PHP脚本mysqli的神秘不工作

php
mysqli

2013-02-07 62 views 0 likes

<?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()) 
    { 
       ....

我得到很奇怪的行为。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

我用这个确切的代码在许多其他的网页和它完美的作品。任何想法为什么我可能会随机得到这个错误？

来源

2013-02-07 SnareChops

回答

我不知道，如果你是电子邮件用户名绑定之前错过了FROM。还ü失踪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

来源

2013-02-07 22:59:29

你的表名

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

来源

2013-02-07 22:53:53 Musa

PHP脚本mysqli的神秘不工作

回答

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