I have a table and I am displaying its contents using PHP and a while(); I have about three fields in the table that are NULL but can be change, but I want them to still display all the results in my table.
But, it only shows the records with data in EVERY field. Anyone how I can display it? I get a count of the table and it gives me 2, but only displays one.
Any thoughts?
UPDATE ADDED CODE:
<h3>Viewing All Updates</h3>
<h4>Below are all active updates for COTC</h4>
<table>
<thead>
<tr>
<th>Site Name</th>
<th>Page</th>
<th>Flag</th>
<th>Date Sent</th>
<th>View</th>
</tr>
</thead>
<tbody>
<?php
$sql = "SELECT sname,page_name,date_submitted,u_id,clients.c_id,flag,completed FROM updates INNER JOIN clients ON updates.c_id = clients.c_id INNER JOIN pages ON updates.page = pages.p_id ORDER BY date_submitted DESC";
$query = mysql_query($sql) or die(mysql_error());
while($row = mysql_fetch_array($query)){
$completed = $row['completed'];
if($completed == 1){
print '<tr class="quiet">';
}else{
print '<tr>';
}
print '<td>'.$row['sname'].'</td>';
print '<td>'.$row['page_name'].'</td>';
print '<td>'.$row['flag'].'</td>';
print '<td>'.$row['date_submitted'].'</td>';
print '<td class="center"><a href="?c=displayupdate&id='.$row['u_id'].'" title="View update for '.$row['sname'].'" id="'.$row['u_id'].'"><img src="images/page_edit.png" alt="Edit entry!" /></a></td>';
print '</tr>';
}
?>
</tbody>
</table>
Thanks,
Ryan