Question

我正在使用登录表单对HTML和PHP界面进行编码，以便可以连接到mySQL数据库并进行SELECT查询。

在表格中输入信息并单击“登录”后，我希望PHP以“ root”身份连接到我的数据库，并查看输入的信息是否属于该数据库的用户。

测试后，我期望mysqli_num_rows()的结果为1，但结果为0。在执行过程中，遇到以下错误：

mysqli_num_rows（）期望参数1为mysqli_result ...

此外，同时使用用户名和密码变量的printf()时，结果为空。

这是我现在正在使用的两个文件的代码段。

auditoria.php

<body>
    <form method="POST" action="login111.php">
        <div id="form_wrapper">

            <div id="form_left">
                <img src="database.png" alt="app icon">
            </div>

            <div id="form_right">
                <h1>Auditor Authentication</h1>

                <div class="input_container">
                    <i class="fas fa-envelope"></i>
                    <input placeholder="Username" type="username" name="Username" id="field_username" class='input_field'>
                </div>

                <div class="input_container">
                    <i class="fas fa-lock"></i>
                    <input  placeholder="Password" type="password" name="Password" id="field_password" class='input_field'>
                </div>

               <input type="submit" value="Login" name="Submit" id='input_submit' class='input_field' >

            </div>

        </div>
    </form> 
</body>

login111.php

<?php
session_start(); // Starting Session 
$error = ''; // Variable To Store Error Message 
if (isset($_POST['Submit'])) { 
    if (empty($_POST['Username']) || empty(MD5($_POST['Password']))) {  
        $error = "Username or Password is invalid"; 
    } 
    else{ 
        // Define $username and $password 
        $username = $_POST['Username']; 
        $passwordMd5 = md5($_POST['Password']); 
        $password = $_POST['Password']; 
        printf("username: .\n",$username);
        printf("pass encriptada: .\n",$passwordMd5);
        printf("pass:.\n",$password);
        // mysqli_connect() function opens a new connection to the MySQL server. 
        $conn = mysqli_connect("localhost", "root", "", "log"); 
        // SQL query to fetch information of registerd users and finds user match. 
        $query = "SELECT IDAuditor, NomeUtilizador, Password FROM auditor";
        printf("Query: \n",$query);
        $result = mysqli_query($conn, $query);
        $num=mysqli_num_rows($result);
        printf("Result set has %d rows.\n",$num);
        // To protect MySQL injection for Security purpose 
        if ($num>0) {
            readfile("CONAN.html");
            $conn2 = mysqli_connect("localhost", $username , $password, "log"); 
        } else {
            mysqli_close($conn); // Closing Connection 
        }
    } 
}
?>

错误：

无法将信息从HTML表单发送到PHP以连接到MySQL数据库

0 个答案: