Finally I decided to use:
--
SELECT id FROM mytable WHERE MATCH(firstname, lastname, comments)
AGAINST ('+"johnie"' IN BOOLEAN MODE) ORDER BY firstname, lastname
It works (except with acute vowel words in UTF8).
Thank you very much.
On 9/20/07, Baron Schwartz <[EMAIL PROTECTED]> wrot
860.674.8796 / FAX: 860.674.8341
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> -Original Message-
> From: thomas Armstrong [mailto:[EMAIL PROTECTED]
> Sent: Thursday, September 20, 2007 3:52 AM
> To: Michael Dykman
> Cc: mysql@lists.mysql.com
> Subject
thomas Armstrong wrote:
If you need something more complicated, such as only ignoring "(", then you
need to get more complicated. You might even need a regular expression.
I'm to browse:
http://dev.mysql.com/doc/refman/5.0/en/regexp.html
http://www.wellho.net/regex/mysql.html
You know, you don
> If you need something more complicated, such as only ignoring "(", then you
> need to get more complicated. You might even need a regular expression.
I'm to browse:
http://dev.mysql.com/doc/refman/5.0/en/regexp.html
http://www.wellho.net/regex/mysql.html
Thank you very much!
>
> Regards,
>
> J
Thank you Chris for your answer.
On 9/19/07, Chris Sansom <[EMAIL PROTECTED]> wrote:
>
> Well I'm hardly the world's greatest expert, but I'm curious as to
> why you're always separating '%' from 'johnie' with a space, because
> that way it will only find Johnie if he has a space before or after
>
Thank you Michael for your answer.
On 9/19/07, Michael Dykman <[EMAIL PROTECTED]> wrote:
> The whitespace counts... try LIKE '%johnie%' (or better '_johnie_' ..
> the underscorematches any single character).
I want to match '(johnie)' and not 'johnies' or 'aljohnier', what it's
the query does wi
You are putting a space between "johnie" and the % wildcards. That space is
not ignored, it is part of the pattern. LIKE "%johnie%" will find every
occurrence of "johnie" no matter what surrounds it.
If you need something more complicated, such as only ignoring "(", then you
need to get more compl
At 19:34 +0200 19/9/07, thomas Armstrong wrote:
I've got this table in mySQL:
item 1:
-- firstname: John (Johnie)
-- phone: 555-600-200
item 2:
-- firstname: Peter
-- phone: 555-300-400
I created this SQL query to find 'johnie':
SELECT friends.id FROM friends WHERE ((f