unzip many files...each zip file needs a directory

linuxhippy · 01-15-2010, 08:31 PM

I have about 50 zip files that are all in 1 directory. How can I unzip these so that each zip file gets it's own directory?

cantab · 01-15-2010, 09:49 PM

Code:

for filename in *.zip ; do mkdir ${filename}.dir ; cd ${filename}.dir ; mv ../${filename} . ; unzip ${filename} ; cd .. ; done

results in ugly directory names I know, but otherwise works.

linuxhippy · 01-15-2010, 09:57 PM

that worked well...any idea how to remove .zip.dir from all those directories?

lwasserm · 01-15-2010, 10:16 PM

"just in case" backup your original zip files tile your done.

cd to the containing directory

for DIR in *.zip.dir; do echo ${DIR##.zip.dir};done

If it produces the names you want, change the do... statement as follows:

for DIR in *.zip.dir; do mv $DIR ${DIR##.zip.dir}; done

linuxhippy · 01-15-2010, 10:30 PM

nope...I got this error:

mv: cannot move `xmms-256.zip.dir' to a subdirectory of itself, `xmms-256.zip.dir/xmms-256.zip.dir'

linuxhippy · 01-15-2010, 11:03 PM

maybe cut -c .zip.dir goes in there??

lwasserm · 01-17-2010, 11:20 AM

Sorry, should have used %% instead of ##. Corrected version would be:

for DIR in *.zip.dir; do mv $DIR ${DIR%%.zip.dir}; done

It still be advisable to test for the desired names first with

for DIR in *.zip.dir; do echo ${DIR%%.zip.dir}; done

before doig the actual name change with "mv"

cantab · 01-17-2010, 01:15 PM

What exactly does the %% do then? It's not something I've come across in bash before.

lwasserm · 01-17-2010, 03:56 PM

From the bash man page:

Code:

 ${parameter%word}
 ${parameter%%word}
              The word is expanded to produce a pattern just  as  in  pathname
              expansion.   If  the  pattern  matches a trailing portion of the
              expanded value of parameter, then the result of the expansion is
              the  expanded value of parameter with the shortest matching pat‐
              tern (the ‘‘%’’ case)  or  the  longest  matching  pattern  (the
              ‘‘%%’’  case)  deleted.   If  parameter  is  @ or *, the pattern
              removal operation is applied to  each  positional  parameter  in
              turn,  and the expansion is the resultant list.  If parameter is
              an array variable subscripted with @ or *, the  pattern  removal
              operation  is  applied  to each member of the array in turn, and
              the expansion is the resultant list.

Look in the "Parameter Expansion" section

cantab · 01-17-2010, 04:36 PM

That extract from the man page is one of the most confusing things I have ever read...I understood it, eventually, after reading through it like 5 times.

lwasserm · 01-17-2010, 05:08 PM

Quote:

Originally Posted by cantab

That extract from the man page is one of the most confusing things I have ever read...I understood it, eventually, after reading through it like 5 times.

Wow, good for you! I'm sure I had to read it 8 or 9 times myself!