Php 如何在引导模式中显示成功或错误消息？

HTML代码：

<li class="llogin">
   <a class="mlogin" data-target="#loginmodal" data-toggle="modal"><span class="glyphicon glyphicon-chevron-down pull-right"></span>Login</a>
   <div class="modal" id="loginmodal" tabindex="-1">
      <div class="modal-dialog">
         <div class="modal-content">
            <div class="modal-header">
               <button class="close" data-dismiss="modal">&times;</button>
            </div>
            <div class="modal-body">
               <div class="form-group">
                  <input type="text" name=username id="username" placeholder="username" value="" class="form-control" />
               </div>
               <div class="form-group">
                  <input type="password" name=password id="password" placeholder="password" value="" class="form-control" />
               </div>
               <div class="form-group">
                  <div class="row">
                     <div class="text-center" class="col-md-6">
                        <input type="submit" name="loginsubmit" id="loginsubmit" class="form-control btn btn-info" value="LogIn" style="width:20%">
                     </div>
                  </div>
               </div>
               <div id="loginresult" class="alert-success"></div>
            </div>
            <div class="modal-footer">
               <button class="btn btn-primary" data-dismiss="modal">Close</button>
               <button class="btn btn-primary">Login</button>
            </div>
         </div>
      </div>
   </div>
</li>

我正在使用PHP构建web商店。我想在

bootstrap

模式中显示消息，但不知道上面的代码出了什么问题

请建议我应该做些什么改变来实现期望的行为

我收到的消息没有模态，但我想在模态中显示消息。

<?php
require_once("config.php"); 
if(isset($_REQUEST['username']))
{
    $username= $_REQUEST['username'];
    $password= $_REQUEST['password'];
//echo $username;

$con = mysqli_connect(DB_HOST,DB_USERNAME,DB_PASSWORD,DB_NAME);
if($con)
{
$sql = "SELECT `username`,`password` FROM `register` WHERE `username`=? AND `password`=?";
$obj = mysqli_prepare($con,$sql);
if(is_object($obj))
{
    mysqli_stmt_bind_param($obj,"ss",$username,$password);
    mysqli_stmt_bind_result($obj,$myuser_name, $mypassword);
    mysqli_stmt_execute($obj);
    mysqli_stmt_store_result($obj);
    mysqli_stmt_fetch($obj);
    if($myuser_name == $username && $mypassword == $password)
    {
        echo "Successfully Logged In";
    }else{
        echo "Logged In failed";
    }
}else{
    echo "Not Object";
}

}else{
    echo "db problem";
}
}else{
    echo "value not set";
}


 ?>

here is PHP Code.

下面是PHP代码。

下面是PHP代码。

在html中找不到id为“loginmodal”的div。可能缺少了？

在html中找不到id为“loginmodal”的div。可能缺少了？

也发布您的PHP代码，这样我们就可以看到服务器返回了什么响应？这将很容易得到帮助。@HikmatSijapati我已经发布了PHP代码。请帮助解决这个问题，同时发布您的PHP代码，这样我们就可以看到服务器返回了什么响应？这将很容易得到帮助。@HikmatSijapati我已经发布了PHP代码。请帮忙解决这个问题

<?php
require_once("config.php"); 
if(isset($_REQUEST['username']))
{
    $username= $_REQUEST['username'];
    $password= $_REQUEST['password'];
//echo $username;

$con = mysqli_connect(DB_HOST,DB_USERNAME,DB_PASSWORD,DB_NAME);
if($con)
{
$sql = "SELECT `username`,`password` FROM `register` WHERE `username`=? AND `password`=?";
$obj = mysqli_prepare($con,$sql);
if(is_object($obj))
{
    mysqli_stmt_bind_param($obj,"ss",$username,$password);
    mysqli_stmt_bind_result($obj,$myuser_name, $mypassword);
    mysqli_stmt_execute($obj);
    mysqli_stmt_store_result($obj);
    mysqli_stmt_fetch($obj);
    if($myuser_name == $username && $mypassword == $password)
    {
        echo "Successfully Logged In";
    }else{
        echo "Logged In failed";
    }
}else{
    echo "Not Object";
}

}else{
    echo "db problem";
}
}else{
    echo "value not set";
}


 ?>

here is PHP Code.