Bash string range and replace












2














To print part of the string, replace , with . I use command:



echo "${q:16:6}" | sed 's/,/./'


Is it possible to use something like:



echo "${q:16:6/,/.}"


because it does not work?










share|improve this question





























    2














    To print part of the string, replace , with . I use command:



    echo "${q:16:6}" | sed 's/,/./'


    Is it possible to use something like:



    echo "${q:16:6/,/.}"


    because it does not work?










    share|improve this question



























      2












      2








      2







      To print part of the string, replace , with . I use command:



      echo "${q:16:6}" | sed 's/,/./'


      Is it possible to use something like:



      echo "${q:16:6/,/.}"


      because it does not work?










      share|improve this question















      To print part of the string, replace , with . I use command:



      echo "${q:16:6}" | sed 's/,/./'


      Is it possible to use something like:



      echo "${q:16:6/,/.}"


      because it does not work?







      bash sed variable replace






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Dec 9 at 14:37









      Jeff Schaller

      38.7k1053125




      38.7k1053125










      asked Dec 8 at 10:21









      pbies

      15110




      15110






















          1 Answer
          1






          active

          oldest

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          10














          You can't stack/nest parameter expansion in Bash, so not even



          echo "${${q:16:6}/,/.}"


          will work. (Nested expansions like that do work in Zsh, though.)



          If you want to stay within Bash, you'll need to use a temporary variable:



          foo="${q:16:6}"
          echo "${foo/,/.}"





          share|improve this answer























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            1 Answer
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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            10














            You can't stack/nest parameter expansion in Bash, so not even



            echo "${${q:16:6}/,/.}"


            will work. (Nested expansions like that do work in Zsh, though.)



            If you want to stay within Bash, you'll need to use a temporary variable:



            foo="${q:16:6}"
            echo "${foo/,/.}"





            share|improve this answer




























              10














              You can't stack/nest parameter expansion in Bash, so not even



              echo "${${q:16:6}/,/.}"


              will work. (Nested expansions like that do work in Zsh, though.)



              If you want to stay within Bash, you'll need to use a temporary variable:



              foo="${q:16:6}"
              echo "${foo/,/.}"





              share|improve this answer


























                10












                10








                10






                You can't stack/nest parameter expansion in Bash, so not even



                echo "${${q:16:6}/,/.}"


                will work. (Nested expansions like that do work in Zsh, though.)



                If you want to stay within Bash, you'll need to use a temporary variable:



                foo="${q:16:6}"
                echo "${foo/,/.}"





                share|improve this answer














                You can't stack/nest parameter expansion in Bash, so not even



                echo "${${q:16:6}/,/.}"


                will work. (Nested expansions like that do work in Zsh, though.)



                If you want to stay within Bash, you'll need to use a temporary variable:



                foo="${q:16:6}"
                echo "${foo/,/.}"






                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited Dec 8 at 14:35









                ilkkachu

                55.9k784155




                55.9k784155










                answered Dec 8 at 10:30









                nohillside

                2,292819




                2,292819






























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