1

我正在开发一个 phonegap 应用程序，我需要将一些数据从 html 文件发送到服务器并获得输出作为响应。当我将所有文件都保存在保存服务器中时，它们就可以工作。当我将文件拆分为客户端和服务器机器时，它们不起作用，据我所知，它与跨域 POST 请求有关。下面是我的代码：

客户端：something.html

$(document).ready(function() {
$("#login").click(function(e){

 var formData = $("#loginForm").serialize();

 $.ajax({
        type: "POST",
        url: "http://testserver.bscheme.com/rss/login.php",
        crossDomain: true,
    cache: false,
        data: formData,
        success: function(txt){
            if(txt !=""){               
                    localStorage.setItem("background", txt);
                    //alert(localStorage.getItem("background"));
                    document.location.href="logged.html";
            }
            else
            alert("Badly Failed");          
            }
        });

        e.preventDefault();
    });   });

服务器端：login.php

<?php

include('inc/db.php');

try
{
$userName       = mysql_real_escape_string($_POST['username']);
$passcode        = mysql_real_escape_string($_POST['passcode']);

if(empty($feedName)!="" && empty($feedUrl)!="") {

    $result = mysql_query("SELECT id,email,password FROM user WHERE email = '".$userName."'");
    $row = mysql_fetch_array($result);
    //echo $userName . $row['email'];
//  echo $passcode . $row['password'];
    if($row['email']== $userName && $row['password'] == $passcode && $row['email']!= "" && $row['password'] != "")  {
        echo $row['id'];

    }else{
        echo "fail";
    }
} else {
echo "All fields are mandatory!";
}
}
catch(Exception $e)
{
// normally we would log this error
echo $e->getMessage();
}

我如何允许跨域 POST，我知道我必须编辑我的服务器 php 文件，并且我添加了以下代码，但我也可以工作

switch ($_SERVER['HTTP_ORIGIN']) {
case 'http://localhost/rss': case 'https://localhost/':
header('Access-Control-Allow-Origin: '.$_SERVER['HTTP_ORIGIN']);
header('Access-Control-Allow-Methods: POST, GET, OPTIONS');
header('Access-Control-Max-Age: 1000');
header('Access-Control-Allow-Headers: Content-Type');
break;
}

使用 Firebug 我说我的 $.POST 这就是我拥有的

ResponseHeaders RequestHeaders Content-Type application/x-www-form-urlencoded; charset=UTF-8 接受 /

参数application/x-www-form-urlencoded passcode asd username qwe Source username=qwe&passcode=asd

回复

响应始终为空白。当所有 html 和 php 都在本地主机或我的测试服务器中时，它可以工作。我的 php 和 jquery 技能缺乏 atm，所以如果其他人已经问过我很抱歉重复的问题，我也很难理解。为你提供帮助

4

1 回答 1

1

我必须做出的改变：客户端：something.html

$(document).ready(function() {
$("#login").click(function(e){      
    var formData = $("#loginForm").serialize();      

var username;
var passcode;
username= document.getElementById('username').value;
passcode = document.getElementById('passcode').value;



    if (window.XMLHttpRequest)
    {// code for IE7+, Firefox, Chrome, Opera, Safari
        xmlhttp=new XMLHttpRequest();
    }
    else
    {// code for IE6, IE5
        xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
    }
        xmlhttp.onreadystatechange=function()
    {
    if (xmlhttp.readyState==4 && xmlhttp.status==200)
    {
        var txt;
        txt=xmlhttp.responseText;
        //alert(txt);
        localStorage.setItem("background", txt);
        document.location.href="logged.html";
    }
}
xmlhttp.open("GET","http://testserver.bscheme.com/rss/login.php?username="+ username +"&passcode="+ passcode,true);
xmlhttp.send();

});




});

服务器上的文件更改：Login.php

<?php


header('Access-Control-Allow-Origin: *');
header('Access-Control-Allow-Methods: POST, GET, OPTIONS');
header('Access-Control-Max-Age: 1000');
header('Access-Control-Allow-Headers: Content-Type');


include('inc/db.php');

try
{
$userName       = mysql_real_escape_string($_GET['username']);
$passcode        = mysql_real_escape_string($_GET['passcode']);

if(empty($feedName)!="" && empty($feedUrl)!="") {

    $result = mysql_query("SELECT id,email,password FROM user WHERE email = '".$userName."'");
    $row = mysql_fetch_array($result);
    //echo $userName . $row['email'];
//  echo $passcode . $row['password'];
    if($row['email']== $userName && $row['password'] == $passcode && $row['email']!= "" && $row['password'] != "")  {
        echo $row['id'];

    }else{
        echo "fail";
    }
} else {
echo "All fields are mandatory!";
}
}
catch(Exception $e)
{
// normally we would log this error
echo $e->getMessage();
}

我所要做的就是确保服务器接受所有类型的标头，并且客户端发送一个 XMLHttpRequest 并获取 html 响应而不是 $.Ajax();

于 2013-05-31T09:53:19.050 回答