Question

我使用以下代码使用YQL执行跨域AJAX请求：

function requestCrossDomain( site, callback ) {

function cbFunc(data) {
// If we have something to work with...
alert("inside call back");
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.');
}   
// If no url was passed, exit.
if ( !site ) {
    alert('No site was passed.');
    return false;
}

// 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=' + encodeURIComponent('select * from html where url="' + site + '"') + '&format=xml&callback=cbFunc';

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

}

并按以下方式调用上述内容：
requestCrossDomain('http://www.cnn.com', function(results) { alert(results); });

当我在firefox中运行上面的代码时，虽然响应（在firebug控制台中）显示回调函数（cbFunc）内部的网站内容但它没有显示任何警告。
也是{{的结果1）}第5行未在firebug控制台中打印。

任何人都可以建议我哪里出错了或者上面有什么解释吗？
顺便说一句，我已经通过了：
http://tek-insight.blogspot.in/2010/05/cross-domain-ajax-request-proxy-json.html http://net.tutsplus.com/tutorials/javascript-ajax/quick-tip-cross-domain-ajax-request-with-yql-and-jquery/ 相关stackoverflow问题中的可能解释。

Answer 1

$.getJSON接受“成功”响应的回调函数。但是如果错误被返回（404,500等），那么它就不会调用这个函数您需要添加额外的功能以捕获其他响应方案：

$.getJSON( yql, cbFunc)
  .done(function() { console.log( "second success" ); })
  .fail(function(jqxhr, textStatus, error) { console.log( "error", textStatus, error ); })
  .always(function() { console.log( "complete" ); });

使用YQL执行跨域AJAX请求时什么也得不到

1 个答案: