Re: SQL question
| From: | Muhammed Mamedov | Date: | Thu, 11 Dec 2003 12:06:12 +0000 |
| Subject: | Re: SQL question | ||
| References: | 1 2 | Groups: | php.db |
| Request: | Send a blank email to php-db+get-31992@lists.php.net to get a copy of this message | ||
I am sorry without equal "=" sign
SELECT * FROM x_table WHERE name Like "%part_of_name%"
there are actually three ways to do this:
"%part_of_name%" part_of_name appears anywhere
"%part_of_name" part_of_name appears at the end
"art_of_name%" part_of_name appears at the begining
M.Mamedov
----- Original Message -----
From: "Muhammed Mamedov" <muhammed@ytm.com.tr>
To: <php-db@lists.php.net>
Sent: Thursday, December 11, 2003 2:00 PM
Subject: Re: [PHP-DB] SQL question
> It should be as follows:
> SELECT * FROM x_table WHERE name Like = "%part_of_name%"
>
> Hope this helps,
> M.Mamedov
>
> ----- Original Message -----
> From: "Constantin Brinzoi" <aurel@irecson.ro>
> To: <php-db@lists.php.net>
> Sent: Thursday, December 11, 2003 1:51 PM
> Subject: [PHP-DB] SQL question
>
>
> > I know it is possible to search a database like this:
> >
> > SELECT * FROM x_table WHERE name="%part_of_name%"
> >
> > but I don't know for sure the correct command.
> >
> > Could you tell me the right syntax?
> >
> > TIA
> > Aurel.
> >
> > --
> > PHP Database Mailing List (http://www.php.net/)
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> >
>
>
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