Extract only a specific file from a zipped archive to a given directory

Question

I need to extract a single file from a ZIP file which I know the path to. Is there a command like the following:

unzip -d . myarchive.zip path/to/zipped/file.txt

Unfortunately, the above command extracts and recreates the entire path to the file at ./path/to/zipped/file.txt. Is there a way for me to simply pull the file out into a specified directory?

score 254 · Accepted Answer · edited Jun 11 '20 at 05:41

254

You can extract just the text to standard output with the -p option:

unzip -p myarchive.zip path/to/zipped/file.txt >file.txt

This won't extract the metadata (date, permissions, …), only the file contents (obviously, it only works for regular files, not symlinks, devices, directories...). That's the price to pay for the convenience of not having to move the file afterwards.

Alternatively, mount the archive as a directory and just copy the file. With AVFS:

mountavfs
cp -p ~/.avfs"$PWD/myarchive.zip#"/path/to/zipped/file.txt .

Or with fuse-zip:

mkdir myarchive.d
fuse-zip myarchive.zip myarchive.d
cp -p myarchive.d/path/to/zipped/file.txt .
fusermount -u myarchive.d; rmdir myarchive.d

edited Jun 11 '20 at 05:41

Stéphane Chazelas

544,893

answered May 31 '11 at 23:43

Gilles 'SO- stop being evil'

829,060

1

Will it work for binary files, say a jar file? – Naftuli Kay Jun 01 '11 at 01:01
1

@TKKocheran: Jar files are zips, so the unzip and fuse-zip methods will obviously work. The AVFS method also works, because AVFS guesses the format based on file names and knows about .jar; if your file is named differently you might need to tell AVFS to use its zip handler, e.g. ~/.avfs$PWD/foo.apk#uzip/META-INF. – Gilles 'SO- stop being evil' Jun 01 '11 at 07:19
1

actually, I meant extracting a JAR/binary file from a ZIP archive. I haven't had the chance to test it out yet, can you see any issues using the first command above to extract binary files? – Naftuli Kay Jun 01 '11 at 18:41
2

@TKKocheran: There's no problem, -p extracts the file as-is (-c does text conversion). – Gilles 'SO- stop being evil' Jun 01 '11 at 20:50
3

I think the answer of @Myles is more elegant, because it doesn't require output redirection and it preserves file attributes. – gertvdijk Dec 04 '12 at 15:42

sMyles · Answer 2 · 2021-12-14T01:14:43.067

244

unzip -j "myarchive.zip" "in/archive/file.txt" -d "/path/to/unzip/to"

Enter full path for zipped file, not just the filename. Be sure to keep the structure as seen from within the zip file.

This will extract the single file file.txt in myarchive.zip to /path/to/unzip/to/file.txt.

-j: junk paths. The archive's directory structure is not recreated; all files are deposited in the extraction directory (by default, the current one)

-d: An optional directory to which to extract files.

https://www.mankier.com/1/unzip

Multiple files:

unzip -j myarchive.zip in/archive/file.txt another/file.ext -d /path/to/unzip/to

Entire Sub-directory

unzip -j archive.zip "sub/dir/*" -d "dest/dir"

edited Dec 14 '21 at 01:14

answered Dec 04 '12 at 06:49

sMyles

2,575

3

It doesn't handle the case where you want the generated filename to be different. – Oleg Vaskevich Oct 27 '14 at 21:21
11

the "-j" paramter: junk paths. The archive's directory structure is not recreated; all files are deposited in the extraction directory (by default, the current one). the "-d" parameter: extract files into exdir – e1i45 Jan 29 '15 at 16:14
2

@OlegVaskevich, the question doesn't request target filename to be different – Tapemaster May 11 '15 at 10:49
1

FYI to find out the path of files in zip, use less: https://superuser.com/questions/216617/view-list-of-files-in-zip-archive-on-linux – Yibo Yang Oct 24 '17 at 06:17
Some unzip binaries don't have the -d flag. I won't work in every linux – neves Dec 06 '19 at 15:02
4

unzip accepts wildcard, so to unzip sub-directory: unzip -j archive.zip "sub/dir/*" -d "dest/dir" – Noam Manos May 21 '20 at 10:27
@YiboYang or mc ;) – jena Nov 17 '21 at 11:11
Thanks guys updated the original answer to include some of these! – sMyles Dec 14 '21 at 01:14

score 54 · Answer 3 · edited Jan 17 '15 at 16:15

54

Simpler version:

unzip ARCHIVE_NAME PATH_OF_FILE_INSIDE_ARCHIVE

This will recreate PATH_OF_FILE_INSIDE_ARCHIVE in current directory but only extracts specified file.

