Quote:
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Originally Posted by Richard M
If you basically need to get a variable from the URL just do it like this:
PHP Code:
$styles = $_GET['styles'];
echo "<link rel=\"stylesheet\" href=\"$styles\" type=\"text/css\" />";
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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
