Question

我正在尝试编写用于创建登录页面的PHP代码。检查，双重检查和三重检查我的代码后，我发现它们没有任何问题。但是，每次执行我的代码时，我都会收到错误消息：

表＆＃39; company.login＆＃39;不存在警告：mysqli_num_rows（）期望参数1为mysqli_result，第26行的C：\ xampp \ htdocs \ login100.php中给出布尔值

我知道当错误被称为＆＃34; login100.php＆＃34;的第26行时，并不一定意味着该行本身有任何问题。这是＆＃34; login100.php＆＃34;：

 <?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);
 // SQL query to fetch information of registerd users and finds user match.
 $sqli = "SELECT * FROM login where password='$password' AND        
 username='$username'";
 $result = mysqli_query($connection, $sqli);
 if ($result === false) {
 echo mysqli_error($connection);
 }
 $rows = mysqli_num_rows($result);
 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
 }
}
?>

第26行对应$ rows = mysqli_num_rows（$ result）;正如您所看到的，没有什么不对，因为很容易看出您只需要一个参数，而不是两个参数。实际的错误与表＆＃34;登录＆＃34;的事实有关。不存在。这让我很烦恼，因为我确信它确实如此。让我告诉你＆＃34; session100.php＆＃34;它创建了一个名为＆＃34; login＆＃34;：

的表

   <?php
   // Establishing Connection with Server by passing server_name, user_id     
   // and password as a parameter
   $connection = mysqli_connect("localhost", "root", "", "Company");
   // Selecting Database
   //$db = mysqli_select_db("company", $connection);
   //CREATE DATABASE company;
   $sql="CREATE TABLE login(
   id int(10) NOT NULL AUTO_INCREMENT,
   username varchar(255) NOT NULL,
   password varchar(255) NOT NULL,
   PRIMARY KEY (id))";

   session_start();// Starting Session
   // Storing Session
   $user_check=$_SESSION['login_user'];
   // SQL Query To Fetch Complete Information Of User
   $ses_sql=mysqli_query("select username from login where      
     username='$user_check'", $connection);
   $row = mysqli_fetch_assoc($ses_sql);
   $login_session =$row['username'];
   if(!isset($login_session)){
   mysqli_close($connection); // Closing Connection
   header('Location: index100.php'); // Redirecting To Home Page
  }
 ?>

正如您所看到的，没有任何故障。显示登录页面的方法是运行＆＃34; index100.php＆＃34;在XAMPP上。＆＃34; index100.php＆＃34;的代码如下所示。我的问题是，你们中的任何一个人都可以发现＆＃34; session100.php＆＃34;如上所示，因为在我的眼中，一切都是桃子的。我希望额外的一双眼睛可能证明我错了，并告诉我哪里错了。

＆＃34;的index.php＆＃34;：

    <?php
    include('login100.php'); // Includes Login Script

    if(isset($_SESSION['login_user'])){
    header("location: profile100.php");
    }
    ?>
   <!DOCTYPE html>
   <html>
   <head>
   <title>Login Form in PHP with Session</title>
   <link href="style100.css" rel="stylesheet" type="text/css">
   </head>
   <body>
   <div id="main">
   <h1>PHP Login Session Example</h1>
   <div id="login">
   <h2>Login Form</h2>
   <form action="" method="post">
   <label>UserName :</label>
   <input id="name" name="username" placeholder="username" type="text">
   <label>Password :</label>
   <input id="password" name="password" placeholder="**********"    
   type="password">
   <input name="submit" type="submit" value=" Login ">
   <span><?php echo $error; ?></span>
   </form>
   </div>
   </div>
   </body>
   </html>

当所有迹象表明它存在时，为什么会出现一条错误消息，表明该表不存在？

0 个答案: