wav2mp3

ncsuapex · 10-03-2006, 11:46 PM

I found a good script that converts wav files into mp3s. But I would like to know how to modify it so that it does each wav file one by one in a directory until they are all done. Right now I am having to copy each wav file to a new directory and running the script. I will post the script below and if someone could help I would appreciate it.

Code:

#!/bin/bash
#
# wav2mp3
#
for i in *.wav; do
    #out=$(ls $i | sed -e 's/.wav//g')
    #out=$(echo $i | sed -e 's/.wav$//')
    #lame -h -b 96 "$i" "$out.mp3"
    lame -h -b 96 "$i" "${i%.wav}.mp3"
done

Thanks!

j

pete1234 · 10-04-2006, 12:43 AM

You just about had it figured out. This seems to work for me:

Code:

#!/bin/bash -
# wav2mp3

FILES=`ls *.wav`

for i in $FILES; do
   OUT=`echo $i | sed -e 's/.wav//'`
   lame -h -b 96 $i $OUT.mp3
done

jens · 10-04-2006, 12:50 AM

"lame -h -b 96" will not give you good quality though.
Consider using a higher bitrate.

pda_h4x0r · 10-04-2006, 01:02 AM

Try this:

Code:

#!/bin/bash
#
# wav2mp3
#
# The directory to export .mp3 files to is passed as argument 1
# The script should be invoked as wav2mp3 new_dir/
# If there is no argument, a default value is used.
# If the directory doesn't exist, then it is created if the user has enough permissions
#
EXPORT_DIR=
if [ $# -eq 0 ]; then
EXPORT_DIR="wav2mp3_out"
else
EXPORT_DIR="$1"
fi

if ! [ -d $EXPORT_DIR ]; then
mkdir -p $EXPORT_DIR
fi

# now EXPORT_DIR holds the directory to which to export the new files

for i in *.wav; do
    lame -h -b 96 "$i" "$EXPORT_DIR/$i.mp3"
done

ncsuapex · 10-04-2006, 01:10 AM

Thanks Pete1234 that worked perfectly!

pete1234 · 10-04-2006, 01:12 AM

You're welcome.