simple problem

[nomad9]

Can someone help me debugging this ... search an hour and find nothing ...

test.php4?analyse=2

<?
session_start();

// Start DB Connection
$db = mysql_connect('localhost', 'xxxxx', 'xxxxx');
mysql_select_db('visionclient_ca',$db);

if($analyse > "1") {
// Detect DB Info -- Check For Analyse Name
$sql = "SELECT formQuestion FROM f55555 WHERE formName = 'typeTwo'";
$result = mysql_query($sql,$db);
echo $result;
}

// End DB Connection
mysql_close();
?>

the echo return this : Resource id #2

don't know what it mean or where he took this

Steve

[pyfsapple]

You have missed this:

$queryData = mysql_fetch_array($result);
$formQuestionValue = $queryData["formQuestion"];
echo $formQuestionValue;

or, if the result is more than one row:

while(($queryData = mysql_fetch_array($result))) {
$formQuestionValue = $queryData["formQuestion"];
echo $formQuestionValue;
}

[nomad9]

thanx alot pyfsapple