How to use a variable name in a Perl Regex bash script

Question

I am trying to write a bash shell script to rename a bunch of photos to my own numbering system. All images filenames are like "IMG_0000.JPG" and I can get the script to match and rename(overwrite) all the photos with the following Perl-regex code:

#!/bin/bash
rename -f 's/\w{4}\d{4}.JPG/replacement.jpg/' *.JPG;

But when I try to use a variable as the name of the replacement, as I keep seeing on other posts here and elsewhere on the internet, nothing happens:

#!/bin/bash
$replacement = "000.jpg";
rename -f 's/\w{4}\d{4}.JPG/$replacement/' *.JPG;

How can I get such a variable to work correctly in my bash script? (NOTE: I am not looking to simply strip the "IMG_" from the filename)

perreal · Accepted Answer · 2013-06-09 22:21:39Z

1

Take the replacement out of single quotes:

#!/bin/bash
$replacement="000.jpg"
rename -f 's/\w{4}\d{4}.JPG/'$replacement'/' *.JPG

Bash does not inspect single quoted strings for interpolation.

answered Jun 9, 2013 at 22:21

perreal

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speakr · Accepted Answer · 2013-06-09 22:33:10Z

0

Using double quotes and correct variable assignment:

#!/bin/bash
replacement="000.jpg"
rename -f "s/\w{4}\d{4}\.JPG/$replacement/" *.JPG

Note that this can cause trouble, e.g. when renaming two files with names like IMG_0001.JPG and FOO_9352.JPG: The first file will be renamed to 000.jpg, then the second file will also be renamed to 000.jpg, overwriting the first.

edited Jun 9, 2013 at 22:33

answered Jun 9, 2013 at 22:21

speakr

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