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Argument list too long when zipping large list of certain files in a folder
The Next CEO of Stack OverflowSolving “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
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
asked Mar 24 at 3:54
ZackZack
234
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
asked Mar 24 at 3:54
ZackZack
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
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
asked Mar 24 at 3:54
ZackZack
234
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
asked Mar 24 at 3:54
ZackZack
234
asked Mar 24 at 3:54
ZackZack
234
edited Mar 24 at 3:59
Zack
asked Mar 24 at 3:54
ZackZack
234
asked Mar 24 at 3:54
ZackZack
234
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
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 :
build a list off all files to be archive , format must one file name by line
run
zipcommandcat BIG_FILENAME_LIST.txt | zip thebigziparchive -@
answered Mar 24 at 4:16
EchoMike444EchoMike444
1,0506
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
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 :
build a list off all files to be archive , format must one file name by line
run
zipcommandcat BIG_FILENAME_LIST.txt | zip thebigziparchive -@
answered Mar 24 at 4:16
EchoMike444EchoMike444
1,0506
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 :
build a list off all files to be archive , format must one file name by line
run
zipcommandcat BIG_FILENAME_LIST.txt | zip thebigziparchive -@
answered Mar 24 at 4:16
EchoMike444EchoMike444
1,0506
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 :
build a list off all files to be archive , format must one file name by line
run
zipcommandcat BIG_FILENAME_LIST.txt | zip thebigziparchive -@
answered Mar 24 at 4:16
EchoMike444EchoMike444
1,0506
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 :
build a list off all files to be archive , format must one file name by line
run
zipcommandcat BIG_FILENAME_LIST.txt | zip thebigziparchive -@
answered Mar 24 at 4:16
EchoMike444EchoMike444
1,0506
answered Mar 24 at 4:16
EchoMike444EchoMike444
1,0506
answered Mar 24 at 4:16
EchoMike444EchoMike444
1,0506
answered Mar 24 at 4:16
EchoMike444EchoMike444
1,0506
1,0506
add a comment |
add a comment |
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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