BigQuery：按日期将子选择合并为一行

Posted 2023-03-25

【中文标题】BigQuery：按日期将子选择合并为一行

【英文标题】：BigQuery: Consolidate subselect into one row by date

【发布时间】：2015-07-01 21:23:51

【问题描述】：

我正在尝试让包含子查询的 BigQuery 查询结果返回一行而不是两行。我正在查询日志文件，所以我需要的所有数据都在同一个字段中。字段中的数据示例如下：

/?cv=p15.0.9350&ctyp=sp&bits=64&os_bits=64&hl=fr&hl=fr&os=win&osv=6.2   

我一直在处理的查询如下：

SELECT day, Win, Mac 
  FROM
    (SELECT DATE(metadata.timestamp) AS day, COUNT(DISTINCT protoPayload.resource) AS Win
     FROM [su_dashboard_streamed_logs.appengine_googleapis_com_request_log_20150424]
     WHERE protoPayload.resource CONTAINS 'ctyp=sp'
     GROUP BY day),
    (SELECT DATE(metadata.timestamp) AS day, COUNT(DISTINCT protoPayload.resource) AS Mac
     FROM [request_log_20150424]
     WHERE protoPayload.resource CONTAINS 'ctyp=sm'
     GROUP BY day)
ORDER BY day

目前上面的查询返回：

Row day Win Mac  
1   2015-04-24  160516  null     
2   2015-04-24  null    109547  

我希望结果是：

Row day Win Mac
1 2015-04-24 160516 109547

有没有办法做到这一点？如果是这样，任何帮助将不胜感激。

谢谢

【参考方案1】：

你想JOIN 两个子选择而不是联合它们。在 BigQuery 中，a comma within a FROM clause indicates a union：

注意：与许多其他基于 SQL 的系统不同，BigQuery 使用逗号语法来表示表联合，而不是联合。

如果您在日期字段上JOIN，那么您可以将两行压缩为一行，如下所示：

SELECT table_1.day as day, table_1.Win as Win, table_2.Mac AS Mac
  FROM
    (SELECT DATE(metadata.timestamp) AS day, COUNT(DISTINCT protoPayload.resource) AS Win
     FROM [su_dashboard_streamed_logs.appengine_googleapis_com_request_log_20150424]
     WHERE protoPayload.resource CONTAINS 'ctyp=sp'
     GROUP BY day) AS table_1
  JOIN
    (SELECT DATE(metadata.timestamp) AS day, COUNT(DISTINCT protoPayload.resource) AS Mac
     FROM [request_log_20150424]
     WHERE protoPayload.resource CONTAINS 'ctyp=sm'
     GROUP BY day) AS table_2
  ON table_1.day = table_2.day
ORDER BY day

【讨论】：

非常感谢丹尼，我已经尝试了一天多的时间。你解决了我的一个主要问题。再次感谢！！