Argument list too long when zipping large list of certain files in a folder2019 Community Moderator...

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Argument list too long when zipping large list of certain files in a folder



2019 Community Moderator ElectionSolving “mv: Argument list too long”?How to print a range of IP addresses with Linux seq commandbash: /usr/bin/perl: Argument list too long/usr/bin/awk: Argument list too longArgument list too long with just 5000 filesAWK Using a Variable in an Equality Expressionbash array with variable in the nameReplace a long string with the sed command: Argument list too long errorAdd text to each value while looping thru and printing them in a array?Moving random files using shuf and mv - Argument list too long












1















I'm trying to zip a certain large number of files by creating an array of them. Then I tried to run the zip command but met with "Argument list too long" error.



declare -a arr=()       

fixed=5
for i in `seq 10 1 200`; do
for j in `seq $((i+fixed)) 1 200`; do
arr+=("${i}_${j}.xxx")
done
done

new_arr=$(printf ",%s" "${arr[@]}")
new_arr=${new_arr:1}

zip all_data.zip {$new_arr}









share|improve this question









New contributor




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

























    1















    I'm trying to zip a certain large number of files by creating an array of them. Then I tried to run the zip command but met with "Argument list too long" error.



    declare -a arr=()       

    fixed=5
    for i in `seq 10 1 200`; do
    for j in `seq $((i+fixed)) 1 200`; do
    arr+=("${i}_${j}.xxx")
    done
    done

    new_arr=$(printf ",%s" "${arr[@]}")
    new_arr=${new_arr:1}

    zip all_data.zip {$new_arr}









    share|improve this question









    New contributor




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























      1












      1








      1








      I'm trying to zip a certain large number of files by creating an array of them. Then I tried to run the zip command but met with "Argument list too long" error.



      declare -a arr=()       

      fixed=5
      for i in `seq 10 1 200`; do
      for j in `seq $((i+fixed)) 1 200`; do
      arr+=("${i}_${j}.xxx")
      done
      done

      new_arr=$(printf ",%s" "${arr[@]}")
      new_arr=${new_arr:1}

      zip all_data.zip {$new_arr}









      share|improve this question









      New contributor




      Zack 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 zip a certain large number of files by creating an array of them. Then I tried to run the zip command but met with "Argument list too long" error.



      declare -a arr=()       

      fixed=5
      for i in `seq 10 1 200`; do
      for j in `seq $((i+fixed)) 1 200`; do
      arr+=("${i}_${j}.xxx")
      done
      done

      new_arr=$(printf ",%s" "${arr[@]}")
      new_arr=${new_arr:1}

      zip all_data.zip {$new_arr}






      bash






      share|improve this question









      New contributor




      Zack 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




      Zack 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




      share|improve this question








      edited 34 mins ago







      Zack













      New contributor




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









      asked 39 mins ago









      ZackZack

      63




      63




      New contributor




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





      New contributor





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






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






















          1 Answer
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          extract from man zip ( linux version )



             zip -@ foo
          will store the files listed one per line on stdin in foo.zip.


          example from the same man page



             find . -name "*.[ch]" -print | zip source -@


          So steps will be :




          1. build a list off all files to be archive , format must one file name by line



          2. run zip command



            cat BIG_FILENAME_LIST.txt | zip thebigziparchive -@








          share|improve this answer























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            1 Answer
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            active

            oldest

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3














            extract from man zip ( linux version )



               zip -@ foo
            will store the files listed one per line on stdin in foo.zip.


            example from the same man page



               find . -name "*.[ch]" -print | zip source -@


            So steps will be :




            1. build a list off all files to be archive , format must one file name by line



            2. run zip command



              cat BIG_FILENAME_LIST.txt | zip thebigziparchive -@








            share|improve this answer




























              3














              extract from man zip ( linux version )



                 zip -@ foo
              will store the files listed one per line on stdin in foo.zip.


              example from the same man page



                 find . -name "*.[ch]" -print | zip source -@


              So steps will be :




              1. build a list off all files to be archive , format must one file name by line



              2. run zip command



                cat BIG_FILENAME_LIST.txt | zip thebigziparchive -@








              share|improve this answer


























                3












                3








                3







                extract from man zip ( linux version )



                   zip -@ foo
                will store the files listed one per line on stdin in foo.zip.


                example from the same man page



                   find . -name "*.[ch]" -print | zip source -@


                So steps will be :




                1. build a list off all files to be archive , format must one file name by line



                2. run zip command



                  cat BIG_FILENAME_LIST.txt | zip thebigziparchive -@








                share|improve this answer













                extract from man zip ( linux version )



                   zip -@ foo
                will store the files listed one per line on stdin in foo.zip.


                example from the same man page



                   find . -name "*.[ch]" -print | zip source -@


                So steps will be :




                1. build a list off all files to be archive , format must one file name by line



                2. run zip command



                  cat BIG_FILENAME_LIST.txt | zip thebigziparchive -@









                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 18 mins ago









                EchoMike444EchoMike444

                9525




                9525






















                    Zack is a new contributor. Be nice, and check out our Code of Conduct.










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                    Zack is a new contributor. Be nice, and check out our Code of Conduct.













                    Zack is a new contributor. Be nice, and check out our Code of Conduct.












                    Zack is a new contributor. Be nice, and check out our Code of Conduct.
















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