Bash command -cut (oops last message truncated)

[neidorff]

Hi,

This baffles me, but should be very simple.

I have a directory with names of audio files.  I want to convert from one
audio format to another.  No problem there.

I want to create a script to do handle the change of file extension chores.
I came up with this:

#!/bin/bash
FILES=$(ls *.flac | cut -f1 -d\.)
for i in $FILES; do
    echo converting: $i
    flac -sd $i.flac -o  wav/$i.wav
done

The first file is named "01 Introduction.flac"  and the problem that I have
is that the first time I go through the for loop, "i" contains "01" the next
time "i" contains "Introduction".  Since I set the delimiter (with the -d
option in cut) to "." (escaped) why am I not having "i" contain "01
Introduction" the first time through the for loop?  What obvious thing am I
missing here?

Thanks,

Mark
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