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

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}

bash

share|improve this question

edited 1 hour ago

Zack

asked 1 hour ago

ZackZack

63

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add a comment | 

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}

bash

share|improve this question

edited 1 hour ago

Zack

asked 1 hour ago

ZackZack

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.

add a comment | 

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

edited 1 hour ago

Zack

asked 1 hour ago

ZackZack

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.

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

edited 1 hour ago

Zack

asked 1 hour ago

ZackZack

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.

share|improve this question

edited 1 hour ago

Zack

asked 1 hour ago

ZackZack

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.

asked 1 hour ago

ZackZack

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.

asked 1 hour ago

ZackZack

63

1 Answer
1

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

answered 53 mins ago

EchoMike444EchoMike444

9525

add a comment | 

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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

answered 53 mins ago

EchoMike444EchoMike444

9525

add a comment | 

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

answered 53 mins ago

EchoMike444EchoMike444

9525

add a comment | 

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

answered 53 mins ago

EchoMike444EchoMike444

9525

   zip -@ foo

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

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

share|improve this answer

answered 53 mins ago

EchoMike444EchoMike444

9525

answered 53 mins ago

EchoMike444EchoMike444

9525

answered 53 mins ago

EchoMike444EchoMike444

9525