CODEWITHSUNDEEP
linuxbashshellzenity
ceechan
ceechan

Reputation: 3

Display file name of all files with certain extension

What I want is for the user to select a file from anywhere using zenity, and the script will detect the file extension, for example (.tar.gz) or (.zip), and perform actions on them accordingly. Here's an example.

#! /bin/bash

FILE=$(zenity --file-selection --title="Select a file to check")
echo "File: $FILE"

if [ "$FILE" = "*.zip" ] 
then
    echo "File that is a .zip found"
    FILENAMENOEXT="${FILE%.*}"
    echo "Filename with extention $FILENAMENOEXT"
    #Perform xx action to $FILE if it is a zip

elif [ "$FILE" = "*.tar.gz" ]
then
echo "File is a .tar.gz found"
FILENAMENOEXT="${FILE%.tar.*}"
echo "Filename with extention $FILENAMENOEXT"
#Perform xx action to $FILE if it is a t.tar.gz

else
    echo "File is neither .zip nor .tar.gz"
fi

echo "test $FILENAMENOEXT"

Upvotes: 0

Views: 94

Answers (1)

Etan Reisner
Etan Reisner

Reputation: 81012

This is almost correct.

You need to use [[ to do pattern matching and quotes disable pattern matching.

So instead of [ "$FILE" = "*.zip" ] you want [[ "$FILE" = *".zip" ]] and instead of [ "$FILE" = "*.tar.gz" ] you want [[ "$FILE" = *".tar.gz" ]].

You could also use a case statement instead of if/elif.

case "$FILE" in
*.zip)
    echo "File that is a .zip found"
    ;;
*.tar.gz)
    echo "File is a .tar.gz found"
    ;;
*)
    echo "File is neither .zip nor .tar.gz"
    ;;
esac

Upvotes: 2

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