How to treat a command output as text












5














I need to compare a command output with a string.
This is the scenario:



pvs_var=$(pvs | grep "sdb1") 


so pvs var is: /dev/sdb1 vg_name lvm2 a-- 100.00g 0



if [[ $($pvs_var | awk '{ print $2 }') = vg_name ]]; then
do something
fi


The issue is that the output of the if statement is



-bash: /dev/sdb1: Permission denied


I don't understand this behavior.
Thank you










share|improve this question



























    5














    I need to compare a command output with a string.
    This is the scenario:



    pvs_var=$(pvs | grep "sdb1") 


    so pvs var is: /dev/sdb1 vg_name lvm2 a-- 100.00g 0



    if [[ $($pvs_var | awk '{ print $2 }') = vg_name ]]; then
    do something
    fi


    The issue is that the output of the if statement is



    -bash: /dev/sdb1: Permission denied


    I don't understand this behavior.
    Thank you










    share|improve this question

























      5












      5








      5







      I need to compare a command output with a string.
      This is the scenario:



      pvs_var=$(pvs | grep "sdb1") 


      so pvs var is: /dev/sdb1 vg_name lvm2 a-- 100.00g 0



      if [[ $($pvs_var | awk '{ print $2 }') = vg_name ]]; then
      do something
      fi


      The issue is that the output of the if statement is



      -bash: /dev/sdb1: Permission denied


      I don't understand this behavior.
      Thank you










      share|improve this question













      I need to compare a command output with a string.
      This is the scenario:



      pvs_var=$(pvs | grep "sdb1") 


      so pvs var is: /dev/sdb1 vg_name lvm2 a-- 100.00g 0



      if [[ $($pvs_var | awk '{ print $2 }') = vg_name ]]; then
      do something
      fi


      The issue is that the output of the if statement is



      -bash: /dev/sdb1: Permission denied


      I don't understand this behavior.
      Thank you







      bash command string output






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 30 at 15:42









      intore

      10619




      10619






















          4 Answers
          4






          active

          oldest

          votes


















          12














          You are attempting to execute the contents of $pvs_var as a command, rather than passing the string to awk.



          To fix this, add an echo or printf in your if statement:



          if [[ $(echo "$pvs_var" | awk '{ print $2 }') = vg_name ]]; then
          do something
          fi





          share|improve this answer























          • Since the shell is bash, might as well awk '{ print $2 }' <<<"$pvs_var"
            – D. Ben Knoble
            Nov 30 at 23:13










          • @D.BenKnoble: Even better, we can replace the double brackets with single brackets and then it’s POSIX compliant.
            – Peschke
            Nov 30 at 23:22










          • @Peschke With single brackets, you'd want to add double-quotes around the $( ) expression.
            – Gordon Davisson
            Dec 1 at 3:49










          • You definitely want printf in the general case. Echo can do strange things if the variable contains backslashes or starts with a dash.
            – Kevin
            Dec 1 at 17:08



















          4














          Get the output in JSON format, and then you'll be able to extract information in a more reliable way:



          pv_info=$(pvs -o pv_all,vg_all --unit b --nosuffix --reportformat json)
          sdb1_vg=$(
          printf '%sn' "$pv_info" |
          jq -r '.report.pv|select(.pv_name == "/dev/sdb1").vg_name'
          )

          if [ "$sdb1_vg" = vg_name ]; then...


          Or use a proper programming language with a JSON library instead of a shell (ksh93 does have JSON support though in its upcoming version).



          If it's just that one query you want to do, pvs can also do all the work for you:



          sdb1_vg=$(
          pvs -o vg_name -S pv_name=/dev/sdb1 --no-heading --config 'log{prefix=""}'
          )


          Also remember to quote your variables and avoid echo.






          share|improve this answer





























            2














            If you want to compare the VG from the output, then it might be easier to pre-process that:



            # if you still need it
            pvs_var=$(pvs | grep "sdb1")
            vg_name=$(pvs | grep "sdb1" | awk '{print $2}')
            if [ "$vg_name" = "vg_name" ]; then
            echo do something
            fi


            What you were doing with




            $($pvs_var | awk '{ print $2 }')




            was initiating a command substitution $( ... ) whose first command was $pvs_var. Bash dutifully substituted the value of the variable and then attempted to execute it. That's not what you wanted.



