Question

HTML：

<form id="loginForm" data-ajax="false" action="">
            <div data-role="fieldcontain">
                <label for="username">Username:</label>
                <input id="username" type="text" name="usern" placeholder="Username" />
            </div>
            <div data-role="fieldcontain">
                <label for="password">Password:</label>
                <input id="password" type="password" name="passw" placeholder="Password" />
            </div>
            <div data-role="fieldcontain">
                <input type="button" id="login" value="Login" />
            </div>
</form>
<!-- login page closing tags go here -->

PHP：

    <?php
      $data = mysqli_query($conn, "SELECT * FROM users WHERE username='$username' AND password='$password'");

      if($userResult == 0){
        $data_response["status"] = "error";
       }
    ?>

连接正常，数据库已成功查询。我不确定为什么这不起作用，建议将不胜感激。

Answer 1

删除：

var_dump($_POST);

“用户名”必须与$ _POST ['用户名']和名称=“用户名”

相同

//php
$_POST["username"]

//html
<input id="username" type="text" name="username" placeholder="Username" />

将json成功返回到ajax

1 个答案: