How to send output of convert to a writable directory in bash?

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up vote
2
down vote

favorite












I want to convert .tbl files to .csv in /data/ directory which I've got read-only access.
I'm using this command:



for i in `ls *.tbl`; do
sed 's/|$//' $i > $i/tbl/csv
echo $i
done


But I'm getting:



-bash: supplier.csv: Permission denied


And I don't know how to change the script such that it could save the output in a writable directory.
Any idea how to solve it?










share|improve this question



























    up vote
    2
    down vote

    favorite












    I want to convert .tbl files to .csv in /data/ directory which I've got read-only access.
    I'm using this command:



    for i in `ls *.tbl`; do
    sed 's/|$//' $i > $i/tbl/csv
    echo $i
    done


    But I'm getting:



    -bash: supplier.csv: Permission denied


    And I don't know how to change the script such that it could save the output in a writable directory.
    Any idea how to solve it?










    share|improve this question

























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      I want to convert .tbl files to .csv in /data/ directory which I've got read-only access.
      I'm using this command:



      for i in `ls *.tbl`; do
      sed 's/|$//' $i > $i/tbl/csv
      echo $i
      done


      But I'm getting:



      -bash: supplier.csv: Permission denied


      And I don't know how to change the script such that it could save the output in a writable directory.
      Any idea how to solve it?










      share|improve this question















      I want to convert .tbl files to .csv in /data/ directory which I've got read-only access.
      I'm using this command:



      for i in `ls *.tbl`; do
      sed 's/|$//' $i > $i/tbl/csv
      echo $i
      done


      But I'm getting:



      -bash: supplier.csv: Permission denied


      And I don't know how to change the script such that it could save the output in a writable directory.
      Any idea how to solve it?







      command-line bash






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 23 at 16:11









      wjandrea

      7,96042258




      7,96042258










      asked Nov 23 at 15:51









      Marzieh Heidari

      1136




      1136




















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          4
          down vote



          accepted










          Just specify a different path:



          for tbl in *.tbl ; do
          sed 's/|$//' "$tbl" > /another/path/"$tbl%.tbl".csv
          echo "$tbl"
          done


          Few minor fixes:



          1. Don't parse the output of ls, just iterate over an expanded wildcard pattern.


          2. Double quote variables (filenames can contain whitespace).


          3. I used Remove Matching Suffix instead of Substitution.






          share|improve this answer





























            up vote
            4
            down vote















            Just specify the path (here /path/to/writable/dir/) before the filename:



            for i in *.tbl; do
            sed 's/|$//' "$i" >"/path/to/writable/dir/$i/%tbl/csv"
            echo "$i"
            done


            You should never parse the output of ls (Why?) and always quote variables like $i here. Adding % at the pattern’s beginning makes it match against the end of the string so that a filename like tbl.tbl becomes tbl.csv rather than csv.tbl. Read more about Parameter expansion here on bash-hackers.org.






            share|improve this answer






















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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              4
              down vote



              accepted










              Just specify a different path:



              for tbl in *.tbl ; do
              sed 's/|$//' "$tbl" > /another/path/"$tbl%.tbl".csv
              echo "$tbl"
              done


              Few minor fixes:



              1. Don't parse the output of ls, just iterate over an expanded wildcard pattern.


              2. Double quote variables (filenames can contain whitespace).


              3. I used Remove Matching Suffix instead of Substitution.






              share|improve this answer


























                up vote
                4
                down vote



                accepted










                Just specify a different path:



                for tbl in *.tbl ; do
                sed 's/|$//' "$tbl" > /another/path/"$tbl%.tbl".csv
                echo "$tbl"
                done


                Few minor fixes:



                1. Don't parse the output of ls, just iterate over an expanded wildcard pattern.


                2. Double quote variables (filenames can contain whitespace).


                3. I used Remove Matching Suffix instead of Substitution.






                share|improve this answer
























                  up vote
                  4
                  down vote



                  accepted







                  up vote
                  4
                  down vote



                  accepted






                  Just specify a different path:



                  for tbl in *.tbl ; do
                  sed 's/|$//' "$tbl" > /another/path/"$tbl%.tbl".csv
                  echo "$tbl"
                  done


                  Few minor fixes:



                  1. Don't parse the output of ls, just iterate over an expanded wildcard pattern.


                  2. Double quote variables (filenames can contain whitespace).


                  3. I used Remove Matching Suffix instead of Substitution.






                  share|improve this answer














                  Just specify a different path:



                  for tbl in *.tbl ; do
                  sed 's/|$//' "$tbl" > /another/path/"$tbl%.tbl".csv
                  echo "$tbl"
                  done


                  Few minor fixes:



                  1. Don't parse the output of ls, just iterate over an expanded wildcard pattern.


                  2. Double quote variables (filenames can contain whitespace).


                  3. I used Remove Matching Suffix instead of Substitution.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Nov 23 at 16:09

























                  answered Nov 23 at 16:04









                  choroba

                  6,56411730




                  6,56411730






















                      up vote
                      4
                      down vote















                      Just specify the path (here /path/to/writable/dir/) before the filename:



                      for i in *.tbl; do
                      sed 's/|$//' "$i" >"/path/to/writable/dir/$i/%tbl/csv"
                      echo "$i"
                      done


                      You should never parse the output of ls (Why?) and always quote variables like $i here. Adding % at the pattern’s beginning makes it match against the end of the string so that a filename like tbl.tbl becomes tbl.csv rather than csv.tbl. Read more about Parameter expansion here on bash-hackers.org.






                      share|improve this answer


























                        up vote
                        4
                        down vote















                        Just specify the path (here /path/to/writable/dir/) before the filename:



                        for i in *.tbl; do
                        sed 's/|$//' "$i" >"/path/to/writable/dir/$i/%tbl/csv"
                        echo "$i"
                        done


                        You should never parse the output of ls (Why?) and always quote variables like $i here. Adding % at the pattern’s beginning makes it match against the end of the string so that a filename like tbl.tbl becomes tbl.csv rather than csv.tbl. Read more about Parameter expansion here on bash-hackers.org.






                        share|improve this answer
























                          up vote
                          4
                          down vote










                          up vote
                          4
                          down vote











                          Just specify the path (here /path/to/writable/dir/) before the filename:



                          for i in *.tbl; do
                          sed 's/|$//' "$i" >"/path/to/writable/dir/$i/%tbl/csv"
                          echo "$i"
                          done


                          You should never parse the output of ls (Why?) and always quote variables like $i here. Adding % at the pattern’s beginning makes it match against the end of the string so that a filename like tbl.tbl becomes tbl.csv rather than csv.tbl. Read more about Parameter expansion here on bash-hackers.org.






                          share|improve this answer
















                          Just specify the path (here /path/to/writable/dir/) before the filename:



                          for i in *.tbl; do
                          sed 's/|$//' "$i" >"/path/to/writable/dir/$i/%tbl/csv"
                          echo "$i"
                          done


                          You should never parse the output of ls (Why?) and always quote variables like $i here. Adding % at the pattern’s beginning makes it match against the end of the string so that a filename like tbl.tbl becomes tbl.csv rather than csv.tbl. Read more about Parameter expansion here on bash-hackers.org.







                          share|improve this answer














                          share|improve this answer



                          share|improve this answer








                          edited Nov 23 at 16:38

























                          answered Nov 23 at 16:04









                          dessert

                          21.3k55896




                          21.3k55896



























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