My problem is as follows:

I want to display a list grouped by the domain part of the URL from my webblog of referers. Is this possible using sql?

Example of values from the URL field:
h t t p://www.google.com/search?hl=en&lr=&q=Freeroll+Tracker
h t t p://www.google.com/search?hl=en&lr=&q=poker+tournament+tracker+sit+an d+go
h t t p://www.google.bg/search?hl=bg&q=nordicbet+bonus+code&btnG=Google+%D 0%A2%D1%8A%D1%80%D1%81%D0%B5%D0%BD%D0%B5&meta

Should give the result:
h t t p://www.google.com/ 2 hits
h t t p://www.google.bg/ 1 hit

Hope this makes sense =)

You can use regular expressions in mysql to match a pattern. I came up with the following, but its untested.

SELECT *, url REGEXP "/^https?:\/\/(www\.)?[^\.\/]*\.[^\.\/]{2,4}\/.*$/i" AS match FROM table_name GROUP BY match

Thanks for your response after changing MATCH to another Word it works in some way which means i get a selection atleast, not what I want but something atleast =)

Can you please explain the code a bit so that I can figure out how to use it so I get the response I want.

Thanks for taking the time with this

SELECT *, url REGEXP "/^https?:\/\/(www\.)?[^\.\/]*\.[^\.\/]{2,4}\/.*$/i" AS match FROM table_name GROUP BY match

Select all and url column that matches http (or https)://www. (or no www) blah . com (or other extention between 2 and 4 chars) / case insenstive - name result as match from the table selected and group by our match result

Thanks again.

Forgot to mention in the previous post that the result i get in the MATCH variable is either 0 or Null. What I really want as result is the domain names neatly grouped from the log.

Yea - I guess that it does return a boolean instead of a match, and I don't know how you can use the regexp references in a sql command. I don't have any other suggestions right off hand, sorry.

Thanks for taking the time trying to solve my problem.

Useful post

Couldnt solve my problem just using SQL so I split the string at insertion instead into its own field so now I can retrieve what I want using an ordinary GROUP BY selection =).

Thanks for the help trying to solve my initial problem.