[RESOLVED] mysql_fetch_array

[gszauer]

$number = mysql_query("SELECT * $housesTable WHERE house_id='".$select."'");
$row = mysql_fetch_array($number);

returns

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/szauerne/public_html/sites/yourcapecoral/wizard/listings.php on line 364

Any ideas? $select has a value of 12.
Thanks for your time
~Gabor

[NogDog]

The error means that your mysql_query() returned FALSE, i.e. MySQL was unable to process your query and returned an error code. It's probably due to not having a "FROM" in your query. In any case, you can debug by doing something like this:

$sql = "SELECT * $housesTable WHERE house_id='".$select."'";
$number = mysql_query($sql) or die("Query failed ($sql) - " . mysql_error());
$row = mysql_fetch_array($number); 

[gszauer]

You are right NogDog, i was missing the FROM
Thank you!