Quote:
|
Originally Posted by Bifta
Still gives me this error
Code:
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in xxxxxxxxxxxxxxxxxxxxxxxxxxxx
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in xxxxxxxxxxxxxxxxxxxxxxxx
because styles=whatever is in the URL it's somehow trying to find that field in the DB by the look of it, when it can't find it it throws the above  |
I'm just guessing here but you'd need something like this:
PHP Code:
// Open SQL connection.
if (!$dbconnection) {
echo( "<p>Unable to connect to the " .
"database server at this time.</p>" );
exit();
}
// Select DB
@mysql_select_db($dbname) or die("Could not select database");
// The query for getting the $style
$result = @mysql_query("SELECT * FROM " . $table_name . " WHERE styles='" . $styles . "'");
if (!$result) {
echo("<p>Error performing query: " .
mysql_error() . "</p>");
exit();
}
// Loop through the array and echo out the $style retrieved from the DB
while ( $row = @mysql_fetch_array($result) ) {
$styles = $row['styles'];
echo "<link rel=\"stylesheet\" href=\"$styles\" type=\"text/css\" />";
}
// Close SQL connection.
mysql_close($dbconnection);
You could also use the more OOP way by using mysql_fetch_object() instead like so:
PHP Code:
// Open SQL connection.
if (!$dbconnection) {
echo( "<p>Unable to connect to the " .
"database server at this time.</p>" );
exit();
}
// Select DB
@mysql_select_db($dbname) or die("Could not select database");
// The query for getting the $style
$result = @mysql_query("SELECT * FROM " . $table_name . " WHERE styles='" . $styles . "'");
if (!$result) {
echo("<p>Error performing query: " .
mysql_error() . "</p>");
exit();
}
// Get the object from the DB and echo out the $style.
$style_object = @mysql_fetch_object($result);
$styles = $style_object->styles;
echo "<link rel=\"stylesheet\" href=\"$styles\" type=\"text/css\" />";
// Close SQL connection.
mysql_close($dbconnection);