Hi,
I'm desperately in search for a solution of the following problem:
I have a directory full of zip-files. All these zip-files contain a single file with a name that should be used for the name of the zip-container.
Anybody a good idea. I'm an absolute beginner in shell scripting - so please bear with me.
Thanks
Mark
the below two commands will do your task.
example :
one.zip contains abc file, then it will create a abc.zip and abc.zip file contains abc file
for i in `ls *.zip`; do unzip $i; done
for i in `ls | grep -v ".zip"`; do zip $i $i;done
You can do something like that :
$ cat mark.sh
for zip in *.zip
do
file="$(unzip -l -qq "${zip}" | sed 's/.*:[0-9]\+[[:space:]]*//')"
newzip=$(echo "${file}" | sed 's/ /_/g').zip
echo "Zip '${zip}' contains file '${file}', renamed to '${newzip}'"
mv ${zip} ${newzip}
done
exit
Intial zip files
$ ls *zip
m1.zip m2.zip m3.zip
$ unzip -l -qq m1.zip
582 05-10-10 16:37 mark_1
$ unzip -l -qq m2.zip
4249 05-10-10 16:37 mark_2.txt
$ unzip -l -qq m3.zip
631 05-10-10 16:43 mark with spaces
$
Script execution:
$ ./mark.sh
Zip 'm1.zip' contains file 'mark_1', renamed to 'mark_1.zip'
Zip 'm2.zip' contains file 'mark_2.txt', renamed to 'mark_2.txt.zip'
Zip 'm3.zip' contains file 'mark with spaces', renamed to 'mark_with_spaces.zip'
$
New zip files :
$ ls *.zip
mark_1.zip mark_2.txt.zip mark_with_spaces.zip
$ unzip -l -qq mark_1.zip
582 05-10-10 16:37 mark_1
$ unzip -l -qq mark_2.txt.zip
4249 05-10-10 16:37 mark_2.txt
$ unzip -l -qq mark_with_spaces.zip
631 05-10-10 16:43 mark with spaces
$
Jean-Pierre.
Sorry,
my description was not entirely correct. Alle zip-files contain a large amount of files but always a file with the name ########.jdf,
where ######## is always a 8-digit number. That filename should be used as the zipfiles name. I suppose that makes it a little bit more complicated.
Is 'sed' still the way to go or is 'grep' more suitable?
Thanks for your kind help
Mark
New version of the mark.sh script :
for zip in *.zip
do
jdf="$( unzip -l -qq "${zip}" |
sed -n 's/.*:[0-9]\+[[:space:]]*\([0-9]\{8\}\)\.jdf$/\1/p')"
if [ -z "${jdf}" ]
then
echo "Invalid zip file '${zip}', jdf file not found"
continue
fi
newzip=${jdf}.zip
echo "Zip '${zip}' contains file '${jdf}.fdf', renamed to '${newzip}'"
mv ${zip} ${newzip}
done
exit
Zip files:
$ ls *.zip
m1.zip m2.zip m3.zip
$ unzip -l -qq m1.zip
172 05-10-2010 21:27 mark_1a.txt
1285 05-10-2010 21:27 mark_1b.txt
0 05-10-2010 21:27 12345678.jdf
$ unzip -l -qq m2.zip
216 05-10-2010 21:30 mark_2a.txt
1562 05-10-2010 21:31 mark_2b.txt
0 05-10-2010 21:27 87654321.jdf
$ unzip -l -qq m3.zip
268 05-10-2010 21:50 mark_3.txt
$
Script execution :
$ ./mark.sh
Zip 'm1.zip' contains file '12345678.fdf', renamed to '12345678.zip'
Zip 'm2.zip' contains file '87654321.fdf', renamed to '87654321.zip'
Invalid zip file 'm3.zip', jdf file not found
$
New zip files :
$ ls *.zip
12345678.zip 87654321.zip m3.zip
$ unzip -l -qq 12345678.zip
172 05-10-2010 21:27 mark_1a.txt
1285 05-10-2010 21:27 mark_1b.txt
0 05-10-2010 21:27 12345678.jdf
$ unzip -l -qq 87654321.zip
216 05-10-2010 21:30 mark_2a.txt
1562 05-10-2010 21:31 mark_2b.txt
0 05-10-2010 21:27 87654321.jdf
$ unzip -l -qq m3.zip
268 05-10-2010 21:50 mark_3.txt
$
Jean-Pierre.
