unzip all .zip files in a directory

tes15 · 06-30-2005, 08:30 PM

Hi,

I was wondering if anyone is able to help with my query. I am trying to unzip every .zip file in a certain directory. I am using the # unzip command and at present I am unzipping each file singularly.

I have searched the man pages on 'unzip' but cannot find any directions/options outlining if it is possible to globally unzip all .zip files in a directory in one command.

If anyone has any suggestions, comments it would be greatly appreciated.

Thanks,

-T

jrdioko · 06-30-2005, 08:33 PM

How about "unzip *.zip"?

tes15 · 06-30-2005, 08:37 PM

Hi jrdioko.

Initially, I tried the unzip command with that syntax and received the following error for each respective zip file: "caution: filename not matched: <file>.zip"

It does seem strange that global pattern matching does not work with the unzip command (or perhaps it does and I don't know about it?).

Thanks for the suggestion.

-T

mhallbiai · 06-30-2005, 08:42 PM

try:

Code:

for i in `ls -1 | grep .zip$`; do unzip $i; done

hope this helps

jrdioko · 06-30-2005, 08:44 PM

How about this then:

Code:

for i in `ls`; do unzip $i; done

jrdioko · 06-30-2005, 08:45 PM

Beat me to it and I'm glad you did. I was trying to unzip everything. Do you need the -l though?

tes15 · 06-30-2005, 08:48 PM

mhallbiai, jrdioko,

Both your solutions worked fine. Thanks for your help and quick responses.

Much appreciated.

-T

mhallbiai · 06-30-2005, 09:01 PM

jrdioko,

the -1 (number one) would ensure that it only fed one filename at a time without any permissions in case of an alias to ls that did not feed just the filenames to unzip.

i also added the grep to ensure that only files ending in .zip would be read

your solution works under ideal conditions which rarely exist for me

tes15, glad to help

jrdioko · 06-30-2005, 09:29 PM

Thanks for the explanation, it's about time I learn how to use those for loops.

Phaethar · 08-08-2005, 02:05 PM

Hey guys,

I know this topic is a bit old, but I'm running into the same issues, but I can't get the solution here to work either.

First, I'm getting exactly the same thing trying 'unzip *.zip'. I have about 30 files, and I get the "caution: filename not matched: <filename>.zip" error message.

So, I being that there are other files that are not .zip, I tried mhallbiai's solution and ran this:

for i in `ls -1 | grep .zip$`; do unzip $i; done

which, in turn, gave me this:

unzip: cannot find or open ls, ls.zip or ls.ZIP

I tried doing them one at a time and putting it in a simple shell script, but no luck.

So, I figure I'm missing something here, and hoping for an easy explanation.

BTW, this system does not have a GUI installed, so this has to be done through the CLI. Really hoping I don't have to sit there and manually unzip almost 30 zip files that are all almost 700MB!

Thanks!

jrdioko · 08-08-2005, 02:11 PM

What does the following return:

Code:

ls -1 | grep .zip$

IsaacKuo · 08-08-2005, 02:18 PM

Make sure the quotes you're typing in are "backquotes" (on my keyboard, it's on the upper left corner underneath "Esc"). Backquotes tell the shell to execute the contents, rather than just quoting the contents.

I personally find it an outrageously ugly syntax "feature".

Also, I'd put double quotes around $i, like so:

unzip "$i"

That's needed to work if some of the zip files have spaces.

Matir · 08-08-2005, 02:25 PM

Not to be difficult, but why not just:

Code:

ls -1 *.zip

Rather than going through grep?

Phaethar, make sure you are using backticks (same key as the tilde: ~), not single quotes or anything.

rdeep · 08-29-2014, 11:34 AM

unzip "*.zip" -d <directory_path>;

PS: this would unzip all the .zip files into the mentioned directory.
Note the difference in placing the commas

frieza · 08-29-2014, 11:46 AM

Quote:

Originally Posted by mhallbiai

try:

Code:

for i in `ls -1 | grep .zip$`; do unzip $i; done

hope this helps

a bit excessive, but works

Quote:

Originally Posted by jrdioko

How about this then:

Code:

for i in `ls`; do unzip $i; done

works if only zip files in directory

though technically `ls` is depricated in favor of $(ls)

a better way would be

Code:

for i in $(ls *.zip)
do
unzip $i
done

but an even more efficient way would be simply

Code:

for zipfile in *.zip
do
unzip $zipfile
done