            Another alternative would be to send the variable as a here-string to the awk command:



            # ...
            if [ $(awk '{print $2}' <<< "$pvs_var") = "vg_name" ]; then
            # ...


            Here, the command substitution is calling awk and passing it input on stdin -- the contents of the $pvs_var variable.






            share|improve this answer





























              0














              You can try only with awk :



              pvs | awk -v search='vg_name' '/sdb1/&&$2==search{exit 1}' || echo "ok"





              share|improve this answer





















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                4 Answers
                4






                active

                oldest

                votes








                4 Answers
                4






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                12














                You are attempting to execute the contents of $pvs_var as a command, rather than passing the string to awk.



                To fix this, add an echo or printf in your if statement:



                if [[ $(echo "$pvs_var" | awk '{ print $2 }') = vg_name ]]; then
                do something
                fi





                share|improve this answer























                • Since the shell is bash, might as well awk '{ print $2 }' <<<"$pvs_var"
                  – D. Ben Knoble
                  Nov 30 at 23:13










                • @D.BenKnoble: Even better, we can replace the double brackets with single brackets and then it’s POSIX compliant.
                  – Peschke
                  Nov 30 at 23:22










                • @Peschke With single brackets, you'd want to add double-quotes around the $( ) expression.
                  – Gordon Davisson
                  Dec 1 at 3:49










                • You definitely want printf in the general case. Echo can do strange things if the variable contains backslashes or starts with a dash.
                  – Kevin
                  Dec 1 at 17:08
















                12














                You are attempting to execute the contents of $pvs_var as a command, rather than passing the string to awk.



                To fix this, add an echo or printf in your if statement:



                if [[ $(echo "$pvs_var" | awk '{ print $2 }') = vg_name ]]; then
                do something
                fi





                share|improve this answer























                • Since the shell is bash, might as well awk '{ print $2 }' <<<"$pvs_var"
                  – D. Ben Knoble
                  Nov 30 at 23:13










                • @D.BenKnoble: Even better, we can replace the double brackets with single brackets and then it’s POSIX compliant.
                  – Peschke
                  Nov 30 at 23:22










                • @Peschke With single brackets, you'd want to add double-quotes around the $( ) expression.
                  – Gordon Davisson
                  Dec 1 at 3:49










                • You definitely want printf in the general case. Echo can do strange things if the variable contains backslashes or starts with a dash.
                  – Kevin
                  Dec 1 at 17:08














                12












                12








                12






                You are attempting to execute the contents of $pvs_var as a command, rather than passing the string to awk.



                To fix this, add an echo or printf in your if statement:



                if [[ $(echo "$pvs_var" | awk '{ print $2 }') = vg_name ]]; then
                do something
                fi





                share|improve this answer














                You are attempting to execute the contents of $pvs_var as a command, rather than passing the string to awk.



                To fix this, add an echo or printf in your if statement:



                if [[ $(echo "$pvs_var" | awk '{ print $2 }') = vg_name ]]; then
                do something
                fi






                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited Dec 1 at 0:50

























                answered Nov 30 at 15:58









                Peschke

                2,455924




                2,455924












                • Since the shell is bash, might as well awk '{ print $2 }' <<<"$pvs_var"
                  – D. Ben Knoble
                  Nov 30 at 23:13










                • @D.BenKnoble: Even better, we can replace the double brackets with single brackets and then it’s POSIX compliant.
                  – Peschke
                  Nov 30 at 23:22










                • @Peschke With single brackets, you'd want to add double-quotes around the $( ) expression.
                  – Gordon Davisson
                  Dec 1 at 3:49










                • You definitely want printf in the general case. Echo can do strange things if the variable contains backslashes or starts with a dash.
                  – Kevin
                  Dec 1 at 17:08


















                • Since the shell is bash, might as well awk '{ print $2 }' <<<"$pvs_var"
                  – D. Ben Knoble
                  Nov 30 at 23:13










                • @D.BenKnoble: Even better, we can replace the double brackets with single brackets and then it’s POSIX compliant.
                  – Peschke
                  Nov 30 at 23:22










                • @Peschke With single brackets, you'd want to add double-quotes around the $( ) expression.
                  – Gordon Davisson
                  Dec 1 at 3:49










                • You definitely want printf in the general case. Echo can do strange things if the variable contains backslashes or starts with a dash.
                  – Kevin
                  Dec 1 at 17:08
















                Since the shell is bash, might as well awk '{ print $2 }' <<<"$pvs_var"
                – D. Ben Knoble
                Nov 30 at 23:13




                Since the shell is bash, might as well awk '{ print $2 }' <<<"$pvs_var"
                – D. Ben Knoble
                Nov 30 at 23:13












