Extract zip files contents and rename the directory dynamically

Question

I have an application zip file created using Play Framework. It create the zip file with name A-1.0.zip. This zip file contains the directory with name A-1.0. (1.0 changes according to the version)

I wanted to extract the zip file and rename the folder from A-1.0 to A. So that my application init.d script finds the directory to start the application. This shuld be done dynamically using shell script.

Is there a way where i can extract all the zip files into A folder instead of extracting into A-1.0 and renaming?? Please help!

The following is what I tried....

unzip A-1.0.zip -d ~/A

(I know that it is very dumb of me to do this !!)

This extracted the file into ~/A/A-1.0/[contents]

I need to extract all the [contents] into ~/A instead of ~/A/A-1.0/. I dunno how to do this using command line.....

My init.d script searched for ~/A/bin/A -Dhttp.port=6565 -Dconfig.file=~/A/conf/application.conf to start the Play! application.

To make this script working, I extract all into A-1.0/ then I rename with mv ~/A-1.0 ~/A manually.

Answer 1

I didn't find any specific unzip option to perform this automatically, but managed to achieve this goal by creating a temporary symbolic link in order to artificially redirect the extracted files this way

ln -s A A-1.0
unzip A-1.0.zip
rm A-1.0

Answer 2

From the unzip man page it boils down to:

unzip A-1.0.zip 'A-1.0/*' -d /the/output/dir
      ^         ^
      |         |
      |         +- files to extract (note the quotes: unzip shall parse the wildcard instd of sh)
      +- The archive