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" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
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" <[EMAIL PROTECTED]>
> To: <[EMAIL PROTECTED]>
> 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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