How do I remove the last characters from a string?

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I have a variable set with var='type_cardio_10-11-2017'. I need to remove the last 10 letters from the variable, and append the remaining value to var2.



I tried with the following script, but it doesn't work as expected.



var=type_cardio_10-11-2017
var2=$var | cut -f1 -d "_"
echo $var2


The output I want is type_cardio.







share|improve this question


























    up vote
    6
    down vote

    favorite
    4












    I have a variable set with var='type_cardio_10-11-2017'. I need to remove the last 10 letters from the variable, and append the remaining value to var2.



    I tried with the following script, but it doesn't work as expected.



    var=type_cardio_10-11-2017
    var2=$var | cut -f1 -d "_"
    echo $var2


    The output I want is type_cardio.







    share|improve this question
























      up vote
      6
      down vote

      favorite
      4









      up vote
      6
      down vote

      favorite
      4






      4





      I have a variable set with var='type_cardio_10-11-2017'. I need to remove the last 10 letters from the variable, and append the remaining value to var2.



      I tried with the following script, but it doesn't work as expected.



      var=type_cardio_10-11-2017
      var2=$var | cut -f1 -d "_"
      echo $var2


      The output I want is type_cardio.







      share|improve this question














      I have a variable set with var='type_cardio_10-11-2017'. I need to remove the last 10 letters from the variable, and append the remaining value to var2.



      I tried with the following script, but it doesn't work as expected.



      var=type_cardio_10-11-2017
      var2=$var | cut -f1 -d "_"
      echo $var2


      The output I want is type_cardio.









      share|improve this question













      share|improve this question




      share|improve this question








      edited Oct 20 '17 at 17:57









      Jeff Schaller

      32.1k849109




      32.1k849109










      asked Oct 20 '17 at 17:38









      Rak kundra

      141110




      141110




















          5 Answers
          5






          active

          oldest

          votes

















          up vote
          14
          down vote



          accepted










          To remove everything from after the last _ in var and assign the result to var2:



          var2=$var%_*


          The parameter expansion $parameter%word removes the pattern word (_* in this case) from the end of the value of the given variable.



          The POSIX standard calls this a "Remove Smallest Suffix Pattern" parameter expansion.






          share|improve this answer





























            up vote
            9
            down vote













            You actually want to remove the trailing 11 characters from the string; here's another way to do it:



            $ var=type_cardio_10-11-2017
            $ var2=$var%???????????
            $ echo "$var2"
            type_cardio





            share|improve this answer



























              up vote
              5
              down vote













              Another approach in bash:



              echo "$var::-10"


              Or in older versions:



              echo "$var::$#var-10" #or
              echo "$var: : -10"





              share|improve this answer





























                up vote
                4
                down vote













                1) bash solution:



                var1="type_cardio_10-11-2017"
                var2=$var1%_*


                2) cut solution:



                var1="type_cardio_10-11-2017"
                var2=$(cut -d'_' -f1,2 <<<"$var1")
                echo "$var2"


                The output:



                type_cardio





                share|improve this answer






















                • Thank you sir, But the numeric values will be changing because it is a date string .That is the reason i was using cut .This answer is prefect but can i know how cut can be applied ?
                  – Rak kundra
                  Oct 20 '17 at 17:44






                • 1




                  @SanthoshPogaku, that was not clear from your initial question, you have my update
                  – RomanPerekhrest
                  Oct 20 '17 at 17:52










                • Yeah sir, I got what am i looking for . Thank's for your valuable time.
                  – Rak kundra
                  Oct 20 '17 at 17:53

















                up vote
                1
                down vote













                Appending var and var2 is the easy part - you can just join them like new_var="$var$var2". If what you really meant to say is you want to store cropped var into var2,then it's just var2=$(...) and you can put the commands that other users and my answer presented here inside the $() portion.



