Question

我正在尝试创建一个页面，该页面使用会话数据在数据库中查找用户，然后发送此用户已注册的事件。我是一个新手，并且对我所处的地方非常困惑。我使用两个不同的表来获取数据，这是我感到困惑的地方，我相信错误发生的地方。在此先感谢。

    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<?php
session_start();
?>
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Untitled Document</title>
</head>
<?php
$username = $_SESSION['username'];
$email = $_SESSION['user_email'];
$con=mysqli_connect("localhost","emuas","******","EMUAS_signUp");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }
echo "<table>
<tr>
<td> Logged in as:</td>
</tr>
<tr>
<th>" . $username . "</th>
</tr>
<tr>
<td> 
<form action='logout.php' method='post'>
<input type='submit' value='Logout' >
</form>
</td>
</tr>
<tr>
<th>Events Attending:</th>
</tr>";

$find = mysqli_query($con,"SELECT * FROM SIGN_UP_TEST WHERE User = '$username'");
while($find_row = mysqli_fetch_array($find)){
    //Get Event ID
    $eventId = $find_row['EventID'];
    //Use Event ID to get Event Name
   $result = mysqli_query($con,"SELECT * TEST WHERE EventID = '$eventId'");
//Insert Event Name into table with link from Page Name
while($row = mysqli_fetch_array($result))
  {
  echo "<tr>";
  echo "<td> <a href='http://www.emuas.co.uk/members/sign_up_sheets/S" . $row['PageName'] . ".php'>" . $row["EventName"] . "</a> </td>";
  echo "</tr>";
  }
  }
echo "</table>";
?>
<body>
</body>
</html>

从两个表中获取相关数据

0 个答案: