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 46 mins 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 46 mins 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 46 mins 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 46 mins 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 46 mins 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 42 mins ago
Zack
asked 46 mins 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 46 mins ago
ZackZack
63
asked 46 mins 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 25 mins ago
EchoMike444EchoMike444
9525
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 25 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
zipcommand
cat BIG_FILENAME_LIST.txt | zip thebigziparchive -@
answered 25 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
zipcommand
cat BIG_FILENAME_LIST.txt | zip thebigziparchive -@
answered 25 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
zipcommand
cat BIG_FILENAME_LIST.txt | zip thebigziparchive -@
answered 25 mins ago
EchoMike444EchoMike444
9525
answered 25 mins ago
EchoMike444EchoMike444
9525
answered 25 mins ago
EchoMike444EchoMike444
9525
answered 25 mins ago
EchoMike444EchoMike444
9525
9525
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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