Question

我正在尝试学习PHP和MySQL，我想创建一个登录页面，它将检查数据库中的信息，如果信息正确，将让用户通过。

我有以下PHP代码：

    if(isset($_POST['submit'])) {
        $username = $_POST['username'];
        $password = $_POST['password'];

        $sql = "SELECT * FROM users WHERE username ='".$username."' AND password='".$password."' LIMIT 1";
        $res = mysqli_query($sql);
        if(mysqli_num_rows($res) == 1 ) {
            echo "You have successfully logged in.";
        } else {
            echo "Invalid login information."; 
        }
    } else {
        echo "Not working.";
    }

在我的主文件中，我有2个包含。一个是连接，另一个是这个php文件。在同一个文件（主文件）中，我有一个包含3个输入字段的表单：用户名，密码和提交类型。

问题是当我包含上面的php文件时，页面显示为空白而且我什么也得不回来，甚至没有＆＃34;不工作＆＃34;串。这是为什么？我的语法有什么不好的？

更新

我的主要文件代码：

<?php

    include 'includes/connection.php';
    include 'includes/login.php';

?>


<html>
<head>
    <title>Login</title>
    <link rel="stylesheet" href="css/login_page.css">
</head>
<body>
    <div class="flex">
        <div class="container">
            <h3>Login to the site...</h3>
            <p>The awesome site we created!</p>
            <form method="post" action="login.php">
                <input type="text" name="username" placeholder="Username">
                <input type="password" name="password" placeholder="Password"><br /><br />
                <input type="submit" name="submit" class="btn-submit" value="Login">
            </form>
        </div>
    </div>
</body>
</html>

连接PHP文件：

<?php

$server = "localhost";
$user = "Flopet17";
$pass = "1234";
$db = "listing";


$conn = new mysqli($server, $user, $pass);
mysqli_select_db($db);


if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}
echo "Connection succeed";
?>

Answer 1

行

需要分号

echo "Invalid login information.";

提示的

另请查看mysqli和prepare statements以获取数据库连接并构建save sql语句。

Answer 2

您应该在查询值$res时查询变量$sql 将以下代码更改为：

$sql = "SELECT * FROM users WHERE username ='".$username."' AND password='".$password."' LIMIT 1";
$res = mysql_query($sql);

另外，@Barry在答案中说，你的分号丢失了

Answer 3

<?php

$server = "localhost";
$user = "Flopet17";
$pass = "1234";
$db = "listing";


//$conn = new mysqli($server, $user, $pass);
//mysql_select_db($db);//Invalid with my sql
$conn = new mysqli($server, $user, $pass,$db);

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}
echo "Connection succeed";
?>

从表单中删除操作属性。行动将在同一页面上。如果您必须在另一个页面上创建操作，则应在其中包含数据库连接。

在查询部分：

 if(isset($_POST['submit'])) {
        $username = $_POST['username'];
        $password = $_POST['password'];

        $sql = "SELECT * FROM users WHERE username ='".$username."' AND password='".$password."' LIMIT 1";
       // $res = mysqli_query($sql);
        $res = $conn->query($sql);
        if($conn->num_rows($res) == 1 ) {
            echo "You have successfully logged in.";
        } else {
            echo "Invalid login information."; 
        }
    } else {
        echo "Not working.";
    }

Answer 4

你有错误报告吗？这将显示任何错误或通知邮件用户名的回显将检查表单是否正常发布

$connect = mysql_pconnect($hostname, $username, $password) or
trigger_error(mysql_error(),E_USER_ERROR);

mysql_select_db($database, $connect;
$query_Recordset1 = "SELECT * FROM users WHERE users.username=$username ORDER BY
user.username LIMIT 1";
//
$Recordset1 = mysql_query($query_Recordset1, $connect) or die();
$row_Recordset1 = mysql_fetch_assoc($Recordset1);
$totalRows_Recordset1 = mysql_num_rows($Recordset1);
$password_check=$row_Recordset1['password'];
if ($password_check==$password){echo"log in ok":}else{ what to do if fail

这非常不安全但是以http://www.wikihow.com/Create-a-Secure-Login-Script-in-PHP-and-MySQL为例

PHP登录检查用户名和密码代码不起作用

4 个答案: