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November 24th, 2003, 09:59 PM

Anonymous

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Order By help!

I have a kindof stupid question that would seem very obvious to anybody but me Xx Anyway here we go.. Here is my piece of code:

php Code:

Original - php Code
<?php $db = mysql_connect("localhost", "*", "**"); mysql_select_db("database",$db); $query = mysql_query("SELECT horsename from horsedata ORDER BY points DESC"); $rows = mysql_fetch_array($query); echo "$rows[horsename]"; ?>

 
 
<?php
$db = mysql_connect("localhost", "***", "****"); 
mysql_select_db("database",$db);
$query = mysql_query("SELECT horsename from horsedata ORDER BY points DESC"); 
$rows = mysql_fetch_array($query);
echo "$rows[horsename]";
?>

But I get this error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/arabq/public_html/tophorses.php on line 14

Any ideas?

November 24th, 2003, 11:30 PM

honcho

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Codewalkers Beginner (1000 - 1499 posts)

Join Date: Apr 2007

Location: Cape Cod

Posts: 1,347

Time spent in forums: 3 h 52 m 2 sec
Reputation Power: 3

RE: Order By help!

Looks like your query failed for some reason.
Try this and see where the error is:

php Code:

Original - php Code
$db = mysql_connect("localhost", "*", "**"); if(!$db) exit("Could not connect: ".mysql_error()); if(!mysql_select_db("database",$db)) exit("Could not select DB: ".mysql_error()); $query = mysql_query("SELECT horsename from horsedata ORDER BY points DESC"); if(!$query) exit("Error in query: ".mysql_error()); $rows = mysql_fetch_array($query); echo "$rows[horsename]";

 
$db = mysql_connect("localhost", "***", "****");
if(!$db) exit("Could not connect: ".mysql_error());
 
if(!mysql_select_db("database",$db)) exit("Could not select DB: ".mysql_error());
 
$query = mysql_query("SELECT horsename from horsedata ORDER BY points DESC");
if(!$query) exit("Error in query: ".mysql_error());
 
$rows = mysql_fetch_array($query);
echo "$rows[horsename]";

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