$ _POST可以读取，但值不能

Question

我使用$ _POST形式输入，代码如下所示

       <form class="form-signin" method="POST" action="">
          <div class="form-label-group">
            <label for="username">User Name</label>
            <input type="text" name="username" class="form-control" placeholder="username" required autofocus>
          </div>

          <div class="form-label-group">
            <label for="password">Password</label>
            <input type="password" name="password" class="form-control" placeholder="Password" required>
          </div>

          <button class="btn btn-primary mt-3" type="submit" name="SubmitLogin">Submit</button>
        </form>

这只是一个简单的用户名和密码，然后我使用此代码检查用户名和密码

if(isset($_POST["SubmitLogin"])){
$data = $_POST['SubmitLogin'];
var_dump($data);
$username = $data['username'];
$password = $data['password'];
$result = mysqli_query($mysqli, "SELECT COUNT(*) AS Exist FROM user WHERE username = '$username' AND pwd = '$password' ");
$login = mysqli_fetch_array($result);

if($login['Exist'] == 1){
    $_SESSION["Login"] = true;
    header('Location:index.php');
}else{
    echo '<script language="javascript">';
    echo 'alert("Username/Password wrong")';
    echo '</script>';
  } 
}

不幸的是，如代码中所写，我想通过将$ _POST [“ SubmitLogin”]包含在$ data变量中来检查其中的内容，然后返回“”。...什么可以解决此问题？

Answer 1

if(isset($_POST["SubmitLogin"])){
    $data = $_POST;
    $username = $data['username'];
    $password = $data['password'];
    $result = mysqli_query($mysqli, "SELECT COUNT(*) AS Exist FROM user WHERE username = '$username' AND pwd = '$password' ");
    $login = mysqli_fetch_array($result);

    if($login['Exist'] == 1){
     $_SESSION["Login"] = true;
        header('Location:index.php');
    }else{
        echo '<script language="javascript">';
        echo 'alert("Username/Password wrong")';
        echo '</script>';
    } 
}