[SOLVED] running a zip command in a bash script

onuhwt · 07-07-2010, 12:00 AM

Hello all, I'm very new to writing bash scripts (wrote my first one this Sunday) and I'm trying to zip a group of files. It has to be in zip format so alternatives like tar won't work here. I have my script in a folder which has a bunch of Sub-directories in the format "Lab\ 3" "Lab\ 5" "Lab\ 6" etc

What I'd like to be able to do is take all of the files (just the contents not the folder itself) in the "Lab\ 3" Folder and put them in Lab3.zip. I'm really close but no matter what I try I keep getting the folders put into the zip file instead of the Folders contents

Code:

#!/bin/bash

LabNum=(3 5 6)

for ARG in ${LabNum[@]};
    do
	` zip -r Lab$ARG Lab\ $ARG/*`
    done

Thanks in advance

Elv13 · 07-07-2010, 12:11 AM

you could add a second level of loop and use zip -g (grow/append) to add each files one after the other. You can "cd" into the directory just before and use

Code:

for FILE in `ls`;do
  zip -g Lab$ARG.zip $FILE
done

I did not tested (you may have to create the archive before?) but it might help you.

onuhwt · 07-07-2010, 12:45 AM

Quote:

Originally Posted by Elv13

you could add a second level of loop and use zip -g (grow/append) to add each files one after the other. You can "cd" into the directory just before and use

Code:

for FILE in `ls`;do
  zip -g Lab$ARG.zip $FILE
done

I did not tested (you may have to create the archive before?) but it might help you.

That did exactly what I wanted Thank you very much Elv13. If anyone else stumbles on this thread looking for what I did I'll include the code below. Thanks again

Code:

#!/bin/bash

LabNum=(3 5 6)

for ARG in ${LabNum[@]};
    do
	#zip -r Lab$ARG "Lab\ $ARG/*"
	cd Lab\ $ARG	
	for FILE in `ls`;do
	  zip -g ../Lab$ARG.zip $FILE
	done
        cd ..


    done