How to get only names from find command without path
Use find trunk/messages/ -name "*.po" -exec basename {} .po \;
Example and explanations:
Create some test files:
$ touch test1.po
$ touch test2.po
$ find . -name "*.po" -print
./test1.po
./test2.po
Ok, files get found, including path.
For each result execute basename
, and strip the .po part of the name
$ find . -name "*.po" -exec basename \{} .po \;
test1
test2
You can use -execdir
parameter which would print the file without path, e.g.:
find . -name "*.po" -execdir echo {} ';'
Files without extensions:
find . -name "*.txt" -execdir basename {} .po ';'
Note: Since it's not POSIX, BSD find
will print clean filenames, however using GNU find
will print extra ./
.
See: Why GNU find -execdir command behave differently than BSD find?
Based on a command I found here: http://www.unixcl.com/2009/08/remove-path-from-find-result-in-bash.html
You could do this using sed
(which executed faster for me, than the basename
approach):
find trunk/messages/ -name "*.po" | sed 's!.*/!!' | sed 's!.po!!'