                The main part is removing those 10 characters. There's number of ways to extract the part you want from var:




                • printf (pretty portable, doesn't rely on specific shell)



                  $ printf "%.11sn" "$var" 
                  type_cardio



                • awk



                  $ var=type_cardio_10-11-2017 
                  $ awk -v awk_var="$var" 'BEGINprint substr(awk_var,0,length(awk_var)-11)'
                  type_cardio


                  or you can use printf trick here as well:



                  $ echo "$var" | awk 'printf "%.11sn",$0' 
                  type_cardio



                • egrep



                  $ echo "$var" | egrep -o '^.0,11' 
                  type_cardio



                • perl:



                  $ echo "$var" | perl -lne 'print substr($_,0,11)' 
                  type_cardio



                • python:



                  $ python -c 'import sys;print sys.argv[1][0:11] ' "$var" 
                  type_cardio






                share|improve this answer






















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                  5 Answers
                  5






                  active

                  oldest

                  votes








                  5 Answers
                  5






                  active

                  oldest

                  votes









                  active

                  oldest

                  votes






                  active

                  oldest

                  votes








                  up vote
                  14
                  down vote



                  accepted










                  To remove everything from after the last _ in var and assign the result to var2:



                  var2=$var%_*


                  The parameter expansion $parameter%word removes the pattern word (_* in this case) from the end of the value of the given variable.



                  The POSIX standard calls this a "Remove Smallest Suffix Pattern" parameter expansion.






                  share|improve this answer


























                    up vote
                    14
                    down vote



                    accepted










                    To remove everything from after the last _ in var and assign the result to var2:



                    var2=$var%_*


                    The parameter expansion $parameter%word removes the pattern word (_* in this case) from the end of the value of the given variable.



                    The POSIX standard calls this a "Remove Smallest Suffix Pattern" parameter expansion.






                    share|improve this answer
























                      up vote
                      14
                      down vote



                      accepted







                      up vote
                      14
                      down vote



                      accepted






                      To remove everything from after the last _ in var and assign the result to var2:



                      var2=$var%_*


                      The parameter expansion $parameter%word removes the pattern word (_* in this case) from the end of the value of the given variable.



                      The POSIX standard calls this a "Remove Smallest Suffix Pattern" parameter expansion.






                      share|improve this answer














                      To remove everything from after the last _ in var and assign the result to var2:



                      var2=$var%_*


                      The parameter expansion $parameter%word removes the pattern word (_* in this case) from the end of the value of the given variable.



                      The POSIX standard calls this a "Remove Smallest Suffix Pattern" parameter expansion.







                      share|improve this answer














                      share|improve this answer



                      share|improve this answer








                      edited Oct 20 '17 at 17:55

























                      answered Oct 20 '17 at 17:49









                      Kusalananda

                      105k14209326




                      105k14209326






















                          up vote
                          9
                          down vote













                          You actually want to remove the trailing 11 characters from the string; here's another way to do it:



                          $ var=type_cardio_10-11-2017
                          $ var2=$var%???????????
                          $ echo "$var2"
                          type_cardio





                          share|improve this answer
























                            up vote
                            9
                            down vote













                            You actually want to remove the trailing 11 characters from the string; here's another way to do it:



                            $ var=type_cardio_10-11-2017
                            $ var2=$var%???????????
                            $ echo "$var2"
                            type_cardio





                            share|improve this answer






















                              up vote
                              9
                              down vote










                              up vote
                              9
                              down vote









                              You actually want to remove the trailing 11 characters from the string; here's another way to do it:



                              $ var=type_cardio_10-11-2017
                              $ var2=$var%???????????
                              $ echo "$var2"
                              type_cardio





                              share|improve this answer












                              You actually want to remove the trailing 11 characters from the string; here's another way to do it:



                              $ var=type_cardio_10-11-2017
                              $ var2=$var%???????????
                              $ echo "$var2"
                              type_cardio