To list all files in a Zip archive:

unzip -l ARCHIVE_NAME

edited Jan 17 '15 at 16:15

Anthon

79,293

answered Jul 08 '13 at 10:31

Taukir

549

This works with wildcards too, at least in CentOS - e.g. unzip foo.zip *.bar – aland May 21 '20 at 14:32
1

this is all i needed. very succinct. don't really see any benefit in using other methods unless you need to extract many files at once. – Ilia Sidorenko Aug 17 '21 at 01:59

Cameron Lowell Palmer · Answer 4 · 2020-06-11T03:55:53.803

On macOS, which by default uses Info-Zip

First list off the files to find what you want

unzip -l my.zip

Then extract file from the archive

unzip my.zip annoying/path/to/file/in/zip

Combine with -p for stdout

unzip -p my.zip annoying/path/to/file/in/zip >./file

or -j for extracting to the current directory (discard junk path)

unzip -j my.zip annoying/path/to/file/in/zip

with -d you can specify to create an arbitrary directory

unzip -d /path/to/dir my.zip annoying/path/to/file/in/zip

If you want the file in the -d directory you probably want to combine it with the -j option.

score 5 · Answer 5 · edited Jun 11 '20 at 14:16

5

simple use:

unzip zipfile.zip path/inside/zip/file.txt

and it will inflate the file.

$ unzip -l ./../html.zip | grep wp-config
 3328  07-22-2019 15:10   html/wp-config.php

 2898  01-07-2019 23:30   html/wp-config-sample.php


$ unzip ./../html.zip html/wp-config.php
 Archive:  ./../html.zip
 inflating: html/wp-config.php


$ ls -lrth
 total 4.0K
 drwxr-sr-x 2 apache apache 4.0K Jul 26 14:41 html


$ ls -lrth html/*
 total 4.0K
 -rw-rw-rw- 1 apache apache 3.3K Jul 22 15:10 wp-config.php

edited Jun 11 '20 at 14:16

Community

1

answered Jul 26 '19 at 18:58

Cesar Alejandro Bravo Anaya

51

1

Can also add the -d outpath argument to specify where the file is put in. – Efren Sep 18 '19 at 02:45
1

And the -j argument to avoid the compressed paths – Efren Sep 18 '19 at 02:56

score 1 · Answer 6 · answered Jun 11 '20 at 06:19

If you want to restore the file's metadata from the archive, be able to restore files of any type (including symlinks) and with arbitrary names, and also choose the new name of the file, you could use libarchive's bsdtar (which also supports ZIP format archives) instead of unzip as:

bsdtar -nqvvxpf file.zip -'s/.*/newname/S' some/dir/originalname

Note: like in unzip, the some/dir/originalname there is taken as a shell wildcard pattern. So if the path of archive member you want to extract does contain *, ?, \ or [ characters, you'll need to escape them as in:

bsdtar -nqvvxpf file.zip -'s/.*/newname/S' 'som[?]/dir/[*]riginalname'

or:

bsdtar -nqvvxpf file.zip -'s/.*/newname/S' 'som\?/dir/\*riginalname'

if the archive member is called som?/dir/*riginalname.

Note that that approach would work with any of the archive formats supported by libarchive, not just zip.

score 0 · Answer 7 · answered Apr 01 '20 at 10:21

Extract to a relative dir

unzip -j -d relativedir archive.zip path/in/archive/file.ext

Extract to the current dir

unzip -j -d . archive.zip path/in/archive/file.ext

Extract to absolute dir

unzip -j -d /absolutedir archive.zip path/in/archive/file.ext

score -1 · Answer 8 · answered Aug 25 '22 at 13:53

-1

unzip myarchive.zip `zipinfo -2 myarchive.zip | head -1`

answered Aug 25 '22 at 13:53

camel

1

Welcome to Unix & Linux! Brevity is acceptable, but fuller explanations are better. – Kusalananda Aug 25 '22 at 14:56

Extract only a specific file from a zipped archive to a given directory

8 Answers8

simple use:

Linked