Supplied argument is not a valid MySQL result resource???

Question

I get this error when I use this code for people to input data. The submit form won't be listed cause it's not useful in this circumstance:

function some_more_custom_content() {

$output="<BR>";

ob_start();

if ($_REQUEST['code'] != "") {
    $code = $_REQUEST['code'];
    $query="INSERT INTO `fc` (`code`,`datetime`) values ('" . mysql_real_escape_string($code) . "', now())";
    $result=mysql_query($query);

    while ($fetch_array = mysql_fetch_array($result)) {
        $seconds = time() - strtotime($fetch_array["datetime"]);

        if ((time() - $entry['datetime']) < 60*60) {
            echo ("The code " . htmlentities($code) ." was updated less than an hour ago.");
        } else {
            echo ("Inserted " . htmlentities($code) ." into the top.");
        }
 }
}

Any idea why? ;\

Answer 1

INSERT statements don't return a resource, they return TRUE on success or FALSE on failure. Thus there is no resource for mysql_fetch_array() to operate on.

(This is one of the main reasons why people complain about PHP -- its semantics are inconsistent at best.)

Answer 2

Adding to Meredith Answer. You shoul use

if($result!==FALSE) {
    // the insert was ok
} else {
    //the inser failed
}

Answer 3

change:

$result=mysql_query($query);

with:

$result=mysql_query($query) or die(mysql_error());

you will see the error.