                @D.BenKnoble: Even better, we can replace the double brackets with single brackets and then it’s POSIX compliant.
                – Peschke
                Nov 30 at 23:22




                @D.BenKnoble: Even better, we can replace the double brackets with single brackets and then it’s POSIX compliant.
                – Peschke
                Nov 30 at 23:22












                @Peschke With single brackets, you'd want to add double-quotes around the $( ) expression.
                – Gordon Davisson
                Dec 1 at 3:49




                @Peschke With single brackets, you'd want to add double-quotes around the $( ) expression.
                – Gordon Davisson
                Dec 1 at 3:49












                You definitely want printf in the general case. Echo can do strange things if the variable contains backslashes or starts with a dash.
                – Kevin
                Dec 1 at 17:08




                You definitely want printf in the general case. Echo can do strange things if the variable contains backslashes or starts with a dash.
                – Kevin
                Dec 1 at 17:08













                4














                Get the output in JSON format, and then you'll be able to extract information in a more reliable way:



                pv_info=$(pvs -o pv_all,vg_all --unit b --nosuffix --reportformat json)
                sdb1_vg=$(
                printf '%sn' "$pv_info" |
                jq -r '.report.pv|select(.pv_name == "/dev/sdb1").vg_name'
                )

                if [ "$sdb1_vg" = vg_name ]; then...


                Or use a proper programming language with a JSON library instead of a shell (ksh93 does have JSON support though in its upcoming version).



                If it's just that one query you want to do, pvs can also do all the work for you:



                sdb1_vg=$(
                pvs -o vg_name -S pv_name=/dev/sdb1 --no-heading --config 'log{prefix=""}'
                )


                Also remember to quote your variables and avoid echo.






                share|improve this answer


























                  4














                  Get the output in JSON format, and then you'll be able to extract information in a more reliable way:



                  pv_info=$(pvs -o pv_all,vg_all --unit b --nosuffix --reportformat json)
                  sdb1_vg=$(
                  printf '%sn' "$pv_info" |
                  jq -r '.report.pv|select(.pv_name == "/dev/sdb1").vg_name'
                  )

                  if [ "$sdb1_vg" = vg_name ]; then...


                  Or use a proper programming language with a JSON library instead of a shell (ksh93 does have JSON support though in its upcoming version).



                  If it's just that one query you want to do, pvs can also do all the work for you:



                  sdb1_vg=$(
                  pvs -o vg_name -S pv_name=/dev/sdb1 --no-heading --config 'log{prefix=""}'
                  )


                  Also remember to quote your variables and avoid echo.






                  share|improve this answer
























                    4












                    4








                    4






                    Get the output in JSON format, and then you'll be able to extract information in a more reliable way:



                    pv_info=$(pvs -o pv_all,vg_all --unit b --nosuffix --reportformat json)
                    sdb1_vg=$(
                    printf '%sn' "$pv_info" |
                    jq -r '.report.pv|select(.pv_name == "/dev/sdb1").vg_name'
                    )

                    if [ "$sdb1_vg" = vg_name ]; then...


                    Or use a proper programming language with a JSON library instead of a shell (ksh93 does have JSON support though in its upcoming version).



                    If it's just that one query you want to do, pvs can also do all the work for you:



                    sdb1_vg=$(
                    pvs -o vg_name -S pv_name=/dev/sdb1 --no-heading --config 'log{prefix=""}'
                    )


                    Also remember to quote your variables and avoid echo.






                    share|improve this answer












                    Get the output in JSON format, and then you'll be able to extract information in a more reliable way:



                    pv_info=$(pvs -o pv_all,vg_all --unit b --nosuffix --reportformat json)
                    sdb1_vg=$(
                    printf '%sn' "$pv_info" |
                    jq -r '.report.pv|select(.pv_name == "/dev/sdb1").vg_name'
                    )

                    if [ "$sdb1_vg" = vg_name ]; then...


                    Or use a proper programming language with a JSON library instead of a shell (ksh93 does have JSON support though in its upcoming version).



                    If it's just that one query you want to do, pvs can also do all the work for you:



                    sdb1_vg=$(
                    pvs -o vg_name -S pv_name=/dev/sdb1 --no-heading --config 'log{prefix=""}'
                    )


                    Also remember to quote your variables and avoid echo.







