YQL中的xml XPath？

Question

所以，我正在尝试执行以下YQL查询：

http://developer.yahoo.com/yql/console/?q=select%20*%20from%20xml%20where%20url%3D%22http%3A%2F%2Fws.audioscrobbler.com%2F2.0%2F%3Fmethod%3Dartist.getsimilar%26artist%3Djayz%26api_key%3Dd47ca2514e350dd0dac6fc46a507585a%22%20#h=select%20*%20from%20xml%20where%20url%3D%22http%3A//ws.audioscrobbler.com/2.0/%3Fmethod%3Dartist.getsimilar%26artist%3Djayz%26api_key%3Dd47ca2514e350dd0dac6fc46a507585a%22

我希望结果只有艺术家的名字。如果有人可以帮助编写XPath以仅产生艺术家名称，那将非常感激！

Answer 1

喜欢这个吗？

select * from xml where url="http://ws.audioscrobbler.com/2.0/?method=artist.getsimilar&artist=jayz&api_key=d47ca2514e350dd0dac6fc46a507585a" and itemPath="//similarartists/artist/name"