Question

我正在尝试使用YQL和Jquery做一些跨域的东西，但是我在使这个yql查询正常工作时遇到了一些麻烦。

好的，这是我正在尝试开始工作的YQL语句：

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=?';

根据我的阅读，为了能够选择和选择css选择器，我需要调用一个开放数据表。

我不知道是什么或如何更改示例yql语句以将USE和AS包含在查询中。

任何帮助都可以。

Answer 1

好吧，我明白了......如果其他人需要这样的东西看看：

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

function requestCrossDomain（site，callback）{

// 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

请参阅以下有关如何从YQL JSONP获取新闻的示例

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

如何为jquery jsonp编写这个YQL语句

2 个答案: