What am I doing wrong? [RESOLVED]

[espylacopa]

Why am I getting the error:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/content/b/i/k/mydirectory/html/ax0!calRe367/view.php on line 13

but it still shows me the data underneath it

<?php
include("cn.php");
include("commonview.php");
checklogin();
 
$sql = "SELECT * FROM calendar";
$calendar = mysql_query($sql) or die(mysql_error());
 
 
echo '<table>
    <tr>';
 
while ($calendar = mysql_fetch_array($calendar)) {
 
    if ($i == 2) {
        echo '</tr><tr>';
        $i = 0;
    }
 
    echo '<td><b>' . $calendar['dept'] . '</b><br>';
    echo $calendar['mon'] . '<br>';
    echo $calendar['tues'] . '<br>';
    echo $calendar['wed'] . '<br>';
    echo $calendar['thurs'] . '<br>';
    echo $calendar['fri'] . '<br>';
    echo $calendar['sat'] . '<br>';
    echo $calendar['sun'] . '<br><br></td>';
 
    $i++;
}
 
echo '</tr></table><br>';
 
?>

[sql_lall]

I've had this before when the SQL statement was incorrect (i.e. DELEE instead of DELETE).
Possibly, is it because your database doesn't have a table called 'calendar', that's the only thing i can think of. That errors usually tends to mean incorrect sql statement.

[ALL]

well, correct my if i am wrong, but shouldnt

while ($calendar = mysql_fetch_array($calendar)) { 

be:

while ($calendar == mysql_fetch_array($calendar)) { 

???

[visualAd]

$calendar first stores a reference to the result of the query. The statement $calendar = mysql_fetch_array($calendar) overwrites this reference. Hence, the next time it loops, it fails because $calendar no longer contains a reference to the result of the query, it contains an array.

To solve this change the name of the variable. I.e:

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

/* other code */

while ($calendar= mysql_fetch_array($result)) { 

[espylacopa]

thanks visual ad. I am used to asp, but the server doesnt support it. It worked great.