regex to get part of string from end

Question

I have some URLs like

http://wac.0E23.ed2323sdf.net/800E13/www.myexample.gr/

http://wac.0E22.ed2323sdf.net/812E13/www.my1example.gr/

http://wac.0E23.ed2323sdf.net/802E13/www.my2example.gr/

I have to write regular expression to get the site name i.e www.myexample.gr, www.my1example.gr and www.my2example.gr

I wrote an expression to find the last occurrence of / but I am not able to capture the string between second last /.

Any idea how to achieve this?.

Regards,

can you post the regex that you have so far?

EkcenierK
– EkcenierK

2011-09-15 09:30:57 +00:00
Commented Sep 15, 2011 at 9:30 — EkcenierK
– EkcenierK, Commented Sep 15, 2011 at 9:30

Diego Sevilla · Accepted Answer · 2011-09-15 09:32:56Z

3

It depends on what language you're using (perl? elisp?). Usually, that would be extracted by the following:

"/([^/]+)/$"

where I put the regex between ". I do that because most languages use // as a search pattern definition in itself. And, depending on the language, the result would be in the first parenthesis value (in some languages \1, in others $1, in others (match-string 1)...)

answered Sep 15, 2011 at 9:32

Diego Sevilla

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4 Comments

user269867 Over a year ago

I was trying ([^/])+/$....Can you please explain It will help me to understand what I was doing wrong.....

Diego Sevilla Over a year ago

Well, the + has to be inside the parenthesis to match several characters inside the group.

user269867 Over a year ago

you are matching string between / from end right?...can you please tell me [^/] is useful here..

Diego Sevilla Over a year ago

[^/] means a character not /, so what you do is try to get all the characters that are consecutive, and not /, and that are followed by a / and and end of line (or end of string).

Toto · Accepted Answer · 2011-09-15 09:32:46Z

2

How about this expression:

~([^/]+)/$~

answered Sep 15, 2011 at 9:32

Toto

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