bash script to sort mp3 bitrates above and below 128k

Damoek · 03-25-2008, 09:10 AM

I posted a question to Ubuntuforums.org yesterday about how one would go about authoring a script to sort several gigs of mp3's into two top level directories (better128) (less128) while preserving current parent directory structure.

I found out that in order to be able to read bitrate from mp3 files in bash you need a tool called mp3info

Code:

mp3info -r m -p %r file.mp3

will output the bitrate 128 on the resulting line

a friendly member showed me a sample script which i modified slightly with my own directory structures that would do almost exactly what i wanted

Code:

#!/bin/bash
music="/media/sda2/Music"
good="/media/sda2/Music/better128"
bad="/media/sda2/Music/less128"

mkdir "$good" "$bad"

find "$music" -type f -name '*.mp3' -print0 | while read -d  $'/0' file; do
	if [ "$(mp3info -r m -p %r  "$file")" -ls 129 ]; then
		cp --parents "$file" "$bad"
	else
		cp --parents "$file" "$good"
	fi
done

when I run this script I get an error

Code:

sh goodbad.sh
read: 12: Illegal option -d

Now, I'm very new to scripting, so most of this is theory to me, but from what I see the find command is searching my music directory structure for files of type 'file' and named '.mp3' and then prints a null character at the end, which the while loop is looking for, once that happens all that data is scanned by mp3info and then sorted into directories based on the -lt statement

what i don't understand is why the delimiter -d command is giving an illegal output, or what the syntax error is in the loop statement. Can someone with more knowledge and expertise advise me if this is the correct way to accomplish what i want, and tell me where i'm going wrong in the syntax?

I appreciate it!

makyo · 03-25-2008, 09:39 AM

Hi.

Without looking in detail at your script, the easiest thing to try is to run it as:

Code:

bash goodbad.sh

I don't know anything about the data with which you are working, but the $'/0' in the read looks odd to me -- are you sure that that is correct? ... cheers, makyo

Damoek · 03-25-2008, 09:59 AM

i think sh goodbad.sh is the same as bash goodbad.sh

and no... haha i'm absolutely not sure that the $'/0' is correct

weibullguy · 03-25-2008, 10:15 AM

Are you using Bash and not a different shell? Have you tried removing the -d option to read? The error is telling you that your shell's version of read doesn't recognize the -d option.

I assume the $'/0' is supposed to represent the null character that is the delimiter being passed to -d. It should use a backward slash (\) IIRC.