Jean-Pierre,
wow, what a fast reaction. I really appreciate your efforts. I still try to figure out the sed command. Unfortunatley it doesn't work for me.
My output of unzip -l -qq is like this:
Archive: ./test.zip
Length Date Time Name
-------- ---- ---- ----
44145 12-29-09 11:33 01090230.jdf
-------- -------
44145 1 file
sed -n 's/.*:[0-9]\+[[:space:]]*\([0-9]\{8\}\)\.jdf$/\1/p'
comes up empty.: Invalid zip file 'test.zip', jdf file not found
BTW, what happens when the *.jdf is stored in a directory tree like this:
42262 04-16-10 11:13 System/Connector/JDF-Import/01100850.jdf
0 04-16-10 11:47 System/Connector/JDF-Import/
0 04-16-10 11:13 System/Connector/
0 04-16-10 11:13 System/ContentHotfolder/Saved/
0 04-16-10 11:13 System/ContentHotfolder/
0 04-16-10 11:13 System/Documents/Current/
Mark
The folllowing new version of the script handles file names with directory specification. The output of your unzip command isn't a problem.
for zip in *.zip
do
jdf=$( unzip -l -qq "${zip}" |
sed -n 's=.*[/ ]\([0-9]\{8\}\)\.jdf[ ]*$=\1=p' )
if [ -z "${jdf}" ]
then
echo "Invalid zip file '${zip}', jdf file not found"
continue
fi
newzip=${jdf}.zip
echo "Zip '${zip}' contains file '${jdf}.jdf', renamed to '${newzip}'"
mv ${zip} ${newzip}
done
exit
If the sed command doesn't work, try this modification :
jdf=$( unzip -l -qq "${zip}" |
awk -F'[ /]' '{ if (sub(/\.jdf$/, "", $NF)) print $NF}' )
Zip files :
$ ls *.zip
m1.zip m2.zip m3.zip
$ unzip -l -qq m1.zip
0 05-11-10 09:02 dir1/
4 05-11-10 09:02 dir1/12345678.jdf
467 05-11-10 09:01 dir1/mark_1a.txt
24 05-11-10 09:02 dir1/mark_1b.txt
$ unzip -l -qq m2.zip
0 05-11-10 09:21 dir2/
37 05-11-10 09:21 dir2/87654321.jdf
29 05-11-10 09:21 dir2/mark_2a.txt
770 05-11-10 09:21 dir2/mark_2b.txt
$ unzip -l -qq m3.zip
0 05-11-10 09:28 dir3/
125 05-11-10 09:28 dir3/mark_3a.txt
$
Script execution :
$ ./mark2.sh
Zip 'm1.zip' contains file '12345678.jdf', renamed to '12345678.zip'
Zip 'm2.zip' contains file '87654321.jdf', renamed to '87654321.zip'
Invalid zip file 'm3.zip', jdf file not found
$
New zip files :
$ ls *.zip
12345678.zip 87654321.zip m3.zip
$ unzip -l -qq 12345678.zip
0 05-11-10 09:02 dir1/
4 05-11-10 09:02 dir1/12345678.jdf
467 05-11-10 09:01 dir1/mark_1a.txt
24 05-11-10 09:02 dir1/mark_1b.txt
$ unzip -l -qq 87654321.zip
0 05-11-10 09:21 dir2/
37 05-11-10 09:21 dir2/87654321.jdf
29 05-11-10 09:21 dir2/mark_2a.txt
770 05-11-10 09:21 dir2/mark_2b.txt
$ unzip -l -qq m3.zip
0 05-11-10 09:28 dir3/
125 05-11-10 09:28 dir3/mark_3a.txt
$
Jean-Pierre.
Thank you Jean-Pierre,
that's the solution. I should have described the problem better in the first place.
Mark