jquery 的跨域.getJSON 本地调试正常,整合到 Greasemonkey 怎么就提示 is not defined 了?

代码本地调试可以正确回调alert弹窗显示JSONP中的内容 可是一旦整合在Greasemonkey中就变成函数没有被定义了.这是什么问题?

以下是本地代码,测试中毫无问题

<script src="http://upcdn.b0.upaiyun.com/libs/jquery/jquery-2.0.3.min.js"></script>

<script type="text/javascript">$.getJSON("JSONP网络路径", function(json){alert("JSON Data: " + json.translation);})</script>

改成Greasemonkey脚本就出错了

// ==UserScript==
// @name Hello World
// @description 暂无描述
// @require http://upcdn.b0.upaiyun.com/libs/jquery/jquery-2.0.3.min.js
// @include *
// ==/UserScript==

$.getJSON("JSONP网络路径", function(json){alert("JSON Data: " + json.translation);})

Greasemonkey脚本firebug调试时提示错误如下:
ReferenceError: jQuery203016154597985419905_1435415657339 is not defined
jQuery203016154597985419905_1435415657339({"translation":["酷"],"basic":{"us-phonetic":"kul","phonetic":"kuːl","uk-phonetic":"kuːl","explains":["n. 凉爽；凉爽的空气","adj. 凉爽的；冷静的；出色的","vt. 使\u2026冷却；使\u2026平静下来","vi. 变凉；平息","adv. 冷静地","n. (Cool)人名；(法)科尔；(英)库尔"]},"query":"cool","errorCode":0,"web":[{"value":["酷","凉爽的","冷静"],"key":"Cool"},{"value":["冷爵士乐","冷爵士","酷派爵士"],"key":"Cool jazz"},{"value":["用烘干机低温烘干","打冷风","用烘焙机低温烘焙"],"key":"Tumble Cool"}]});