Question

我正在尝试使用jQuery获取外部URL的HTML，但我无法让下面的代码工作。

任何人都可以告诉我如何使这项工作吗？

<script language="javascript" type="text/javascript">
$.ajax({
    url: 'http://news.bbc.co.uk',
    type: 'GET',
    success: function(res) {
        var headline = $(res.responseText).find('a.tsh').text();
        alert(headline);
    }
});
</script>

Answer 1

// 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.');
}
}

调用函数

requestCrossDomain('http://www.cnn.com', function(results) {
   $('#container').html(results);
});

如何使用JS或jQuery获取外部HTML

1 个答案: