current community

your communities

more stack exchange communities

company blog
Stack Exchange Inbox Reputation and Badges
tour help
Stack Overflow is a community of 4.7 million programmers, just like you, helping each other.

Join them; it only takes a minute:

Sign up
Join the Stack Overflow community to:
  1. Ask programming questions
  2. Answer and help your peers
  3. Get recognized for your expertise
up vote 0 down vote favorite

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)

share|improve this question
up vote 0 down vote

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.

share|improve this answer
up vote 0 down vote

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.