Question

我正在制作一个单页的网络应用程序，其中包括注册，登录和搜索表单。目前，搜索和注册功能正在运行，但登录表单无法正常运行。代码如下：

//Login code
if(isset($_POST['Name'])){
$Name = $_POST['Name'];
$password = $_POST['password'];
$message= '';   
//Handle errors
if((!$Name) || (!$password)){
    $message = 'Please fill in all the fields';
}else{
    //Secure the data
    $Name = preg_replace("#[0-9a-z]#i", "", $Name);
    $password = sha1($password);
$login_query = mysql_query("SELECT * FROM users WHERE name='$Name' AND password='$password' LIMIT 1") or die("Could not check your details");
    $login_count = mysql_num_rows($login_query);
    if($login_count != 0){//If account does exist
        //Get the id
        //while($row = mysql_fetch_array($login_query)){
            //$id = $row['id'];
    //  }
        $logged = 1;
        $message = 'Logged in';
    }else{
        $message = 'Your details are incorrect';
        $logged = 0;
    }
}
}//Big if statement for login

我感谢任何建议或意见。提前谢谢

Answer 1

尝试将{}添加到变量中，并将``添加到MySQL查询中的名称

mysql_query("SELECT * FROM `users` WHERE `name`='{$Name}' AND `password`='{$password}' LIMIT 1")

为什么我的登录表单不起作用

1 个答案: