php - 将 $_SESSION[] 数据制作成 php 变量，但它仍然无法在 mysql 查询中工作

Question

我在检查 mysql 查询的 $_SESSION 变量时遇到了问题。我想做的是获取登录用户的详细信息，但它似乎无法正常工作。

我有 $user = mysql_real_escape_string($_SESSION['username']); 将代码放入常规变量中，然后我查询数据库是:$sql = "SELECT * FROM admin WHERE username='$user' LIMIT 1";

为了计算用户是否存在，我使用代码:$userCount = mysql_num_rows($sql);//统计输出量

这似乎行不通。我不断收到此错误:“警告:mysql_num_rows() 期望参数 1 为资源，第 18 行/home/alexartl/public_html/CRM/headercode.php 中给出的字符串”

顺便说一句，当我测试这个时，用户帐户确实存在并且已登录下面是完整的代码

  // 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 to grab the users name
        $sql = "SELECT * FROM admin WHERE username='$user' LIMIT 1";
        $userCount = mysql_num_rows($sql); // count the output amount
        if ($userCount == 1) {
        while($row = mysql_fetch_array($sql)){
//just the array that grabs all the users info
            $username = $row["username"];
            $password = $row["password"];
            $first_name = $row["first_name"];
            $last_name = $row["last_name"];
            $gender = $row["gender"];
            $birthdate = $row["birthdate"];
            $email_address = $row["email_address"];
            $city = $row["city"];
            $state = $row["state"];
            $retrieval = $row["retrieval"];
            $isAdmin = $row["isAdmin"];
            $join_date = $row["join_date"];


//if the user has "isAdmin" as "Yes", then this link to a "manage Users" page will appear
            if($isAdmin == "Yes"){
                $ifAdmin = '<li><a href="manageUsers.php">Manage Users</a></li>';
                }
            }       
        }
     }      
?>

Answer 1

我不会进入“不要使用 mysql_* 命令”，但不要 :P

你错过了:

 $result = mysql_query($sql);  //Actually execute the query

然后用作

$userCount = mysql_num_rows($result); // count the output amount

---

另外，您似乎也没有连接到或使用您希望查询的数据库。

$link = mysql_connect('localhost', 'user', 'pass') or die('Could not connect to mysql server.' );
mysql_select_db('databaseName');