YQL - 无法返回 yql 结果
我正在尝试使用其余查询返回
我正在查询的页面上的所有超链接。
这是我使用的 yql 查询
select * from html where url="http://www.stickam.com/videoPlaylist.do?uId=182005497" and xpath="//*[@class='mediaThum']/a"
这是代码，
<script src="jquery.1.6.1.js"></script>
<script>
$(document).ready(function(){
var yql = "http://query.yahooapis.com/v1/public/yql?q=%20SELECT%20*%20FROM%20html%20WHERE%20url%3D%22http%3A%2F%2Fwww.stickam.com%2FvideoPlaylist.do%3FuId%3D182005497%22%20and%20xpath%3D%22%2F%2F*%5B%40class%3D'mediaThum'%5D%2Fa%22%20";
$.get( yql, cbFunc );
function cbFunc(data) {
alert(data.query.results.a[0].href);
}//END FUNC
});//end document.ready
</script>
谢谢安东尼。
I am trying to return back all the
hyperlinks on the page that I am querying using the rest query.
this is the yql query I used
select * from html where url="http://www.stickam.com/videoPlaylist.do?uId=182005497" and xpath="//*[@class='mediaThum']/a"
here is the code
<script src="jquery.1.6.1.js"></script>
<script>
$(document).ready(function(){
var yql = "http://query.yahooapis.com/v1/public/yql?q=%20SELECT%20*%20FROM%20html%20WHERE%20url%3D%22http%3A%2F%2Fwww.stickam.com%2FvideoPlaylist.do%3FuId%3D182005497%22%20and%20xpath%3D%22%2F%2F*%5B%40class%3D'mediaThum'%5D%2Fa%22%20";
$.get( yql, cbFunc );
function cbFunc(data) {
alert(data.query.results.a[0].href);
}//END FUNC
});//end document.ready
</script>
Thanks Anthony.
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从 YQL 请求 JSON
您需要告诉 YQL 您希望查询得到 JSON 格式的响应。 YQL 网址必须包含
format=json。从 jQuery 请求 JSON
使用
$.getJSON函数代替$.get。There are several minor issues, I'll only cover a couple of them.
Ask for JSON from YQL
You need to tell YQL that you're expecting a JSON formatted response from your query. The YQL url must contain
format=json.Ask for JSON from jQuery
Use the
$.getJSONfunction in place of$.get.