I have files in a directory. The file names are constructed using a date timestamp notation, the file names are of the form:
name_YYMM.csv
I want to rename all such files using the following naming convention:
name_YYYYMM.csv
I am not sure if I can use a command line utility like grep to do this, or if I need to write a bash script to do this.
Any help on how to solve this prob will be appreciated.
I am running on Ubuntu
A very simple way to do it
ls name* | while read a; do mv $a `echo $a | sed s/name_/name_20/`; done
or
ls name* | while read a; do mv $a $(echo $a | sed s/name_/name_20/); done
(since use of back-quotes is depreciated (or hard to read anyway))
it lists all files matching name_ then for each file it finds it replaces name_ with name_20. You can change 20 to 19 if you are working with files from the last millennium.
name_ for _ in the above example and replace ls name* with ls *.csv
for loop without ls. Variables that contain filenames (and, similarly, command substitution that processes filenames) should always be quoted.
Dec 10, 2010 at 16:48
It is likely that you have a program or Perl script called rename on your system.
For the Perl script version:
rename 's/_/_20/' *_*.csv
For the util-linux-ng version:
rename _ _20 *_*.csv
Or, using Bash:
for f in name*; do mv "$f" "${f/_/_20}"; done
In zsh:
autoload zmv
zmv '(*_[0-9][0-9])([0-9][0-9].csv)' '${1}20${2}'
You can make the pattern less precise, e.g. '(*_??)(??.csv)' or even '(*)(??).csv' if there's no risk of other files matching.
If you don't want to use zsh, see Dennis Williamson's answer.
