如何显示经过身份验证的用户数据

时间:2017-07-20 13:30:09

标签: php angularjs authentication

我正在使用AngularJS和PHP后端,我正在尝试显示经过身份验证的用户的数据,我有2个视图,一个是'login.html',另一个是'info.html',2个控制器(一个用于登录功能,另一个用于选择用户的数据)。我实现了身份验证阶段,但对于第二步,我希望当用户进行身份验证时,它将被重定向到其他视图(info.html),其中将显示该用户的所有信息。

我尝试使用会话,但它不起作用,我不知道为什么。如何从登录功能存储用户数据并在第二个控制器(第二个Web服务)中使用它?

的login.php

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

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

 if(count($data) > 0)  

 { 

$Email=mysqli_real_escape_string($connect, $data->Email);
$password=mysqli_real_escape_string($connect, $data->password);

$query = 'SELECT * FROM `client` WHERE (EmailClient = "'.$Email.'" AND   password= "'.$password.'")';

$q = mysqli_query($connect , $query);

if(mysqli_num_rows($q) > 0 )
  { 
       $token = md5($Email.time()."51395+81519851");
       $query = "UPDATE client SET token = '".$token."' WHERE EmailClient = '".$Email."'";
       mysqli_query($connect , $query);
       $_SESSION["logged_in"] = true; 
       $_SESSION["token"] = $token; 
       $result['message'] ='Logged In';
       $result['email'] =$Email;
       $result['token'] = $token;

       $resultstring=json_encode($result);
       $resultstring=str_replace("null", '""', $resultstring);
       echo $resultstring;
       exit;
  }

       $result['message'] ='The username or password are incorrect!';

$resultstring = json_encode($result);
$resultstring = str_replace("null",'""',$resultstring);
echo $resultstring;
exit;
}

?>

info.php的

<?php  
session_start();
 $connect = mysqli_connect("localhost", "root", "", "test");  

 $output = array();  
 $query = "SELECT Name,Adresse FROM client WHERE token = '".$_SESSION['token']."'";
 $result = mysqli_query($connect, $query);  
 if(mysqli_num_rows($result) > 0)  
 {  
      while($row = mysqli_fetch_array($result))  
      {  
           $output[] = $row;  
      }  
      echo json_encode($output);  
 }  

 ?>

0 个答案:

没有答案