                              share|improve this answer












                              share|improve this answer



                              share|improve this answer










                              answered Oct 20 '17 at 17:56









                              Jeff Schaller

                              32.1k849109




                              32.1k849109




















                                  up vote
                                  5
                                  down vote













                                  Another approach in bash:



                                  echo "$var::-10"


                                  Or in older versions:



                                  echo "$var::$#var-10" #or
                                  echo "$var: : -10"





                                  share|improve this answer


























                                    up vote
                                    5
                                    down vote













                                    Another approach in bash:



                                    echo "$var::-10"


                                    Or in older versions:



                                    echo "$var::$#var-10" #or
                                    echo "$var: : -10"





                                    share|improve this answer
























                                      up vote
                                      5
                                      down vote










                                      up vote
                                      5
                                      down vote









                                      Another approach in bash:



                                      echo "$var::-10"


                                      Or in older versions:



                                      echo "$var::$#var-10" #or
                                      echo "$var: : -10"





                                      share|improve this answer














                                      Another approach in bash:



                                      echo "$var::-10"


                                      Or in older versions:



                                      echo "$var::$#var-10" #or
                                      echo "$var: : -10"






                                      share|improve this answer














                                      share|improve this answer



                                      share|improve this answer








                                      edited Oct 20 '17 at 18:07

























                                      answered Oct 20 '17 at 18:00









                                      αғsнιη

                                      15.6k92563




                                      15.6k92563




















                                          up vote
                                          4
                                          down vote













                                          1) bash solution:



                                          var1="type_cardio_10-11-2017"
                                          var2=$var1%_*


                                          2) cut solution:



                                          var1="type_cardio_10-11-2017"
                                          var2=$(cut -d'_' -f1,2 <<<"$var1")
                                          echo "$var2"


                                          The output:



                                          type_cardio





                                          share|improve this answer






















                                          • Thank you sir, But the numeric values will be changing because it is a date string .That is the reason i was using cut .This answer is prefect but can i know how cut can be applied ?
                                            – Rak kundra
                                            Oct 20 '17 at 17:44






                                          • 1




                                            @SanthoshPogaku, that was not clear from your initial question, you have my update
                                            – RomanPerekhrest
                                            Oct 20 '17 at 17:52










                                          • Yeah sir, I got what am i looking for . Thank's for your valuable time.
                                            – Rak kundra
                                            Oct 20 '17 at 17:53














                                          up vote
                                          4
                                          down vote













                                          1) bash solution:



                                          var1="type_cardio_10-11-2017"
                                          var2=$var1%_*


                                          2) cut solution:



                                          var1="type_cardio_10-11-2017"
                                          var2=$(cut -d'_' -f1,2 <<<"$var1")
                                          echo "$var2"


                                          The output:



                                          type_cardio





                                          share|improve this answer






















                                          • Thank you sir, But the numeric values will be changing because it is a date string .That is the reason i was using cut .This answer is prefect but can i know how cut can be applied ?
                                            – Rak kundra
                                            Oct 20 '17 at 17:44






                                          • 1




                                            @SanthoshPogaku, that was not clear from your initial question, you have my update
                                            – RomanPerekhrest
                                            Oct 20 '17 at 17:52










                                          • Yeah sir, I got what am i looking for . Thank's for your valuable time.
                                            – Rak kundra
                                            Oct 20 '17 at 17:53












                                          up vote
                                          4
                                          down vote










                                          up vote
                                          4
                                          down vote









                                          1) bash solution:



                                          var1="type_cardio_10-11-2017"
                                          var2=$var1%_*


                                          2) cut solution:



                                          var1="type_cardio_10-11-2017"
                                          var2=$(cut -d'_' -f1,2 <<<"$var1")
                                          echo "$var2"


                                          The output:



                                          type_cardio





                                          share|improve this answer














                                          1) bash solution:



                                          var1="type_cardio_10-11-2017"
                                          var2=$var1%_*


                                          2) cut solution:



