jquery YQL变量不返回值

时间:2013-04-25 20:40:54

标签: jquery variables syntax yql

为什么这对avgRent不起作用? YQL控制台使用以下语句提供值:

select * from html where url="http://www.zillow.com/widgets/zestimate/ZestimateSmallWidget.htm?did=zillow-shv-small-iframe-widget&type=iframe&forRent=true&address='5894+Dogwood+Cir+35111'" and xpath='//span[@id="zestimate-value"]'

但我的代码不会返回值。我可以对变量avgRent做错了吗?

var postcode = $('#postcode').val();
            var address = $('#address').val();


            $.get(makeUrl('http://www.zillow.com/widgets/zestimate/ZestimateSmallWidget.htm?did=zillow-shv-small-iframe-widget&type=iframe&forRent=true&address=5894+Dogwood+Cir&csz=35111'),
                    function(data) {
                        var context = $('<div />').html($(data).find('content').text  ());

                        $('#avgRent').html($(context).find('span.zestimate-value').text());

                    });

                <tr>
    <td><strong>Average Monthly Rent: </strong></td>
                <td><div id="avgRent"></div></td>
            </tr>

0 个答案:

没有答案