file , execute it - exit if 'error' and continue if ' no error'

I have to write a shell script and i don't know how to go about it.
Basically i have to write a script where i'd find a file ( it could be possibly named differently). If either file exists then it must be executed, if it returns a 0 ( no error), it should continue the build, if it's not equal to 0 ( returns with error), it should exit. If either file is not found it should continue the build.
the file i have to find could be either file.1 or file.2 so it could be either named (file.1), or (file.2).
some of the conditions to make it more clear.
1) if either file exists , it should execute - if it has any errors it should exit, if no errors it should continue.
2) none could exist, if that's the case then it should continue the build.
3) both files will not be present at the same time ( additional info)
I have tried to write a script but i doubt it's even closer to what i am looking for.
if [-f /home/(file.1) or (file.2)]
then
-exec /home/(file.1) or (file.2)
if [ $! -eq 0]; then
echo "no errors continuing build"
fi
else
if [ $! -ne 0] ; then
exit ;
fi
else
echo "/home/(file.1) or (file.2) not found, continuing build"
fi
any help is much appreciated.
Thanks in advance
DOIT=""
for f in file1.sh file2.sh; do
if [ -x /home/$f ]; then DOIT="/home/$f"; break; fi
done
if [ -z "$DOIT" ]; then echo "Files not found, continuing build"; fi
if [ -n "$DOIT" ]; then $DOIT && echo "No Errors" || exit 1; fi
For those confused about my syntax, try running this:
true && echo "is true" || echo "is false"
false && echo "is true" || echo "is false"
Just putting the line
file.sh
in your script should work, if you set up your script to exit on errors.
For example, if your script was
#!/bin/bash -e
echo one
./file.sh
echo two
Then if file.sh exists and is executable it would run and your whole script would run. If not, the script would fail when it tried to execute the non-existing file.
If you want to execute one file or the other, extend the idea to the following:
#!/bin/bash -e
echo one
./file1.sh || ./file2.sh
echo two
This means if file1.sh does not exist, it will try file2.sh and if that is there it will run and your whole script will run.
This give preference to file1 of course, meaning if they both exist, then only file1 will run.

Shell script continues to run even after exit command

My shell script is as shown below:
#!/bin/bash
# Make sure only root can run our script
[ $EUID -ne 0 ] && (echo "This script must be run as root" 1>&2) || (exit 1)
# other script continues here...
When I run above script with non-root user, it prints message "This script..." but it doe not exit there, it continues with the remaining script. What am I doing wrong?
Note: I don't want to use if condition.
You're running echo and exit in subshells. The exit call will only leave that subshell, which is a bit pointless.
Try with:
#! /bin/sh
if [ $EUID -ne 0 ] ; then
echo "This script must be run as root" 1>&2
exit 1
fi
echo hello
If for some reason you don't want an if condition, just use:
#! /bin/sh
[ $EUID -ne 0 ] && echo "This script must be run as root" 1>&2 && exit 1
echo hello
Note: no () and fixed boolean condition. Warning: if echo fails, that test will also fail to exit. The if version is safer (and more readable, easier to maintain IMO).
I think you need && rather than||, since you want to echo and exit (not echo or exit).
In addition (exit 1) will run a sub-shell that exits rather than exiting your current shell.
The following script shows what you need:
#!/bin/bash
[ $1 -ne 0 ] && (echo "This script must be run as root." 1>&2) && exit 1
echo Continuing...
Running this with ./myscript 0 gives you:
Continuing...
while ./myscript 1 gives you:
This script must be run as root.
I believe that's what you were looking for.
I would write that as:
(( $EUID != 0 )) && { echo "This script must be run as root" 1>&2; exit 1; }
Using { } for grouping, which executes in the current shell. Note that the spaces around the braces and the ending semi-colon are required.

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