如何在 SQL 中将来自不同变量的数据组合在一起？

Question

假设我有这样的数据：

USER_ID                 TIMESTAMP   data  data2
   0001   2021-05-09 12:13:03.445            44
   0001   2021-05-09 13:13:03.445    rob    
   0001   2021-05-09 11:13:03.445       
   0002   2021-05-09 09:13:03.445  perry    333
   0002   2021-05-09 12:13:03.445   carl    333
   0003   2021-05-09 16:13:03.445  mitch      1
   0003   2021-05-09 17:13:03.445
   0002   2021-05-09 16:13:03.445  mitch      5

我想做的就是从每一列中收集最新的非空值，并将它们压缩成一个 table，每一行都是一个条目。

最终结果：

USER_ID   data  data2
   0001    rob     44 
   0003  mitch      1
   0002  mitch      5

这是我的资料，但还不完整：

WITH form AS (
    select b.*,
        rank() over (
            partition by user_id
            order by timestamp DESC
        ) as num
    FROM b
SELECT *
FROM b
WHERE num = 1

Answer 1

嗯。 . .这是 ignore nulls 真正有用的地方——但 Postgres 不支持它（还？？）。

相反，您可以使用数组先对非 NULL 值排序，然后再按时间戳排序：

select user_id,
       (array_agg(data order by (data is not null) desc, timestamp desc))[1],
       (array_agg(data2 order by (data2 is not null) desc, timestamp desc))[1]
from t
group by user_id;

Here 是一个 db<>fiddle.

Answer 2

您可以使用 LAST_VALUE 或 FIRST_VALUE 函数来使用 IGNORE NULL。对于您的数据集：

WITH x AS (
SELECT *
FROM (VALUES ('0001','2021-05-09 12:13:03.445'::timestamp,NULL,44),
   ('0001','2021-05-09 13:13:03.445'::timestamp,'rob',NULL),
   ('0001','2021-05-09 11:13:03.445'::timestamp,NULL,NULL),
   ('0002','2021-05-09 09:13:03.445'::timestamp,'perry',333),
   ('0002','2021-05-09 12:13:03.445'::timestamp,'carl',333),
   ('0003','2021-05-09 16:13:03.445'::timestamp,'mitch',1),
   ('0003','2021-05-09 17:13:03.445'::timestamp,NULL,NULL),
   ('0002','2021-05-09 16:13:03.445'::timestamp,'mitch',5)
  ) x (id, ts, data, data2)
)

你会这样做：

SELECT id,
       LAST_VALUE(data) IGNORE NULLS OVER (PARTITION BY ID ORDER BY ts) as data_last,
       LAST_VALUE(data2) IGNORE NULLS OVER (PARTITION BY ID ORDER BY ts) as data2_last
FROM x
QUALIFY ROW_NUMBER() OVER (PARTITION BY id ORDER BY ts) = 1;

Answer 3

相关：Equivalent for Keep in Snowflake：

可以通过以下方式实现：

WITH cte(user_id, timestamp, "data", data2) AS (
  SELECT *
  FROM (VALUES ('0001','2021-05-09 12:13:03.445'::timestamp,NULL,44),
   ('0001','2021-05-09 13:13:03.445'::timestamp,'rob',NULL),
   ('0001','2021-05-09 11:13:03.445'::timestamp,NULL,NULL),
   ('0002','2021-05-09 09:13:03.445'::timestamp,'perry',333),
   ('0002','2021-05-09 12:13:03.445'::timestamp,'carl',333),
   ('0003','2021-05-09 16:13:03.445'::timestamp,'mitch',1),
   ('0003','2021-05-09 17:13:03.445'::timestamp,NULL,NULL),
   ('0002','2021-05-09 16:13:03.445'::timestamp,'mitch',5)
  ) 
)
SELECT user_id,
  (ARRAY_AGG("data") WITHIN GROUP (ORDER BY timestamp DESC))[0]::STRING AS "data",
  (ARRAY_AGG(data2)  WITHIN GROUP (ORDER BY timestamp DESC))[0] AS data2
FROM cte
GROUP BY user_id
ORDER BY user_id;

输出：

+---------+----------+-------+
| USER_ID |   data   | data2 |
+---------+----------+-------+
|    0001 | rob      |    44 |
|    0002 | mitch    |     5 |
|    0003 | mitch    |     1 |
+---------+----------+-------+

ARRAY_AGG 默认省略 NULL，并按时间戳降序排列。一旦创建了每个 user_id 的数组，它就是访问第一个元素（索引为 [0] 的元素）的问题。

如何在 SQL 中将来自不同变量的数据组合在一起？

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