Argument list too long when zipping large list of certain files in a folder
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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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 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
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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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
zipcommand
cat BIG_FILENAME_LIST.txt | zip thebigziparchive -@
answered 1 hour ago
EchoMike444EchoMike444
9675
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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
zipcommand
cat BIG_FILENAME_LIST.txt | zip thebigziparchive -@
answered 1 hour ago
EchoMike444EchoMike444
9675
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
zipcommand
cat BIG_FILENAME_LIST.txt | zip thebigziparchive -@
answered 1 hour ago
EchoMike444EchoMike444
9675
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
zipcommand
cat BIG_FILENAME_LIST.txt | zip thebigziparchive -@
answered 1 hour ago
EchoMike444EchoMike444
9675
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
zipcommand
cat BIG_FILENAME_LIST.txt | zip thebigziparchive -@
answered 1 hour ago
EchoMike444EchoMike444
9675
answered 1 hour ago
EchoMike444EchoMike444
9675
answered 1 hour ago
EchoMike444EchoMike444
9675
answered 1 hour ago
EchoMike444EchoMike444
9675
9675
add a comment |
add a comment |
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Zack is a new contributor. Be nice, and check out our Code of Conduct.
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