Question

我不确定如何在BigQuery中编写此SQL查询。我有一个包含名称和时间戳的事件表。让我们说我表中只有两个事件：A和B.我想要做的是查询表以获取事件A的所有实例，并获得下一个最接近的B事件并创建一个新列时差。 B将始终发生在A。

之后

例如，如果我的表格如下：

A1 | 1:00 pm
B5 | 2:00 pm
A3 | 3:00 pm
B9 | 5:00 pm

我的结果表是：

A1 | 1 hour
A3 | 2 hours

我提出的查询如下：

SELECT
CAST(TIMESTAMP_DIFF((SELECT MIN(sub.time)
FROM table sub
WHERE sub.time > main.time), main.time, SECOND) AS INT64) duration 
FROM table main

这适用于获取上面我想要的表，但我还想在子查询中包含一个额外的列。看起来像：

A1 | 1 hour  | B5Column
A3 | 2 hours | B9Column

我尝试使用以下查询：

SELECT
(SELECT
 sub.SubQueryColumn
 FROM table sub
 WHERE sub.time > main.time
 ORDER BY sub.time asc
 LIMIT 1) SubColumn,
CAST(TIMESTAMP_DIFF((SELECT MIN(sub.time)
FROM table sub
WHERE sub.time > main.time), main.time, SECOND) AS INT64) duration 
FROM table main

但它不起作用。我得到的错误是

不支持引用其他表的相关子查询，除非它们可以解相关，例如将它们转换为有效的JOIN。

我可以帮忙解决这个问题吗？

Answer 1

这是一种方法：

select m.*,
       timestamp_diff(time, next_b_time, second) as duration
from (select m.*,
             min(case when event like 'B%' then time end) over (order by time desc) as next_b_time
      from main m
     )  m
where event like 'A%';

Answer 2

以下是BigQuery Standard SQL

#standardSQL
SELECT event, TIMESTAMP_DIFF(b_time, time, SECOND) duration, b_event
  FROM (
  SELECT event, time,
    LEAD(time) OVER(PARTITION BY grp ORDER BY time) b_time,
    LEAD(event) OVER(PARTITION BY grp ORDER BY time) b_event
  FROM (
    SELECT *, 
      COUNTIF(STARTS_WITH(event, 'A')) OVER(ORDER BY time) grp
    FROM `project.dataset.your_table` t
  )
)
WHERE STARTS_WITH(event, 'A')
-- ORDER BY time

您可以使用问题中的虚拟数据进行测试/播放，如下所示

#standardSQL
WITH `project.dataset.your_table` AS (
  SELECT 'A1' event, TIMESTAMP '2018-01-01 1:00:00' time UNION ALL
  SELECT 'B5', TIMESTAMP '2018-01-01 2:00:00' UNION ALL
  SELECT 'A3', TIMESTAMP '2018-01-01 3:00:00' UNION ALL
  SELECT 'B9', TIMESTAMP '2018-01-01 5:00:00' 
)
SELECT event, TIMESTAMP_DIFF(b_time, time, SECOND) duration, b_event
  FROM (
  SELECT event, time,
    LEAD(time) OVER(PARTITION BY grp ORDER BY time) b_time,
    LEAD(event) OVER(PARTITION BY grp ORDER BY time) b_event
  FROM (
    SELECT *, 
      COUNTIF(STARTS_WITH(event, 'A')) OVER(ORDER BY time) grp
    FROM `project.dataset.your_table` t
  )
)
WHERE STARTS_WITH(event, 'A')
ORDER BY time

结果为

Row event   duration    b_event  
1   A1      3600        B5   
2   A3      7200        B9

请注意：上述解决方案依赖于您问题中的陈述 - B will always happen after A，如果您的序列如下

WITH `project.dataset.your_table` AS (
  SELECT 'A1' event, TIMESTAMP '2018-01-01 1:00:00' time UNION ALL
  SELECT 'A2', TIMESTAMP '2018-01-01 1:30:00' UNION ALL
  SELECT 'B5', TIMESTAMP '2018-01-01 2:00:00' UNION ALL
  SELECT 'A3', TIMESTAMP '2018-01-01 3:00:00' UNION ALL
  SELECT 'B9', TIMESTAMP '2018-01-01 5:00:00' 
)

结果将是

Row event   duration    b_event  
1   A1      null        null     
2   A2      1800        B5   
3   A3      7200        B9

如果您需要解决此问题，请尝试以下

#standardSQL
WITH `project.dataset.your_table` AS (
  SELECT 'A1' event, TIMESTAMP '2018-01-01 1:00:00' time UNION ALL
  SELECT 'A2', TIMESTAMP '2018-01-01 1:30:00' UNION ALL
  SELECT 'B5', TIMESTAMP '2018-01-01 2:00:00' UNION ALL
  SELECT 'A3', TIMESTAMP '2018-01-01 3:00:00' UNION ALL
  SELECT 'B9', TIMESTAMP '2018-01-01 5:00:00' 
)
SELECT event, TIMESTAMP_DIFF(b_time, time, SECOND) duration, b_event
FROM (
  SELECT event, time, type, grp,
    FIRST_VALUE(event) OVER(ORDER BY grp RANGE BETWEEN 1 FOLLOWING AND 1 FOLLOWING) b_event,
    FIRST_VALUE(time) OVER(ORDER BY grp RANGE BETWEEN 1 FOLLOWING AND 1 FOLLOWING) b_time
  FROM (
    SELECT event, time, SUBSTR(event, 1, 1) type,
      COUNTIF(STARTS_WITH(event, 'B')) OVER(ORDER BY time) grp
    FROM `project.dataset.your_table` t
  )
)
WHERE STARTS_WITH(event, 'A')
ORDER BY time

此版本将返回

Row event   duration    b_event  
1   A1      3600        B5   
2   A2      1800        B5   
3   A3      7200        B9

SQL连接来自同一个表的时间戳差异

2 个答案: