php mysqli脚本神秘地无法正常工作

时间:2013-02-07 22:50:59

标签: php mysqli

我有以下代码:

<?php 
session_start();
echo $_SESSION['Username'];
$mysqli = new mysqli('****','****','****','****');
if($mysqli->errno)
{
    mail("**@yahoo.com", "***/Account.php Connection Error", $mysqli->error . "\nUser: " . $_SESSION['Username']);
}
else 
{
    $stmt = $mysqli->prepare("SELECT FirstName, LastName, Expires, Expires WHERE EMail=?");
    $stmt->bind_param('s', $_SESSION['Username']);
    $stmt->execute();
    $stmt->bind_result($FirstName, $LastName, $Expires);
    $stmt->store_result();
    while($row = $stmt->fetch())
    {
                ....

我的行为非常奇怪。我收到错误Fatal error: Call to a member function bind_param() on a non-object in /home/content/42/7401242/html/****/Account.php on line 12

我在许多其他页面上使用此EXACT代码,它完美无缺。有什么想法我可能会随机收到这个错误吗?

2 个答案:

答案 0 :(得分:1)

您错过了表名

之前的FROM
SELECT FirstName, LastName, Expires FROM Expires WHERE EMail=?

答案 1 :(得分:1)

我不知道您是否使用电子邮件绑定用户名。而且你也缺少FROM条款

   $stmt = $mysqli->prepare("SELECT FirstName, LastName, Expires from Expires WHERE EMail=?");
   $stmt->bind_param('s', $_SESSION['Username']);
                               ^^^^^^^^^^-----------------be sure if its email variable