Question

我正在编写一个代码，该代码应该显示登录，注销页面，所有这些都是用PHP编写的，包括一些嵌入式HTML5和CSS。代码如下

  <?php
   session_start(); // Starting Session
   $error=''; // Variable To Store Error Message
   if (isset($_POST['submit'])) {
   if (empty($_POST['username']) || empty($_POST['password'])) {
   $error = "Username or Password is invalid";
   }
   else
   {
  // Define $username and $password
   $username=$_POST['username'];
   $password=$_POST['password'];
  // Establishing Connection with Server by passing server_name, user_id and    
  // password as a parameter
   $connection = mysqli_connect("localhost", "root", "", "Company");
  // To protect MySQL injection for Security purpose
   $username = stripslashes($username);
   $password = stripslashes($password);
   $username = mysqli_real_escape_string($connection, $username);
   $password = mysqli_real_escape_string($connection, $password);
  // Selecting Database
  // SQL query to fetch information of registered users and finds user    
  //match.
   $query = mysqli_query("select * from login where password='$password' AND    
   username='$username'", $connection);
   $rows = mysqli_num_rows($query);
   if ($rows == 1) {
   $_SESSION['login_user']=$username; // Initializing Session
   header("location: profile100.php"); // Redirecting To Other Page
   } else {
   $error = "Username or Password is invalid";
   }
   mysqli_close($connection); // Closing Connection
   }
   }
   ?>

当我在XAMPP上运行这个PHP文件时，我不断收到错误

      Warning: mysqli_query() expects parameter 1 to be mysqli, string given 
      in C:\xampp\htdocs\FuriousCustomer1\login100.php on line 23

      Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result,        
      null given in C:\xampp\htdocs\FuriousCustomer1\login100.php on line 24

我不明白mysqli_query（）如何为参数提供mysqli，或者mysqli_num_rows（）如何为参数提供mysqli_result。我已经检查过，双重检查过，并且三重检查了我的代码，并参考了几本关于PHP的书籍，其中一本是Kevin Tatroe（一本非常好的书），并没有任何暗示我做错了什么，但显然我是。你们当中有人可以指出我需要做什么吗？

使用mysqli_query（）函数的参数很困难

0 个答案: