RE: [PHP] Difficulty with SQL LIKE clause
| From: | Nathan Tobik | Date: | Fri, 25 Feb 2005 21:22:56 +0000 |
| Subject: | RE: [PHP] Difficulty with SQL LIKE clause | ||
| Groups: | php.general | ||
| Request: | Send a blank email to php-general+get-209452@lists.php.net to get a copy of this message | ||
Not exactly clear what database you're using. In SQL Server the syntax
is
SELECT [options]
FROM [table]
WHERE [field] LIKE '%whatever value you're looking for%'
The % means any string of characters, so it doesn't have to be exact
Best of luck
Nate Tobik
(412)661-5700 x206
VigilantMinds
-----Original Message-----
From: David Freedman [mailto:eatc7402@comcast.net]
Sent: Friday, February 25, 2005 3:28 PM
To: php-general@lists.php.net
Subject: [PHP] Difficulty with SQL LIKE clause
When I use this query in PHP it works, and I get all things with the
YEAR of
1977, as I expected.
$query= "SELECT * FROM my_table WHERE Year LIKE 1977 ";
But, when I use this query it does not work.
$query= "SELECT * FROM my_table WHERE Year LIKE 197* ";
I thought I should get the result of ALL things with the years in the
1970's
(1970-1979). But the asterisk in the LIKE clause does not seem to work.
I have tried various syntax's to try to get the 197* recognized but
without
any luck.
Can the '*' be used? What am I doing wrong.
David F.
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