Batch renaming using shell script

Question

I have a folder with files named as

input (1).txt
input (2).txt
input (3).txt
...
input (207).txt

How do I rename them to

input_1.in
input_2.in
input_3.in
...
input_207.in

I am trying this

for f in *.txt ; do mv $f `echo $f | sed -e 's/input\ (\(\d*\))\.txt/input_\1.in/'` ; done

But it gives me

mv: target `(100).txt' is not a directory
mv: target `(101).txt' is not a directory
mv: target `(102).txt' is not a directory
...

Where did I go wrong?

---

I have put in the quotes now, but I get this now

mv: `input (90).txt' and `input (90).txt' are the same file

It is somehow trying to rename the file to the same name. How is that happening?

Answer 1

You're forgetting to quote your arguments.

... mv "$f" "$(echo "$f" | ... )" ; done

Answer 2

That is because bash for split the element with space ' ' so you are commanding it to move 'input' to '(1)'.

The way to solve this is to tell bash to split by new line using IFS variable.

Like this:

IFS=$'\n'

Then do your command.

However, I suggest you to use find to do this instead using -exec command.

For example:

find *.txt -exec mv "{}" `echo "{}" | sed -e 's/input\ (\([0-9]*\))\.txt/input_\1.in/'` \;

NOTE: I write this from memory and I did test this so let try and adjust it.

Hope this helps.

Answer 3

no need to call external commands

#!/bin/bash
shopt -s nullglob
shopt -s extglob
for file in *.txt
do
  newfile="${file//[)]/}"
  newfile="${file// [(]/_}"
  mv "$file" "${newfile%.txt}.in"
done

Answer 4

for f in *.txt ; do mv "$f" `echo $f | sed -e 's/input\ (\(\d*\))\.txt/input_\1.in/'` ; done

Answer 5

If you have GNU Parallel http://www.gnu.org/software/parallel/ installed you can do this:

seq 1 207 | parallel -q mv 'input ({}).txt' input_{}.in

Watch the intro video for GNU Parallel to learn more: http://www.youtube.com/watch?v=OpaiGYxkSuQ

Answer 6

As you've already fixed, you need to quote the $f argument to mv.

As to your second problem, sed doesn't support \d. You could use [0-9] instead.