- Original Message -
From: "David Freedman" <[EMAIL PROTECTED]>
To:
Sent: Friday, February 25, 2005 11:27 PM
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.
>
;
> /G
> @varupiraten.se
>
> -Ursprungligt meddelande-
> Från: David Freedman [mailto:[EMAIL PROTECTED]
> Skickat: den 25 februari 2005 21:28
> Till: php-general@lists.php.net
> Ämne: [PHP] Difficulty with SQL LIKE clause
>
>
> When I use this query in PHP it
You should use a _ instead.
Please have a look at the mySQL manual:
http://dev.mysql.com/doc/mysql/en/string-comparison-functions.html
Christian
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 ";
L PROTECTED]
Skickat: den 25 februari 2005 21:28
Till: php-general@lists.php.net
Ämne: [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, whe
Hello David,
Friday, February 25, 2005, 1:27:54 PM, you wrote:
D> Can the '*' be used? What am I doing wrong.
It's the % symbol, not the *.
--
Leif (TB lists moderator and fellow end user).
Using The Bat! 3.0.2.3 Rush under Windows XP 5.1
Build 2600 Service Pack 2 on a Pentium 4 2GHz with 51
1-5700 x206
VigilantMinds
-Original Message-
From: David Freedman [mailto:[EMAIL PROTECTED]
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
197
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
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