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Why can't I get pgrep output right to variable on bash script?



2019 Community Moderator ElectionBash script doesn't TEE output to subdirectoryExiting a Bash script when a sudo child quitsConditional execution block with || and parentheses problemGet PID and return code from 1 line bash callWhy is “Doing an exit 130 is not the same as dying of SIGINT”?bash script: capturing tcp traffic on a remote server sometimes works, sometimes fails. No errorsWhat does typing a single exclamation mark do in Bash?Execute command and store everything to variable in bashWhy does bash 'read' exit with status 1?What does it mean when Duplicity exits with exit status 23?










1















I'm trying to make a script to either quit compton if it's running or start it if it's not running. I've read from man that it should exit 1 if process is found, so I've tried to make a script that uses that... However this just doesn't work, It starts if it's closed but doesn't close it. what am I doing wrong ??



#!/bin/bash


status=$(pgrep compton 2>&1)

if [[ $status == 1 ]];
then
killall compton
else
exec compton -b
fi

echo $status









share|improve this question









New contributor




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  • 0 or 1 is exit status, but $(...) captures output.

    – Charles Duffy
    1 min ago















1















I'm trying to make a script to either quit compton if it's running or start it if it's not running. I've read from man that it should exit 1 if process is found, so I've tried to make a script that uses that... However this just doesn't work, It starts if it's closed but doesn't close it. what am I doing wrong ??



#!/bin/bash


status=$(pgrep compton 2>&1)

if [[ $status == 1 ]];
then
killall compton
else
exec compton -b
fi

echo $status









share|improve this question









New contributor




Tube is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • 0 or 1 is exit status, but $(...) captures output.

    – Charles Duffy
    1 min ago













1












1








1








I'm trying to make a script to either quit compton if it's running or start it if it's not running. I've read from man that it should exit 1 if process is found, so I've tried to make a script that uses that... However this just doesn't work, It starts if it's closed but doesn't close it. what am I doing wrong ??



#!/bin/bash


status=$(pgrep compton 2>&1)

if [[ $status == 1 ]];
then
killall compton
else
exec compton -b
fi

echo $status









share|improve this question









New contributor




Tube is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












I'm trying to make a script to either quit compton if it's running or start it if it's not running. I've read from man that it should exit 1 if process is found, so I've tried to make a script that uses that... However this just doesn't work, It starts if it's closed but doesn't close it. what am I doing wrong ??



#!/bin/bash


status=$(pgrep compton 2>&1)

if [[ $status == 1 ]];
then
killall compton
else
exec compton -b
fi

echo $status






bash stdout stderr exit-status pgrep






share|improve this question









New contributor




Tube is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




Tube is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|improve this question




share|improve this question








edited 23 mins ago









Kusalananda

136k17257426




136k17257426






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asked 2 hours ago









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  • 0 or 1 is exit status, but $(...) captures output.

    – Charles Duffy
    1 min ago

















  • 0 or 1 is exit status, but $(...) captures output.

    – Charles Duffy
    1 min ago
















0 or 1 is exit status, but $(...) captures output.

– Charles Duffy
1 min ago





0 or 1 is exit status, but $(...) captures output.

– Charles Duffy
1 min ago










2 Answers
2






active

oldest

votes


















2














You should control exit status of pgrep process which will be in $? variable. Or check if $status variable where you're storing the output of pgrep is f.e. non zero-length string. The script in the question checks whether string in variable status is "1"



so



#!/bin/bash
pgrep compton >/dev/null

if [[ $? -eq 0 ]]
then
killall compton
else
exec compton -b
fi


or



#!/bin/bash
status=$(pgrep compton 2>&1)

if [[ -n "$status" ]]
then
killall compton
else
exec compton -b
fi





share|improve this answer























  • if pgrep compton >/dev/null; then is the better practice, so one doesn't directly inspect $? at all. Among other things, this means that set -e will no longer treat pgrep returning 1 as cause to exit the script (because branching on its exit status marks it as "checked").

    – Charles Duffy
    45 secs ago



















2














You are getting the pgrep output in your status variable. It's just not the output that you expect it to be.



pgrep outputs the process IDs (PIDs) of the processes matching the pattern that you give it. If there is a process whose name matches compton, then $status would be the PID of that process, or of those processes. pgrep also returns an exit status, but an exit status is not captured by a command substitution as a string.



In your test, you compare $status against 1. It is unlikely that compton has PID 1.




If you want to kill any compton process if they exist, and start compton -b if no compton process exists, you may do that with



#!/bin/sh

if ! pkill compton; then
exec compton -b
fi


This uses the exit status of pkill. The pkill tool works in an equivalent way to pgrep (they are usually distributed and installed as a pair) but instead of outputting PIDs of matching processes like pgrep would do, pkill sends the TERM signal (by default) to the matching processes.



