LinuxQuestions.org - trivial: bash, local variable assignment

I made a short script to help me learn bash. The function in question is:

Code:

watchVideo()        {

# Create alphabetical array of video files by recursively scanning $HOME.

video=$(find $HOME/* -name *.flv |sort; find $HOME/* -name *.mpg |sort; find $HOME/* -name *.avi |sort)

PS3=">>> Choose; hit Enter. > "

echo ""

echo " 0) Return to Main Menu."

# Display all video files.

select choice2 in $video

do

    if [ $choice2 ==0 ]; then

        hello # If $USER chooses 0, return to Main Menu.

    else :

        exec mplayer $choice2

        break

    fi

done



exit 0

}

I just want 1 command for the video variable: find's -name parameter accepting a list of filenames prefaced with * instead of the same command 3 times. I've tried

Code:

find $HOME/* -name [*.mpg,*.flv,*.avi] | sort

and many variations, but to no avail. How can this be done? BTW, this is on Ubuntu 8.10.

*****
Update: I tried "find -name $HOME/*/[*.mpg, *.flv, *.avi] | sort" which gave:

Code:

find: warning: Unix filenames usually don't contain slashes (though pathnames do).  That means that '-name `/home/agrestic/*/[*.mpg,'' will probably evaluate to false all the time on this system.  You might find the '-wholename' test more useful, or perhaps '-samefile'.

And "find -wholename $HOME/*/[*.mpg, *.flv, *.avi] | sort" which gives:

Code:

find: paths must precede expression: *.flv,

So still no clue. :)

TIA.