I have a very long file containing file paths, one on each line. I would like to retrieve a list of all directories listed which are only 1-level deep. As such, I would like to extract only those lines which have a single / in them:
I want: ./somedir
I don't want: ./somedir/someotherdir
Can I do this with grep and/or regular expressions?
2 Answers
Sure. That's pretty easy:
find . | grep -P '^\./[^/]*$'... where obviously the 'find' command I'm using just to illustrate what you described. The regex works as follows:
^and$are anchors specifying (respectively) the beginning & end of the line. All content must be between those two characters\./is a literal period followed by a literal forward-slash[^]is a character-class that disallows anything after the^and before the]- The
*after[^/]says to allow zero or more occurrences of that character class
You could also do it like this:
dosomething | grep -P '^[^/]*/[^/]*$'This will allow only a single / on the line, in any position on the line (even first or last character).
@Brians solution works, this is a more generic one that doesn't need a ./ at the beginning:
grep -P '^[^/]*/[^/]*$'For example:
/aa/somedir
test/somedir/someotherdir
somedir/otherdir
--> somedir/otherdir
