Shell Programming and Scripting

Hi,
I need help in unziping input file through shell script.
I had written script, which checks for input file extention.
If Extension is "zip" or "gz", then I want to do unzip/uncompress that file.
Caould you please let me know that, How to unzip a file through shell script (ksh).

Thanks in advance.

Code:

gzip -d  # uncompress gz file
unzip # unzip zip file

Thanks for the reply.

i had written as,

if [ $EXTN == "zip" ]; then
unzip $FILE
elif [ $EXTN == "gz" ]; then
gzip -d $FILE
elif [ $EXTN == "CSV" ]; then
// doing lots of things
// like parsing file line by line.

I just want to ask one question that,
I need to work on CSV file.If we unzip/uncrompress the file will it be stored in $FILE as extension is "CSV"?

no, it will generate a new file.

for example, a file named as abc.gz, after gzip -d abc.gz, you will get a file named: abc

sometimes, the original file will be deleted automatically.

But if I zip a .CSV file then what happen. After extraction it should be abc.CSV right?

So unzip $FILE will extract file into .CSV, so this changed name should I get into the $FILE or I need to assign it to variable like inputfile=`unzip $FILE` ?

normally gzip will keep the orig file name in new filename, so is the file is abc.csv, after gzip, the file name will be abc.csv.gz, then when uncompress it, the file name will be back to abc.csv

In this case, if you manually rename the gz file from abc.csv.gz to abc.xyz.gz, and uncompress it, you will get file name as abc.xyz

but zip command not, you need provide the dest filename when zip it, after unzip, the orig file name will be always restored.

Thanks a lot

Should I write FILE = unzip $FILE? will it work to store file name in same variable?