The if keyword uses the exit status of the command that you use with it.



The ! inverts the sense of the test so that



  • If pkill compton succeeds, it means that there was one or several compton processes that have now been killed, or at least signalled, and exec compton -b will not be executed.


  • If pkill compton fails (no process matched the name, or there was some internal error in pkill), the body of the if statement would call your exec compton -b, which would replace the shell process with that of compton -b.






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    You should control exit status of pgrep process which will be in $? variable. Or check if $status variable where you're storing the output of pgrep is f.e. non zero-length string. The script in the question checks whether string in variable status is "1"



    so



    #!/bin/bash
    pgrep compton >/dev/null

    if [[ $? -eq 0 ]]
    then
    killall compton
    else
    exec compton -b
    fi


    or



    #!/bin/bash
    status=$(pgrep compton 2>&1)

    if [[ -n "$status" ]]
    then
    killall compton
    else
    exec compton -b
    fi





    share|improve this answer























    • if pgrep compton >/dev/null; then is the better practice, so one doesn't directly inspect $? at all. Among other things, this means that set -e will no longer treat pgrep returning 1 as cause to exit the script (because branching on its exit status marks it as "checked").

      – Charles Duffy
      45 secs ago
















    2














    You should control exit status of pgrep process which will be in $? variable. Or check if $status variable where you're storing the output of pgrep is f.e. non zero-length string. The script in the question checks whether string in variable status is "1"



    so



    #!/bin/bash
    pgrep compton >/dev/null

    if [[ $? -eq 0 ]]
    then
    killall compton
    else
    exec compton -b
    fi


    or



    #!/bin/bash
    status=$(pgrep compton 2>&1)

    if [[ -n "$status" ]]
    then
    killall compton
    else
    exec compton -b
    fi





    share|improve this answer























    • if pgrep compton >/dev/null; then is the better practice, so one doesn't directly inspect $? at all. Among other things, this means that set -e will no longer treat pgrep returning 1 as cause to exit the script (because branching on its exit status marks it as "checked").

      – Charles Duffy
      45 secs ago














    2












    2








    2







    You should control exit status of pgrep process which will be in $? variable. Or check if $status variable where you're storing the output of pgrep is f.e. non zero-length string. The script in the question checks whether string in variable status is "1"



    so



    #!/bin/bash
    pgrep compton >/dev/null

    if [[ $? -eq 0 ]]
    then
    killall compton
    else
    exec compton -b
    fi


    or



    #!/bin/bash
    status=$(pgrep compton 2>&1)

    if [[ -n "$status" ]]
    then
    killall compton
    else
    exec compton -b
    fi





    share|improve this answer













    You should control exit status of pgrep process which will be in $? variable. Or check if $status variable where you're storing the output of pgrep is f.e. non zero-length string. The script in the question checks whether string in variable status is "1"



    so



    #!/bin/bash
    pgrep compton >/dev/null

    if [[ $? -eq 0 ]]
    then
    killall compton
    else
    exec compton -b
    fi


    or



    #!/bin/bash
    status=$(pgrep compton 2>&1)

    if [[ -n "$status" ]]
    then
    killall compton
    else
    exec compton -b
    fi






    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 2 hours ago









    Jakub JindraJakub Jindra

    307310




    307310












    • if pgrep compton >/dev/null; then is the better practice, so one doesn't directly inspect $? at all. Among other things, this means that set -e will no longer treat pgrep returning 1 as cause to exit the script (because branching on its exit status marks it as "checked").

      – Charles Duffy
      45 secs ago


















    • if pgrep compton >/dev/null; then is the better practice, so one doesn't directly inspect $? at all. Among other things, this means that set -e will no longer treat pgrep returning 1 as cause to exit the script (because branching on its exit status marks it as "checked").

      – Charles Duffy
      45 secs ago

















    if pgrep compton >/dev/null; then is the better practice, so one doesn't directly inspect $? at all. Among other things, this means that set -e will no longer treat pgrep returning 1 as cause to exit the script (because branching on its exit status marks it as "checked").

    – Charles Duffy
    45 secs ago






    if pgrep compton >/dev/null; then is the better practice, so one doesn't directly inspect $? at all. Among other things, this means that set -e will no longer treat pgrep returning 1 as cause to exit the script (because branching on its exit status marks it as "checked").

    – Charles Duffy
    45 secs ago














    2














    You are getting the pgrep output in your status variable. It's just not the output that you expect it to be.



    pgrep outputs the process IDs (PIDs) of the processes matching the pattern that you give it. If there is a process whose name matches compton, then $status would be the PID of that process, or of those processes. pgrep also returns an exit status, but an exit status is not captured by a command substitution as a string.



