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regex to select last part of a path

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# 1  
Old 07-31-2009
regex to select last part of a path
Hi all,
I am learning the use of regular expression and I would like to know which regex can be used to select only the last part of a directory path name.

Something like:
/dir1/dir2/dir2
and I want to select the last /dir2 where dir2 can be any kind of string.

Thanks a lot for your help. Cheers,
Elric
# 2  
Old 07-31-2009
This isn't regex (at least I don't think it is) but you could accomplish it like this:

Code:
$ cat test.sh
#!/bin/bash
string="/dir1/dir2/dir2"
echo $string
string=${string##*/}
echo $string

$ ./test.sh
/dir1/dir2/dir2
dir2

Actually, after I posted this I reread your "I am learning the use of regular expression" line. My post isn't relevant
# 3  
Old 07-31-2009
In Perl it would be
Code:
$string = "/dir1/dir2/dir2";
$string =~ m|(/.+?)$|;
print $1;

# 4  
Old 07-31-2009
thanks to everybody but I think there is a problem because the output printed by the perl script is:

/dir1/dir2/dir2

I think it is due to the ".+" that includes the / character so it matches all the string starting from the first /dir1. Or maybe I did something wrong in my script... I am checking.
However, how can I veto the / from .+ ?
# 5  
Old 07-31-2009
Hm, you're right, should have checked that. Change that from (/.+?) to (/[^/]+?) and it should be fine.
# 6  
Old 07-31-2009
Quote:
Originally Posted by mglenney
This isn't regex (at least I don't think it is)

You're right; the shell uses filename expansion patterns, not regular expressions.
Quote:
but you could accomplish it like this:

Code:
$ cat test.sh
#!/bin/bash
string="/dir1/dir2/dir2"
echo $string
string=${string##*/}
echo $string

$ ./test.sh
/dir1/dir2/dir2
dir2

Actually, after I posted this I reread your "I am learning the use of regular expression" line. My post isn't relevant
# 7  
Old 08-01-2009
I works perfectly, thanks to everyone ! Smilie
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