My Sql Problem

sp33d1

It echo's out Problems Occured and i also get this error

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/layoutst/public_html/Graphics/category.php on line 51

can anyone tell me what im doing wrong with the sql fetch?


$fetch = mysql_query ( "SELECT * FROM categories LEFT JOIN parents ON ( categoryParentID = parentID ) WHERE categoryPerm = '".$category."' LIMIT 1" );
if ( mysql_num_rows ( $fetch ) == 1 )
{
	$category = mysql_fetch_assoc ( $fetch );
}
else
{
	echo 'Problems Occured! ';
}


$limit = LIMIT;

$fetch = mysql_query ( "SELECT 1 FROM graphics WHERE graphicCategoryID = '".$category['categoryID']."'");
$totalPages = ceil ( mysql_num_rows ( $fetch ) / $limit );

if ( mysql_num_rows ( $fetch ) == 0 )
{
	header ( "Location: /Graphics/" );
}

echo '<center><span class="catlinks"><a href="/Graphics/"><b>Home</b></a> - <a href="/Graphics/'.$category['parentPerm'].'.html"><b>'.$category['parentName'].'</b></a> - <a href="/Graphics/'.$category['parentPerm'].'-'.$category['categoryPerm'].'.html"><b>'.$category['categoryName'].' Graphics</b></a></span></center>';

if ( $page > $totalPages )
{
	$page = $totalPages;
}

$start = ( $page * $limit ) - $limit;

echo '<div align="center">';

$fetch = mysql_query ( "SELECT * FROM graphics WHERE graphicCategoryID = '".$category['categoryID']."' ORDER BY graphicID DESC LIMIT $start, $limit" );
while ( $row = mysql_fetch_assoc ( $fetch ) )  // this is line 51

It echo's out Problems Occured and i also get this error

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/layoutst/public_html/Graphics/category.php on line 51

you see anything wrong with the say im executing the sql fetch?

bradgrafelman

Rather than trying to guess what the problem is, it's easier to first examine where the SQL server thinks the problem is. Normally, I debug SQL queries like so:

$query = 'SELECT foo FROM bar ....';
$exec = mysql_query($query) or die('MySQL error: ' . mysql_error() . '<hr>Query: ' . $query);

That way you have both the error returned by the SQL server as well as the full query that was sent.

sp33d1

I get this error.

MySQL error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '...' at line 1

what does that mean?

bradgrafelman

Erm, that means you probably copied my code snippet above exactly. :p

The $query variable is supposed to be whatever SQL query you're trying to run...

sp33d1

ok i changed the links still the same error maybe im putting it in the wrong place or putting the wrong sql query

$connect = 'SELECT foo FROM bar ....';
$exec = mysql_query($connect) or die('MySQL error: ' . mysql_error() . '<hr>Query: ' . $connect);

$connect = mysql_connect ( 'localhost', 'User', 'Pass' )
	or die(mysql_error());

$connect = mysql_select_db ( 'Database' )
	or die(mysql_error());

is that right?

bradgrafelman

This:

$connect = 'SELECT foo FROM bar ....';

shouldn't literally appear in your script; I used that as a placeholder in my code snippet. Instead of that nonsensical query, you're supposed to insert your own query.