PhoneGap登录Error

Question

我做了一个PhoneGap应用程序，当记录日志记录是否成功，或者即使不成功时，我也遇到了在else部分中处理的相同错误。 这是我的Javascript Ajax登录代码。

$(document).ready(function() { 
    $("#login").click(function() { 
    var email = $("#email").val(); 
    var password = $("#password").val(); 
    var dataString = "email=" + email + "&password=" + password + "&login="; 
    if ($.trim(email).length > 0 & $.trim(password).length > 0) { 
     $.ajax({ 
     type: "POST", 
     url: "http://192.168.43.173/viamobile/login.php", 
     data: dataString, 
     crossDomain: true, 
     cache: false, 
     beforeSend: function() { 
      $("#login").html('Connecting...'); 
     }, 
     success: function(data) { 
      if (data == "success") { 
      localStorage.login = "true"; 
      localStorage.email = email; 
      window.location.href = "index.html"; 
      } else if (data = "failed") { 
      alert("Login error"); 
      $("#login").html('Login'); 
      } 
     } 
     }); 
    } 
    return false; 
    }); 
}); 

和我的PHP脚本

<?php 
include "db.php"; 
if(isset($_POST['login'])) 
{ 
    $email = $_POST['email']; 
    $password = $_POST['password']; 

    $login = mysqli_num_rows(mysqli_query($con,"select * from `admin` where `username`='$email' and `password`='$password'")); 
    if($login!=0) 
    { 
    echo "success"; 
    } 
    else 
    { 
    echo "failed"; 
    } 
} 
?> 


<?php 
header("Access-Control-Allow-Origin: *"); 
$con= mysqli_connect("localhost","root","","phonegap"); 
?> 

我真的需要验证用户，并根据用户的访问重定向到不同的页面。但是，既然这个错误我不能这样做。可以有人告诉我我错了哪里？

Answer 1

我可以建议你像我的工作代码。客户端JS代码：

$.ajax({ 
    type: /* GET or POST */, 
    url: /* your url */, 
    dataType: 'json', 
    contentType: "application/json", 
    beforeSend: function (event) { 
    /* before funcs */ 
    } 
}) 
.done(function (data) { 
    if (data.status == "ok" || !data.error) { 
    /* good */ 
    } else { 
    /* manage error */ 
    } 
}) 
.fail(function (data) { 
    /* manage error */ 
});