                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered Nov 30 at 17:40









                    Stéphane Chazelas

                    298k54563910




                    298k54563910























                        2














                        If you want to compare the VG from the output, then it might be easier to pre-process that:



                        # if you still need it
                        pvs_var=$(pvs | grep "sdb1")
                        vg_name=$(pvs | grep "sdb1" | awk '{print $2}')
                        if [ "$vg_name" = "vg_name" ]; then
                        echo do something
                        fi


                        What you were doing with




                        $($pvs_var | awk '{ print $2 }')




                        was initiating a command substitution $( ... ) whose first command was $pvs_var. Bash dutifully substituted the value of the variable and then attempted to execute it. That's not what you wanted.



                        Another alternative would be to send the variable as a here-string to the awk command:



                        # ...
                        if [ $(awk '{print $2}' <<< "$pvs_var") = "vg_name" ]; then
                        # ...


                        Here, the command substitution is calling awk and passing it input on stdin -- the contents of the $pvs_var variable.






                        share|improve this answer


























                          2














                          If you want to compare the VG from the output, then it might be easier to pre-process that:



                          # if you still need it
                          pvs_var=$(pvs | grep "sdb1")
                          vg_name=$(pvs | grep "sdb1" | awk '{print $2}')
                          if [ "$vg_name" = "vg_name" ]; then
                          echo do something
                          fi


                          What you were doing with




                          $($pvs_var | awk '{ print $2 }')




                          was initiating a command substitution $( ... ) whose first command was $pvs_var. Bash dutifully substituted the value of the variable and then attempted to execute it. That's not what you wanted.



                          Another alternative would be to send the variable as a here-string to the awk command:



                          # ...
                          if [ $(awk '{print $2}' <<< "$pvs_var") = "vg_name" ]; then
                          # ...


                          Here, the command substitution is calling awk and passing it input on stdin -- the contents of the $pvs_var variable.






                          share|improve this answer
























                            2












                            2








                            2






                            If you want to compare the VG from the output, then it might be easier to pre-process that:



                            # if you still need it
                            pvs_var=$(pvs | grep "sdb1")
                            vg_name=$(pvs | grep "sdb1" | awk '{print $2}')
                            if [ "$vg_name" = "vg_name" ]; then
                            echo do something
                            fi


                            What you were doing with




                            $($pvs_var | awk '{ print $2 }')




                            was initiating a command substitution $( ... ) whose first command was $pvs_var. Bash dutifully substituted the value of the variable and then attempted to execute it. That's not what you wanted.



                            Another alternative would be to send the variable as a here-string to the awk command:



                            # ...
                            if [ $(awk '{print $2}' <<< "$pvs_var") = "vg_name" ]; then
                            # ...


                            Here, the command substitution is calling awk and passing it input on stdin -- the contents of the $pvs_var variable.






                            share|improve this answer












                            If you want to compare the VG from the output, then it might be easier to pre-process that:



                            # if you still need it
                            pvs_var=$(pvs | grep "sdb1")
                            vg_name=$(pvs | grep "sdb1" | awk '{print $2}')
                            if [ "$vg_name" = "vg_name" ]; then
                            echo do something
                            fi


                            What you were doing with




                            $($pvs_var | awk '{ print $2 }')




                            was initiating a command substitution $( ... ) whose first command was $pvs_var. Bash dutifully substituted the value of the variable and then attempted to execute it. That's not what you wanted.



                            Another alternative would be to send the variable as a here-string to the awk command:



                            # ...
                            if [ $(awk '{print $2}' <<< "$pvs_var") = "vg_name" ]; then
                            # ...


                            Here, the command substitution is calling awk and passing it input on stdin -- the contents of the $pvs_var variable.







                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered Nov 30 at 15:55









                            Jeff Schaller

                            38.4k1053125




                            38.4k1053125























                                0














                                You can try only with awk :



                                pvs | awk -v search='vg_name' '/sdb1/&&$2==search{exit 1}' || echo "ok"





                                share|improve this answer


























                                  0














                                  You can try only with awk :



                                  pvs | awk -v search='vg_name' '/sdb1/&&$2==search{exit 1}' || echo "ok"





                                  share|improve this answer
























                                    0












                                    0








                                    0






                                    You can try only with awk :



                                    pvs | awk -v search='vg_name' '/sdb1/&&$2==search{exit 1}' || echo "ok"





                                    share|improve this answer












                                    You can try only with awk :



                                    pvs | awk -v search='vg_name' '/sdb1/&&$2==search{exit 1}' || echo "ok"






                                    share|improve this answer












                                    share|improve this answer



                                    share|improve this answer










                                    answered Nov 30 at 16:23









                                    ctac_

                                    1,344117




                                    1,344117






























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