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












4















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

























  • As an alternative: you can create an empty zip file at the beginning and add a file to that zip file, instead of making an array first and adding them all at once. Might save you a loop, unless you have need for the array for other reasons...

    – Henno Brandsma
    Mar 24 at 13:51


















4















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

























  • As an alternative: you can create an empty zip file at the beginning and add a file to that zip file, instead of making an array first and adding them all at once. Might save you a loop, unless you have need for the array for other reasons...

    – Henno Brandsma
    Mar 24 at 13:51
















4












4








4








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
















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




bash






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 24 at 3:59







Zack

















asked Mar 24 at 3:54









ZackZack

234




234













  • As an alternative: you can create an empty zip file at the beginning and add a file to that zip file, instead of making an array first and adding them all at once. Might save you a loop, unless you have need for the array for other reasons...

    – Henno Brandsma
    Mar 24 at 13:51





















  • As an alternative: you can create an empty zip file at the beginning and add a file to that zip file, instead of making an array first and adding them all at once. Might save you a loop, unless you have need for the array for other reasons...

    – Henno Brandsma
    Mar 24 at 13:51



















As an alternative: you can create an empty zip file at the beginning and add a file to that zip file, instead of making an array first and adding them all at once. Might save you a loop, unless you have need for the array for other reasons...

– Henno Brandsma
Mar 24 at 13:51







As an alternative: you can create an empty zip file at the beginning and add a file to that zip file, instead of making an array first and adding them all at once. Might save you a loop, unless you have need for the array for other reasons...

– Henno Brandsma
Mar 24 at 13:51












1 Answer
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oldest

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6














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 -@








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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









    6














    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




























      6














      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


























        6












        6








        6







        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 Mar 24 at 4:16









        EchoMike444EchoMike444

        1,0506




        1,0506






























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