2016-11-07 84 views
0

我用post方法通過XMLHttpRequest對象從xdk應用程序發送數據(用戶名和密碼)到php文件,但是我遇到了一個問題。 我通過post方法發送的所有數據在php文件中未定義。Post方法不發送數據到php頁面

這個XDK代碼

<!DOCTYPE html> 
<html lang="en-US"> 
<head> 
    <script type="text/javascript"> 
     function login() { 
      var xhr = new XMLHttpRequest(); 
      var username1 = document.getElementById("user_name").value; 
      var pass = document.getElementById("pass_word").value; 
      var params = "username=" + username1 + "&password=" + pass; 
      alert(params); 
      xhr.open("post", "http://localhost/hospital/test.php", false); 
      xhr.onload = function() { 
       if (xhr.status == 200) { 
        var json_string = xhr.responseText; 
        test.innerHTML = json_string; 
       } 
       else if (xhr.status == 404) { 
        alert("Web Service Doesn't Exist", "Error"); 
       } 
       else { 
        alert("Unknown error occured while connecting to server", "Error"); 
       } 
      } 
      xhr.send(params); 
     } 
    </script> 
</head> 
<body> 
<div class="container"> 
    <div id="login"> 
     <img src="img/logo.png" width="100 px" height="100 px"/> 
     <form onsubmit="login(); return false;"> 
      <fieldset class="clearfix"> 
       <p><span class="fontawesome-user"></span><input type="text" id="user_name"/></p> 
       <p><span class="fontawesome-lock"></span><input type="password" id="pass_word"/></p> 
       <p><input type="submit" value="تسجيل الدخول"/></p> 
      </fieldset> 
     </form> 
     <div id="test"></div> 
</body> 
</html> 

這是PHP代碼

<?php 
$con = mysqli_connect("localhost", "root", "", "interactive"); 
if (mysqli_connect_errno()) { 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
} 
if (isset($_POST["username"])) { 
    $username = $_POST["username"]; 
    $password = $_POST["password"]; 
} 

$sql = "SELECT * FROM Caregiver WHERE UserName = $username && password =  $password "; 
$result = mysqli_query($con, $sql); 
$num = $result->num_rows; 
if ($num) { 
    $row = $result->fetch_array(MYSQLI_BOTH); 
    $data = array('status' => 'success', 'username' => $row['UserName'], 'password' => $row['password']); 
} else { 
    $data = array('status' => 'failure', 'Error: ' . mysqli_error($con)); 
} 
echo json_encode($data); 
mysqli_close($con); 
?> 
+0

'WHERE UserName = $ username && password = $ password'那些是字符串值,不是整數。 –

+0

是的,這些都是字符串數據,但是當我使用GET方法時,在命令中沒有錯誤。 – Malak

+0

沒有人幫助我:( – Malak

回答

0

這是一個小我的駕駛室外面,但我想你可能需要在這裏端口:

xhr.open("post", "http://localhost/hospital/test.php", false); 

xhr.open("post", "http://localhost:port/hospital/test.php", false); 
+0

我也看看這篇文章:http://stackoverflow.com/questions/32084571/why-is-an-object-in-an-xmlhttprequest-sent-to-a-node-express - 服務器空 – user3634054

+0

感謝您的建議,堅果仍然不適合我:( – Malak