小编ran*_*r91的帖子

我在检查mysql查询上的$ _SESSION变量时遇到了麻烦.我想要做的是获取用户登录的详细信息,但它似乎无法正常工作.

我$user = mysql_real_escape_string($_SESSION['username']);把代码放入常规变量,然后我对数据库进行查询:$sql = "SELECT * FROM admin WHERE username='$user' LIMIT 1";

并计算用户是否存在我使用代码: $userCount = mysql_num_rows($sql); // count the output amount

这似乎不起作用.我一直收到这个错误:"警告:mysql_num_rows()期望参数1是资源,在第18行的/home/alexartl/public_html/CRM/headercode.php中给出的字符串"

顺便说一句,用户帐户确实存在并在我测试时登录.下面是完整代码

  // If the session vars aren't set, try to set them with a cookie
  if (!isset($_SESSION['user_id'])) {
    if (isset($_COOKIE['user_id']) && isset($_COOKIE['username'])) {
      $_SESSION['user_id'] = $_COOKIE['user_id'];
      $_SESSION['username'] = $_COOKIE['username'];   
    }
  }
?>
<?php
//if the username is set
  if (isset($_SESSION['username'])) {
//making the username into a php variable 
        $user = mysql_real_escape_string($_SESSION['username']);
//the query …