2

ここにスクリプトがあり、これからほとんど直接コピーされています。以下にリストされているコードが何も返さないのはなぜですか？

ajax.html：

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN"
  "http://www.w3.org/TR/html4/strict.dtd">

<html dir="ltr" lang="en-US">
    <head>
        <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
        <title>Cross-Domain Ajax Demo</title>
    </head>
    <body>
        <div id="container">
            <form>
                <p><label>Type a URL:</label><input type="text" name="sitename" id="sitename"/></p>
                <p><input type="submit" name="submit" id="submit" value="Make Cross Domain Ajax request"</p>
            </form>
        </div>

        <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4/jquery.min.js" charset="utf-8"></script>
        <script type="text/javascript" src="cross-domain-requests.js"></script>

        <script type="text/javascript">
            $('form').submit(function() {
                var path = "www.google.com";
                requestCrossDomain(path, function(results) {
                    $('#container').html(results);
                });
                return false;
            });
        </script>
    </body>
</html>

cross-domain-requests.js：

// Accepts a URL and a callback function to run.
function requestCrossDomain( site, callback ) {

    // 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 );

    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.');
    }
}

（私はスクリプトとAjaxに比較的慣れていないので、愚かなことをした場合は事前に謝罪します。）

4

1 に答える 1

3

varyqlのコールバックをcallback=？に変更してみてください。そして、次のように'fromxml'へのselectステートメント：

var yql = 'http://query.yahooapis.com/v1/public/yql?q=' + encodeURIComponent('select * from xml where url="' + site + '"') + '&format=xml&callback=?';
于 2012-04-13T20:16:23.353 に答える