AngularJS：使用php后端身份验证的基本示例

我想做什么 我正在开发一个后台使用angularJS和php的应用程序，我想在我的应用程序中实现身份验证，我尝试了下面粘贴的代码，但它不会生成任何错误、任何消息，当我尝试输入用户名和密码时，我会进入“管理员”视图“无论我输入什么用户，这意味着它都不会检查用户和密码。在“client”表的数据库中，我有：NomClient和mdp（密码）

我尝试的

。。。。。这么多的例子都是徒劳的

代码：

login.html

<ion-content class="padding" ng-controller="loginCtrl">
<div class="list list-inset" >
<label class="item item-input">
      <input type="text" placeholder="nom" required="" ng-model="NomClient"> 
</label> 
<label class="item item-input">
      <input type="password" placeholder="Password" ng-model="mdp"> 
</label> 
    <button class="button button-block button-positive" ng-click="submit()">Login</button>       
 </ion-content>

login.php

<?php  

$data = json_decode(file_get_contents("php://input"));

 $connect = mysqli_connect("localhost", "root", "", "tem");  


$response['status'] = 0;
$response['message'] = '';
$NomClient=mysqli_real_escape_string($connect, $data->NomClient);
$mdp=mysqli_real_escape_string($connect, $data->mdp);


$query = 'SELECT * FROM `client` WHERE NomClient = "'.$NomClient.'" AND   mdp= "'.$mdp.'"';


if(mysqli_connect_errno()){
    $response['status'] = 0;
    $response['message'] = "Failed to connect to MySQL: ".mysqli_connect_error();
    echo jsone_encode($response);exit;
}

$result = mysqli_query($connect, $query);
$rowcount=mysqli_num_rows($result);
if($rowcount>0){
    $response['status'] = 1;
    $response['message'] = 'Login successful';
}
else{
    $response['status'] = 0;
    $response['message'] = 'Invalid username of password';
}

echo json_encode($response);exit;
 ?>

---

提前感谢。

数据是从ionic传递到php的？是的，我的应用程序是基于此传输的，它可以工作，但我现在的问题是如何检查登录名和密码。非常感谢。

<?php  

$data = json_decode(file_get_contents("php://input"));

 $connect = mysqli_connect("localhost", "root", "", "tem");  


$response['status'] = 0;
$response['message'] = '';
$NomClient=mysqli_real_escape_string($connect, $data->NomClient);
$mdp=mysqli_real_escape_string($connect, $data->mdp);


$query = 'SELECT * FROM `client` WHERE NomClient = "'.$NomClient.'" AND   mdp= "'.$mdp.'"';


if(mysqli_connect_errno()){
    $response['status'] = 0;
    $response['message'] = "Failed to connect to MySQL: ".mysqli_connect_error();
    echo jsone_encode($response);exit;
}

$result = mysqli_query($connect, $query);
$rowcount=mysqli_num_rows($result);
if($rowcount>0){
    $response['status'] = 1;
    $response['message'] = 'Login successful';
}
else{
    $response['status'] = 0;
    $response['message'] = 'Invalid username of password';
}

echo json_encode($response);exit;
 ?>