Re: MySQL error
| From: | Brian Rue | Date: | Tue, 10 Oct 2000 01:50:53 +0000 |
| Subject: | Re: MySQL error | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-19274@lists.php.net to get a copy of this message | ||
OOPS! I left out part of that code snipet. Here it is again:
$db = mysql_connect("localhost","user","pass");
mysql_select_db("database_name",$db);
$query="SELECT * FROM games WHERE
genre=$gameone_genre&number=$gameone_number";
$info=mysql_query($query,$db);
$row=mysql_fetch_row($info);
----- Original Message -----
From: "Brian Rue" <dwjw@quixnet.net>
To: <php-general@lists.php.net>
Sent: Monday, October 09, 2000 6:45 PM
Subject: [PHP] MySQL error
Hi,
I'm getting the following error:
Warning: Supplied argument is not a valid MySQL result resource .....
Here's the code:
$query="SELECT * FROM games WHERE
genre=$gameone_genre&number=$gameone_number";
$info=mysql_query($query,$db);
$row=mysql_fetch_row($info);
The last line is the one causing the error. What am I doing wrong?
Thanks
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