我知道3个方法来解决你的问题:

First if you have access to both domains you can allow access for all other domain using : header("Access-Control-Allow-Origin: *"); or just a domain by adding code bellow to .htaccess file: <FilesMatch "\.(ttf|otf|eot|woff)$"> <IfModule mod_headers.c> SetEnvIf Origin "http(s)?://(www\.)?(google.com|staging.google.com|development.google.com|otherdomain.net|dev02.otherdomain.net)$" AccessControlAllowOrigin=$0 Header add Access-Control-Allow-Origin %{AccessControlAllowOrigin}e env=AccessControlAllowOrigin </IfModule> </FilesMatch> you can have ajax request to a php file in your server and handle request to another domain using this php file. you can use jsonp , because it doesn't need permission. for this you can read our friend @BGerrissen answer.

我使用Apache服务器，所以我使用mod_proxy模块。使模块:

LoadModule proxy_module modules/mod_proxy.so
LoadModule proxy_http_module modules/mod_proxy_http.so

然后添加:

ProxyPass /your-proxy-url/ http://service-url:serviceport/

最后，将proxy-url传递给脚本。

同源策略确实阻止了JavaScript跨域发起请求，但是CORS规范只允许您正在寻找的那种API访问，并且得到当前一批主要浏览器的支持。

查看如何为客户端和服务器启用跨源资源共享:

http://enable-cors.org/

“跨域资源共享(CORS)是一个实现跨域边界真正开放访问的规范。如果你提供公共内容，请考虑使用CORS将其开放给通用JavaScript/浏览器访问。”

它的工作，所有你需要:

PHP:

header('Access-Control-Allow-Origin: http://www.example.com');
header("Access-Control-Allow-Credentials: true");
header('Access-Control-Allow-Methods: GET, PUT, POST, DELETE, OPTIONS');

JS (jQuery ajax):

var getWBody = $.ajax({ cache: false,
        url: URL,
        dataType : 'json',
        type: 'GET',
        xhrFields: { withCredentials: true }
});

我知道3个方法来解决你的问题:

First if you have access to both domains you can allow access for all other domain using : header("Access-Control-Allow-Origin: *"); or just a domain by adding code bellow to .htaccess file: <FilesMatch "\.(ttf|otf|eot|woff)$"> <IfModule mod_headers.c> SetEnvIf Origin "http(s)?://(www\.)?(google.com|staging.google.com|development.google.com|otherdomain.net|dev02.otherdomain.net)$" AccessControlAllowOrigin=$0 Header add Access-Control-Allow-Origin %{AccessControlAllowOrigin}e env=AccessControlAllowOrigin </IfModule> </FilesMatch> you can have ajax request to a php file in your server and handle request to another domain using this php file. you can use jsonp , because it doesn't need permission. for this you can read our friend @BGerrissen answer.

使用JSONP。

jQuery:

$.ajax({
     url:"testserver.php",
     dataType: 'jsonp', // Notice! JSONP <-- P (lowercase)
     success:function(json){
         // do stuff with json (in this case an array)
         alert("Success");
     },
     error:function(){
         alert("Error");
     }      
});

PHP:

<?php
$arr = array("element1","element2",array("element31","element32"));
$arr['name'] = "response";
echo $_GET['callback']."(".json_encode($arr).");";
?>

echo可能是错误的，我已经有一段时间没用php了。在任何情况下，你都需要输出callbackName('jsonString')，注意引号。jQuery会传递自己的回调名称，所以你需要从get参数中获取。

正如Stefan Kendall所发布的，$.getJSON()是一个简便方法，但随后你需要附加'callback=?'到url作为GET参数(是的，值是?，jQuery用自己生成的回调方法替换它)。