[英]PHP login check username and password code doesn't work
我正在尝试学习PHP和MySQL，我想创建一个登录页面，该页面将检查数据库中的信息，并在信息正确时让用户通过。
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include。 一个是连接，另一个是此php文件。 在同一个文件（主文件）中，我有一个带有3个输入字段的表单：用户名，密码和提交类型。
问题是当我包含上述php文件时，页面显示为空白，并且我什么也收不到，甚至没有“ Not working”字符串。 这是为什么？ 我的语法有什么不好？
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: " . $conn->connect_error);
}
echo "Connection succeed";
?>
您正在查询变量$res而应查询值$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，您缺少分号，如@Barry在回答中所说
<?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: " . $conn->connect_error);
}
echo "Connection succeed";
?>
从表单中删除动作属性。 动作将在同一页面上。 如果必须在另一个页面上创建操作，则应该在其中包含数据库连接。
在查询部分：
if(isset($_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.";
}
您有错误报告吗？ 这将显示任何错误或注意到回帖用户名，这将使您检查表单是否可以正常发布
$connect = mysql_pconnect($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
但是，这种非常不安全的示例以http://www.wikihow.com/Create-a-Secure-Login-Script-in-PHP-and-MySQL为例
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