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[code]
$query = “select wwlands.type as type from wwlands where number1 = $y”;
$result = [b]@[/b]querydatabase ($query);
if (mysql_result($result, 0, “type”) == 1)
echo “<img src=images/stst.jpg alt=M />”;
Heres the code. Now I am looking up alot of things that may exist or may not. I do I do this look up and not have all the warnings not found printed on the page.
tia