BrianM Posted June 17, 2008 Share Posted June 17, 2008 My table wont display anything but the headers. It isn't displaying output from the database... <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>MPS - View Data</title> </head> <?php mysql_connect('localhost', 'brian', '') or die(mysql_error()); mysql_select_db('mps') or die(mysql_error()); if (isset($_COOKIE['username'])) { $username = $_COOKIE['username']; $password = $_COOKIE['password']; $check_one = mysql_query("SELECT * FROM mps_login WHERE username = '$username'") or die(mysql_error()); while($check_two = mysql_fetch_array($check_one)) { if ($password != $check_two['password']) { header('Location: login.php'); } else { ?> <body> <?php mysql_select_db('mps'); $result = mysql_query("SELECT * FROM mps_data"); ?> <table> <tr> <th>Client project number</th> <th>Client project name</th> <th>Client name</th> <th>Client address one</th> <th>Client address two</th> <th>Client office phone</th> <th>Client fax phone</th> <th>Client cell phone</th> <th>Client email</th> </tr> <?php while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['client_project_number'] . "</td>"; echo "<td>" . $row['client_project_name'] . "</td>"; echo "<td>" . $row['client_name'] . "</td>"; echo "<td>" . $row['client_address_one'] . "</td>"; echo "<td>" . $row['client_address_two'] . "</td>"; echo "<td>" . $row['client_office_phone'] . "</td>"; echo "<td>" . $row['client_fax_phone'] . "</td>"; echo "<td>" . $row['client_cell_phone'] . "</td>"; echo "<td>" . $row['client_email'] . "</td>"; echo "</tr>"; } ?> </table> </body> </html> <?php } } } else { header('Location: login.php'); } ?> Link to comment https://forums.phpfreaks.com/topic/110638-any-reason-why/ Share on other sites More sharing options...
hitman6003 Posted June 17, 2008 Share Posted June 17, 2008 change $result = mysql_query("SELECT * FROM mps_data"); to $result = mysql_query("SELECT * FROM mps_data") or die(mysql_error()); And see what the error is. Link to comment https://forums.phpfreaks.com/topic/110638-any-reason-why/#findComment-567595 Share on other sites More sharing options...
BrianM Posted June 17, 2008 Author Share Posted June 17, 2008 Actually, false alarm; no data inside the table But thank you for the quick reply. Link to comment https://forums.phpfreaks.com/topic/110638-any-reason-why/#findComment-567596 Share on other sites More sharing options...
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