I have multiple folders and each one have around 175 files in it. file names are like
1.wav
2.wav
3.wav
......
175.wav
I have to rename them as
A1.wav
A2.wav
A3.wav
......
A175.wav
In other words I have to append Letters in previous file names.
I am wondering if there is a easy way to do this.
Ubuntu Version is 16.10
User prename command:
$ prename -nv 's/^(.*)$/A$1/' *.wav
1.wav renamed as A1.wav
2.wav renamed as A2.wav
3.wav renamed as A3.wav
The way this reads is simple:
*.wav allows the shell to expand the wild card to list of all files that end with .wav. When the shell runs the full command, the computer will see prename -nv 's/^(.*)$/A$1/' 1.wav 2.wav 3.wav and so on as the actual command. 's/^(.*)$/A$1/' is actually is s/PATTERN/REPLACEMENT regular expression with grouping (.*), which allows us to group the whole name of the file from beginning ^ to end $ and refer to it as $1. Note that -nv switches are for verbose -v and dry-run -n. If you're happy with the test run, remove -n to apply actual renaming.
There are many ways, my own choice would be a 'for' loop:
for f in *.wav ; do mv "$f" "A$f" ; done
This is simple and easily modified for other needs...
Using rename:
rename 's/([0-9]+).wav/A\1.wav/' *.wav
s/SEARCH-FOR/REPLACE-WITH/ within-this-files([0-9]+) holds the number section then we can use it again using \1.A\1.wav: A + (numberic section) + .wav
^(.*)makes the whole name to be treated as$1. So you're literally sticking A at the end of whole name. What you want to do is use extra grouping, make$1be everything before.wavand make.wavbe$2. Like this:prename -nv 's/^(.*)(.wav)$/$1A$2/' *.wavSo here first(.*)will be$1and is everything before.wav. The second group(.wav)is$2– Sergiy Kolodyazhnyy Jun 07 '17 at 17:06