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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
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 1 hour ago
ZackZack
63
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Zack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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 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
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.
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 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.
edited 59 mins 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
asked 1 hour 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.
add a comment |
add a comment |
1 Answer
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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 42 mins ago
EchoMike444EchoMike444
9525
add a comment |
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1 Answer
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1 Answer
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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 42 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 :
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 42 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 :
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 42 mins ago
EchoMike444EchoMike444
9525
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 42 mins ago
EchoMike444EchoMike444
9525
answered 42 mins ago
EchoMike444EchoMike444
9525
answered 42 mins ago
EchoMike444EchoMike444
9525
answered 42 mins ago
EchoMike444EchoMike444
9525
9525
add a comment |
add a comment |
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