如何根据第一个表中的非唯一 id 连接两个表（SQLite 3.12.2）

Question

表 A 的序列号是随机生成的。 然后表 B 会根据表 A 的序列号和类别来生成它的声音。

我试过这个命令，但失败了。

SELECT          A.timestamp, B.timestamp, B.SerialID, A.Category, B.Sound
FROM            A 
JOIN            B
ON              B.timestamp > A.timestamp
AND             A.SerialID = B.SerialID
ORDER BY        B.timestamp

表 A

Timestamp	SerialID	Category
1	3	Cat
2	5	Dog
10	44	Cat
13	5	Cat
15	3	Dog

表 B

Timestamp	SerialID	Sound
3	3	Meow
4	5	Bark
11	44	Meow
14	5	Meow
16	3	Bark

Desire表：表B的第一个匹配表A和B的序列id。Timestamp大于A.Timestamp

A.Timestamp	B.Timestamp	Serial ID	Category	Sound
1	3	3	Cat	Meow
2	4	5	Dog	Bark
10	11	44	Cat	Meow
13	14	5	Cat	Meow
15	16	3	Dog	Bark

Answer 1

SELECT
   t1.Timestamp AS TimestampA,
   t2.Timestamp AS TimestampB,
   t1.SerialID,
   t1.Category,
   t2.Sound
FROM t1
JOIN t2 ON t1.SerialID = t2.SerialID AND t2.Timestamp = (
  SELECT MIN(t2.Timestamp)
  FROM t2 
  WHERE t2.SerialID = t1.SerialID AND t2.Timestamp >= t1.Timestamp
)

或者同样使用LEAD窗口函数

WITH cte AS (
  SELECT *,
    LEAD(Timestamp) OVER (PARTITION BY SerialID ORDER BY Timestamp) as next_timestamp
  FROM t1
)
SELECT
   t1.Timestamp AS TimestampA,
   t2.Timestamp AS TimestampB,
   t1.SerialID,
   t1.Category,
   t2.Sound
FROM cte t1
JOIN t2 ON t1.SerialID = t2.SerialID 
       AND t2.Timestamp >= t1.Timestamp 
       AND (t2.Timestamp < t1.next_timestamp OR next_timestamp IS NULL)
ORDER BY t1.Timestamp  

db<>fiddle

Answer 2

我最喜欢的解决方案是横向连接，只需为每个 A 行选择所需的 B 行。但是 SQLite 中还没有横向连接。

您可以通过两个步骤获得所需的 B 行，方法是在 select 子句的子查询中选择 tmestamp：

select
  ab.timestamp as a_timestamp, b.timestamp as b_timestamp,
  b.serialid, ab.category, b.sound
from
(
  select a.*,
    (
      select b.timestamp
      from b
      where b.serialid = a.serialid
      and b.timestamp > a.timestamp
      order by b.timestamp
      limit 1
    ) as best_timestamp
  from a
) ab
join b on b.serialid = ab.serialid
      and b.timestamp = ab.best_timestamp
order by ab.timestamp, ab.serialid;

另一种方法使用窗口函数。在那里，您将加入所有候选人，然后保留最好的候选人。这是您添加了选择的查询。

select a_timestamp, b_timestamp, serialid, category, sound
from
(
  select 
    a.timestamp as a_timestamp, b.timestamp as b_timestamp,
    b.serialid, a.category, b.sound,
    min(b.timestamp) over (partition by a.serialid) as best_timestamp
  from a
  join b on b.serialid = a.serialid and b.timestamp > a.timestamp
) ab
where b_timestamp = best_timestamp
order by a_timestamp, ab.serialid;

Answer 3

您只需要一个连接和聚合，而不需要子查询：

SELECT a.Timestamp a_Timestamp,
       MIN(b.Timestamp) b_Timestamp,
       a.SerialID,
       a.Category,
       b.Sound
FROM TableA a INNER JOIN TableB b
ON b.SerialID = a.SerialID AND b.Timestamp > a.Timestamp
GROUP BY a.Timestamp, a.SerialID;

当使用MIN()聚合函数时，此代码依赖SQLite的特性来返回包含列最小值的行。

如果Timestamp 和SerialID 的组合在TableA 中不是唯一的，则更改为：

GROUP BY a.Timestamp, a.SerialID, a.Category 

请参阅demo。