How to capture the output of one shell script into other in unix

Question

I am running one shell script run_sftp.sh whose output will be either "done" or "failed" and I am calling this script into another script which will execute some command if run_sftp.sh output is "done"

If [ Output(run_sftp.sh) = 'done' ] then
  echo "run"
else
  "Stop running"
fi 

This is the algorithm. Please suggest.

Solution 1

like this?

retval=$(path/to/run_sftp.sh)

now you have done/failed in var retval. you can do your if check with your logic.

Solution 2

You can capture the output using $(), or the inverse `

So:

if [ $(run_sftp.sh) = 'done' ] then

etc

Solution 3

I am assuming that your script run_sftp.sh will call other statements and produce done message only when it is success and no other commands called by run_sftp.sh will produce done message. In that case you capture output of script run and then grep done

MSG=$(run_sftp.sh)
echo $MSG | grep 'done'
if [ $? -eq 0 ]
then
echo "run"
else
exit 9
fi

Thanks

Comments

[Pooja25]

I am executing other commands also so there will be other outputs except Done and Failed, but I want to capture only done, then in that case how should I capture only done from command output?

[Pooja25]

I am executing other commands also so there will be other outputs except Done and Failed, but I want to capture only done, then in that case how should I capture only done from command output?

[Kent]

regex match check @Pooja25

[Pooja25]

I am using retval=$(sh run_sftp.sh) cat $retval | grep "Done" but it is not working

[Kent]

it won't work, if you want to use grep, try grep -qi "done" <<< $retval and check the return code of the grep. @Pooja25