    In your test, you compare $status against 1. It is unlikely that compton has PID 1.




    If you want to kill any compton process if they exist, and start compton -b if no compton process exists, you may do that with



    #!/bin/sh

    if ! pkill compton; then
    exec compton -b
    fi


    This uses the exit status of pkill. The pkill tool works in an equivalent way to pgrep (they are usually distributed and installed as a pair) but instead of outputting PIDs of matching processes like pgrep would do, pkill sends the TERM signal (by default) to the matching processes.



    The if keyword uses the exit status of the command that you use with it.



    The ! inverts the sense of the test so that



    • If pkill compton succeeds, it means that there was one or several compton processes that have now been killed, or at least signalled, and exec compton -b will not be executed.


    • If pkill compton fails (no process matched the name, or there was some internal error in pkill), the body of the if statement would call your exec compton -b, which would replace the shell process with that of compton -b.






    share|improve this answer





























      2














      You are getting the pgrep output in your status variable. It's just not the output that you expect it to be.



      pgrep outputs the process IDs (PIDs) of the processes matching the pattern that you give it. If there is a process whose name matches compton, then $status would be the PID of that process, or of those processes. pgrep also returns an exit status, but an exit status is not captured by a command substitution as a string.



      In your test, you compare $status against 1. It is unlikely that compton has PID 1.




      If you want to kill any compton process if they exist, and start compton -b if no compton process exists, you may do that with



      #!/bin/sh

      if ! pkill compton; then
      exec compton -b
      fi


      This uses the exit status of pkill. The pkill tool works in an equivalent way to pgrep (they are usually distributed and installed as a pair) but instead of outputting PIDs of matching processes like pgrep would do, pkill sends the TERM signal (by default) to the matching processes.



      The if keyword uses the exit status of the command that you use with it.



      The ! inverts the sense of the test so that



      • If pkill compton succeeds, it means that there was one or several compton processes that have now been killed, or at least signalled, and exec compton -b will not be executed.


      • If pkill compton fails (no process matched the name, or there was some internal error in pkill), the body of the if statement would call your exec compton -b, which would replace the shell process with that of compton -b.






      share|improve this answer



























        2












        2








        2







        You are getting the pgrep output in your status variable. It's just not the output that you expect it to be.



        pgrep outputs the process IDs (PIDs) of the processes matching the pattern that you give it. If there is a process whose name matches compton, then $status would be the PID of that process, or of those processes. pgrep also returns an exit status, but an exit status is not captured by a command substitution as a string.



        In your test, you compare $status against 1. It is unlikely that compton has PID 1.




        If you want to kill any compton process if they exist, and start compton -b if no compton process exists, you may do that with



        #!/bin/sh

        if ! pkill compton; then
        exec compton -b
        fi


        This uses the exit status of pkill. The pkill tool works in an equivalent way to pgrep (they are usually distributed and installed as a pair) but instead of outputting PIDs of matching processes like pgrep would do, pkill sends the TERM signal (by default) to the matching processes.



        The if keyword uses the exit status of the command that you use with it.



        The ! inverts the sense of the test so that



        • If pkill compton succeeds, it means that there was one or several compton processes that have now been killed, or at least signalled, and exec compton -b will not be executed.


        • If pkill compton fails (no process matched the name, or there was some internal error in pkill), the body of the if statement would call your exec compton -b, which would replace the shell process with that of compton -b.






        share|improve this answer















        You are getting the pgrep output in your status variable. It's just not the output that you expect it to be.



        pgrep outputs the process IDs (PIDs) of the processes matching the pattern that you give it. If there is a process whose name matches compton, then $status would be the PID of that process, or of those processes. pgrep also returns an exit status, but an exit status is not captured by a command substitution as a string.



        In your test, you compare $status against 1. It is unlikely that compton has PID 1.




        If you want to kill any compton process if they exist, and start compton -b if no compton process exists, you may do that with



        #!/bin/sh

        if ! pkill compton; then
        exec compton -b
        fi


        This uses the exit status of pkill. The pkill tool works in an equivalent way to pgrep (they are usually distributed and installed as a pair) but instead of outputting PIDs of matching processes like pgrep would do, pkill sends the TERM signal (by default) to the matching processes.



        The if keyword uses the exit status of the command that you use with it.



        The ! inverts the sense of the test so that



        • If pkill compton succeeds, it means that there was one or several compton processes that have now been killed, or at least signalled, and exec compton -b will not be executed.


        • If pkill compton fails (no process matched the name, or there was some internal error in pkill), the body of the if statement would call your exec compton -b, which would replace the shell process with that of compton -b.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 31 mins ago

























        answered 1 hour ago









        KusalanandaKusalananda

        136k17257426




        136k17257426




















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