Shell Programming and Scripting

Hi ,
I am checking whether the file is gz file or not. If it's gz file, i need to unzip it.

My code is as below,

Code:

inpfile=$@
echo $inpfile
file1=`echo $inpfile | cut -d "." -f2`
echo &file1
if [ file1="gz" ] 
then
	gunzip $inpfile
	echo "gunzipped"
fi

when i tried to execute, i got the below response,

Code:

$ . ./script.sh File.gz
File.gz

bash: file1: command not found
gunzipped

Eventhough the file is unzipped, i need to know why the file1 value is not getting printed? Pls solve my above query
Thanks

Try this

Code:

$cat unzip.sh
inpfile=$@
echo $inpfile
file1=`echo $inpfile | cut -d "." -f2`
echo $file1
if [ "$file1" = "zip" ]
then
        unzip $inpfile
        echo "unzipped"
fi
$
$sh unzip.sh no.zip
no.zip
zip
Archive:  no.zip
  inflating: no1
  inflating: no2
  inflating: no3
unzipped

It works perfectly..... :-) thanks a lot

I realize you've already solved your problem, but I share this in hopes that it may be helpful anyway.

The problem with your original code is in:

Code:

echo &file1

The '&' terminates the echo command and causes it to run in the background. The shell then tries to execute the command named "file1". I assume what you were going for was "echo $file1", to echo the filename.

In this case, if you want to be verbose, instead of manually echoing filenames and the like, you can just use gunzip's -v option, which, in addition to the filename, provides information on compression efficiency.

Also, there is no need to bother with an external tool such as cut when you can use the case statement's inherent pattern matching.

Code:

case $1 in
    *.gz) gunzip -v "$1";;
esac

Regards,
Alister