【发布时间】:2014-08-18 05:10:07
【问题描述】:
从人中选择 *
<root>
<row>
<name>a</name>
<address>address1</address>
<loc_id>1</loc_id>
</row>
<row>
<name>b</name>
<address>address2</address>
<loc_id>2</loc_id>
</row>
</root>
从位置中选择 *
<root>
<row>
<id>1</id>
<name>location1</name>
<details>locationdetails1</details>
</row>
<row>
<id>2</id>
<name>location2</name>
<details>locationdetails2</details>
</row>
</root>
无论如何,在 YQL 中我们可以在 people.loc_idlocation.id 上连接这两个数据源并返回包含所有值的结果。我知道有命名冲突的可能性,但无论如何也解决这个问题?所以基本上任何可能有助于返回如下结果或类似结果的 yql 查询。
<root>
<row>
<name>a</name>
<address>address1</address>
<loc_id>1</loc_id>
<location.id>1</location.id>
<location.name>location1</location.name>
<location.details>locationdetails1</location.details>
</row>
<row>
<name>b</name>
<address>address2</address>
<loc_id>2</loc_id>
<location.id>2</location.id>
<location.name>location2</location.name>
<location.details>locationdetails2</location.details>
</row>
</root>
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标签: yql