                                          var1="type_cardio_10-11-2017"
                                          var2=$(cut -d'_' -f1,2 <<<"$var1")
                                          echo "$var2"


                                          The output:



                                          type_cardio






                                          share|improve this answer














                                          share|improve this answer



                                          share|improve this answer








                                          edited Oct 20 '17 at 17:54

























                                          answered Oct 20 '17 at 17:42









                                          RomanPerekhrest

                                          22.5k12145




                                          22.5k12145











                                          • Thank you sir, But the numeric values will be changing because it is a date string .That is the reason i was using cut .This answer is prefect but can i know how cut can be applied ?
                                            – Rak kundra
                                            Oct 20 '17 at 17:44






                                          • 1




                                            @SanthoshPogaku, that was not clear from your initial question, you have my update
                                            – RomanPerekhrest
                                            Oct 20 '17 at 17:52










                                          • Yeah sir, I got what am i looking for . Thank's for your valuable time.
                                            – Rak kundra
                                            Oct 20 '17 at 17:53
















                                          • Thank you sir, But the numeric values will be changing because it is a date string .That is the reason i was using cut .This answer is prefect but can i know how cut can be applied ?
                                            – Rak kundra
                                            Oct 20 '17 at 17:44






                                          • 1




                                            @SanthoshPogaku, that was not clear from your initial question, you have my update
                                            – RomanPerekhrest
                                            Oct 20 '17 at 17:52










                                          • Yeah sir, I got what am i looking for . Thank's for your valuable time.
                                            – Rak kundra
                                            Oct 20 '17 at 17:53















                                          Thank you sir, But the numeric values will be changing because it is a date string .That is the reason i was using cut .This answer is prefect but can i know how cut can be applied ?
                                          – Rak kundra
                                          Oct 20 '17 at 17:44




                                          Thank you sir, But the numeric values will be changing because it is a date string .That is the reason i was using cut .This answer is prefect but can i know how cut can be applied ?
                                          – Rak kundra
                                          Oct 20 '17 at 17:44




                                          1




                                          1




                                          @SanthoshPogaku, that was not clear from your initial question, you have my update
                                          – RomanPerekhrest
                                          Oct 20 '17 at 17:52




                                          @SanthoshPogaku, that was not clear from your initial question, you have my update
                                          – RomanPerekhrest
                                          Oct 20 '17 at 17:52












                                          Yeah sir, I got what am i looking for . Thank's for your valuable time.
                                          – Rak kundra
                                          Oct 20 '17 at 17:53




                                          Yeah sir, I got what am i looking for . Thank's for your valuable time.
                                          – Rak kundra
                                          Oct 20 '17 at 17:53










                                          up vote
                                          1
                                          down vote













                                          Appending var and var2 is the easy part - you can just join them like new_var="$var$var2". If what you really meant to say is you want to store cropped var into var2,then it's just var2=$(...) and you can put the commands that other users and my answer presented here inside the $() portion.



                                          The main part is removing those 10 characters. There's number of ways to extract the part you want from var:




                                          • printf (pretty portable, doesn't rely on specific shell)



                                            $ printf "%.11sn" "$var" 
                                            type_cardio



                                          • awk



                                            $ var=type_cardio_10-11-2017 
                                            $ awk -v awk_var="$var" 'BEGINprint substr(awk_var,0,length(awk_var)-11)'
                                            type_cardio


                                            or you can use printf trick here as well:



                                            $ echo "$var" | awk 'printf "%.11sn",$0' 
                                            type_cardio



                                          • egrep



                                            $ echo "$var" | egrep -o '^.0,11' 
                                            type_cardio



                                          • perl:



                                            $ echo "$var" | perl -lne 'print substr($_,0,11)' 
                                            type_cardio



                                          • python:



                                            $ python -c 'import sys;print sys.argv[1][0:11] ' "$var" 
                                            type_cardio






                                          share|improve this answer


























                                            up vote
                                            1
                                            down vote













                                            Appending var and var2 is the easy part - you can just join them like new_var="$var$var2". If what you really meant to say is you want to store cropped var into var2,then it's just var2=$(...) and you can put the commands that other users and my answer presented here inside the $() portion.



