YQL 中 xml 的 XPath？

Question

因此，我尝试执行以下 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%3Dd47ca25 14e350dd0dac6fc46a507585a%22%20#h=select%20*%20from%20xml%20where%20url%3D%22http%3A//ws.audiocrobbler。 com/2.0/%3Fmethod%3Dartist.getsimilar%26artist%3Djayz%26api_key%3Dd47ca2514e350dd0dac6fc46a507585a%22

我希望结果只包含艺术家姓名。如果有人可以帮助编写 XPath 以仅生成艺术家姓名，我们将不胜感激！

Answer 1

像这样？

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