我怎样才能写这个YQL语句为jquery jsonp

Question

我想与YQL和JQuery做一些跨域的东西，但我有点麻烦让这个yql查询正常工作。

好了，所以这是YQL语句我试图去上班：

use 'http://yqlblog.net/samples/data.html.cssselect.xml' as data.html.cssselect; select * from data.html.cssselect where url="www.holylandmoments.org/devotionals/the-sabbath-experience" and css="#main-content p" 

现在，我试图改变，以适应我想要做的YQL语句是：

var yql = 'http://query.yahooapis.com/v1/public/yql?q=' + encodeURIComponent('select * from html where url="' + site + '"') + '&format=xml&callback=?'; 

从我为了读过我能够挑选CSS选择我需要调用一个开放的数据表。

我不知道我可以如何更改示例yql语句以将USE和AS包含到查询中。

任何帮助都可以。

Answer 1

好，我知道了......如果任何人都需要这样的东西来看看：

// Accepts a url and a callback function to run. 

功能requestCrossDomain（网站，回调）{

// Take the provided url, and add it to a YQL query. Make sure you encode it! 
var yql = 'http://query.yahooapis.com/v1/public/yql?q=use' + "%20'http%3A%2F%2Fyqlblog.net%2Fsamples%2Fdata.html.cssselect.xml'%20as%20data.html.cssselect%3B%20" + encodeURIComponent('select * from data.html.cssselect where url="' + site + '"') + "%20and%20css%3D%22%23main-content%20p%22" + '&format=xml&callback=?'; 

// Request that YSQL string, and run a callback function. 
// Pass a defined function to prevent cache-busting. 
$.getJSON(yql, cbFunc); 

function cbFunc(data) { 
// If we have something to work with... 
if (data.results[0]) { 
    // Strip out all script tags, for security reasons. 
    // BE VERY CAREFUL. This helps, but we should do more. 
    data = data.results[0].replace(/<script[^>]*>[\s\S]*?<\/script>/gi, ''); 

    // If the user passed a callback, and it 
    // is a function, call it, and send through the data var. 
    if (typeof callback === 'function') { 
     callback(data); 
    } 
} 
// Else, Maybe we requested a site that doesn't exist, and nothing returned. 
else throw new Error('Nothing returned from getJSON.'); 
} 
} 

Answer 2

查看如何下面的例子从YQL获取消息JSONP

http://www.techtricky.com/jquery-code-to-display-top-news-using-yql-jsonp-api-service/