                                            The main part is removing those 10 characters. There's number of ways to extract the part you want from var:




                                            • printf (pretty portable, doesn't rely on specific shell)



                                              $ printf "%.11sn" "$var" 
                                              type_cardio



                                            • awk



                                              $ var=type_cardio_10-11-2017 
                                              $ awk -v awk_var="$var" 'BEGINprint substr(awk_var,0,length(awk_var)-11)'
                                              type_cardio


                                              or you can use printf trick here as well:



                                              $ echo "$var" | awk 'printf "%.11sn",$0' 
                                              type_cardio



                                            • egrep



                                              $ echo "$var" | egrep -o '^.0,11' 
                                              type_cardio



                                            • perl:



                                              $ echo "$var" | perl -lne 'print substr($_,0,11)' 
                                              type_cardio



                                            • python:



                                              $ python -c 'import sys;print sys.argv[1][0:11] ' "$var" 
                                              type_cardio






                                            share|improve this answer
























                                              up vote
                                              1
                                              down vote










                                              up vote
                                              1
                                              down vote









                                              Appending var and var2 is the easy part - you can just join them like new_var="$var$var2". If what you really meant to say is you want to store cropped var into var2,then it's just var2=$(...) and you can put the commands that other users and my answer presented here inside the $() portion.



                                              The main part is removing those 10 characters. There's number of ways to extract the part you want from var:




                                              • printf (pretty portable, doesn't rely on specific shell)



                                                $ printf "%.11sn" "$var" 
                                                type_cardio



                                              • awk



                                                $ var=type_cardio_10-11-2017 
                                                $ awk -v awk_var="$var" 'BEGINprint substr(awk_var,0,length(awk_var)-11)'
                                                type_cardio


                                                or you can use printf trick here as well:



                                                $ echo "$var" | awk 'printf "%.11sn",$0' 
                                                type_cardio



                                              • egrep



                                                $ echo "$var" | egrep -o '^.0,11' 
                                                type_cardio



                                              • perl:



                                                $ echo "$var" | perl -lne 'print substr($_,0,11)' 
                                                type_cardio



                                              • python:



                                                $ python -c 'import sys;print sys.argv[1][0:11] ' "$var" 
                                                type_cardio






                                              share|improve this answer














                                              Appending var and var2 is the easy part - you can just join them like new_var="$var$var2". If what you really meant to say is you want to store cropped var into var2,then it's just var2=$(...) and you can put the commands that other users and my answer presented here inside the $() portion.



                                              The main part is removing those 10 characters. There's number of ways to extract the part you want from var:




                                              • printf (pretty portable, doesn't rely on specific shell)



                                                $ printf "%.11sn" "$var" 
                                                type_cardio



                                              • awk



                                                $ var=type_cardio_10-11-2017 
                                                $ awk -v awk_var="$var" 'BEGINprint substr(awk_var,0,length(awk_var)-11)'
                                                type_cardio


                                                or you can use printf trick here as well:



                                                $ echo "$var" | awk 'printf "%.11sn",$0' 
                                                type_cardio



                                              • egrep



                                                $ echo "$var" | egrep -o '^.0,11' 
                                                type_cardio



                                              • perl:



                                                $ echo "$var" | perl -lne 'print substr($_,0,11)' 
                                                type_cardio



                                              • python:



                                                $ python -c 'import sys;print sys.argv[1][0:11] ' "$var" 
                                                type_cardio







                                              share|improve this answer














                                              share|improve this answer



                                              share|improve this answer








                                              edited Oct 21 '17 at 17:20

























                                              answered Oct 21 '17 at 17:15









                                              Sergiy Kolodyazhnyy

                                              7,91011548




                                              7,91011548



























                                                   

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