Question

Batch Mp3 Converter Shell Script

Asked by: bootforce

Hai Experts

Im Using Linux Server CentOs5.3

I Looking For A Shell Script To Converting Mp3 126kbs bitrate to 64 Kbs......

It Should Convert Folder And Sub Folder Where Mp3 Is There....Batch Convertion

Im Newbie To Linux....SO Please Explain Me Well....How To Do It....

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Asked On
2009-07-05 at 05:16:05ID24544760
Tags

Batch Mp3 Converter Shell Script

Topics

Linux Administration

,

Linux Programming

Participating Experts
2
Points
500
Comments
2

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Answers

 

by: elf_binPosted on 2009-07-07 at 01:23:35ID: 24792313

Hi,

Use the lame encoder.  The command you need is:
lame --mp3input -b 256 <file_in.mp3> <file_out.mp3>
So from the command line you would cd to the directory where the files are and type:
for FILE in *
do
   lame --mp3input -b 256 "$FILE" <path_to_output_directory>/"$FILE"
done
What this means is for every file located in this directory, pass the file-name to lame, using the variable FILE.  We place the variable in quotes to "escape" the spaces, so that the shell does not recognize the spaces in the file-names as additional parameters.

So now to place this in a shell script, create a file with the following:
#!/bin/bash

INPUT_DIRECTORY=$1
OUTPUT_DIRECTORY=$2

for FILE in $INPUT_DIRECTORY/*
do
  lame --mp3input -b 256 "$FILE" OUTPUT_DIRECTORY/"$FILE"
done

This file is a shell script that starts off identifying the shell it'll use (#!/bin/bash) - this is mandatory for shell scripts and Linux supports many, many different shells.
Next, we take the parameters passed to the shell script, the first one ($1) being the source directory that you user has provided on the command line and assign it to the local variable INPUT_DIRECTORY (I do this for clarity).  We then do the same for loop as we did before on the command line.  There's a lot of things I've left out, such as checking that the user did provide arguments and that they are legal arguments (i.e.: real directories) and that they are readable and/or writable, or if the user swaps the location of the read and write positional arguments.  Another thing to note is that if the user provides an argument such as /home/me/srcmp3/ (note the tail ending /) that would expand in the shell to for FILE in /home/me/srcmp3//* which is probably a bad idea.  Another problem could be where there is a space in the directory name(s), we could just place a quote around that, but the user could pass the arguments escaping the space already and the shell would interpret as a literal (with the escaping loosing it's special meaning).  Like I said, I've kept this *very* simple.

Anyway, once you've created the file, you need to make it executable, which you can do by typing chmod u+x <name_of_file>.  So say we created our file as convMp3.sh we would type
chmod u+x convMp3.sh

To execute the file we would call the file (by the full path - there are other ways of doing it but I am trying to keep it simple) and pass the arguments as required, such as:
/home/me/convMp3.sh /home/me/srcmp3 /home/me/dstnmp3
This would call the script convMp3.sh located in /home/me/ with the arguments input directory: /home/me/srcmp3 and output directory: /home/me/dstnmp3

The best place (in my opinion) is to start at the bash how to, which can be accessed here: http://tldp.org/HOWTO/Bash-Prog-Intro-HOWTO.html

Hope this helps

 

by: notmistakenPosted on 2009-07-08 at 08:13:03ID: 24804641

another tool you can use is ffmpeg. The command to convert one file is:
ffmepg -i in.mp3 -ab 64000 out.mp3

so to convert a whole directory tree of mp3 files, the command is:

find /inputdir -name '*.mp3' -exec ffmpeg -i {} -ab 64000 /outputdir/{} \;

20120131-EE-VQP-002

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