CODEWITHSUNDEEP
linuxunixksh
Thang Pham
Thang Pham

Reputation: 38705

Extract list of file names in a zip archive when `unzip -l`

When I do unzip -l zipfilename, I see

1295627  08-22-11 07:10   A.pdf
473980  08-22-11 07:10   B.pdf
...

I only want to see the filenames. I try this 

unzip -l zipFilename | cut -f4 -d" "

but I don't think the delimiter is just " ".

Upvotes: 39

Views: 40131

Answers (4)

glenn jackman
glenn jackman

Reputation: 247022

Assuming none of the files have spaces in names:

unzip -l filename.zip | awk '{print $NF}'

My unzip output has both a header and footer, so the awk script becomes:

unzip -l filename.zip | awk '/-----/ {p = ++p % 2; next} p {print $NF}'

A version that handles filenames with spaces:

unzip -l filename.zip | awk '
    /----/ {p = ++p % 2; next}
    $NF == "Name" {pos = index($0,"Name")}
    p {print substr($0,pos)}
'

Upvotes: 28

amicitas
amicitas

Reputation: 13671

The easiest way to do this is to use the following command:

unzip -Z -1 archive.zip

or

zipinfo -1 archive.zip

This will list only the file names, one on each line.

The two commands are exactly equivalent. The -Z option tells unzip to treat the rest of the options as zipinfo options. See the man pages for unzip and zipinfo.

Upvotes: 106

peteretep
peteretep

Reputation: 317

If you need to cater for filenames with spaces, try:

unzip -l zipfilename.zip | awk -v f=4  ' /-----/ {p = ++p % 2; next} p { for (i=f; i<=NF;i++) printf("%s%s", $i,(i==NF) ? "\n" : OFS) }'

Upvotes: 6

Manny D
Manny D

Reputation: 20724

Use awk:

unzip -l zipfilename | awk '{print $4}'

Upvotes: 3

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