Not sure what the isue is but when i use the following code nothing is displayed: <?php $result = mysql_query("SELECT * FROM playerstats WHERE name='$playername'"); while($row = mysql_fetch_array($result)) { echo "<table border=\"0\" width=\"800\" cellpadding=\"0\" cellspacing=\"0\">"; echo "<tr>"; echo "<td>Health: " . $row['health'] . "/100</td>"; echo "<td>Energy: " . $row['energy'] . "/10</td>"; echo "<td>Stamina: " . $row['stamina'] . "/5</td>"; echo "<td>Fame: " . $row['fame'] . "</td>"; echo "<td>Cash: $" . $row['gold'] . "</td>"; echo "<td>Experience: " . $row['experience'] . "/500</td>"; echo "<td>Level: " . $row['level'] . "</td>"; echo "</tr>"; echo "</table>"; } ?> Code (markup): It seems the problem is with the: $result = mysql_query("SELECT * FROM playerstats WHERE name='$playername'"); while($row = mysql_fetch_array($result)) Because if take that out the table will display (without db info of course) but with those lines nothing shows at all. Any help would be appriciated thank you
$result = mysql_query("SELECT * FROM playerstats WHERE name='$playername'") or die(mysql_error()); PHP: Does it return any error? Is the $playername safe?
No error ... it just doesn't display anything at all $playername = "test" and there is a name called test in the playerstats table
it seems to work fine that way ... I just checked the variable and it seems i had 2 = signs when saving the member name .... thank you for your help it was a stupid overlook on my part Thank you