Question

我正在尝试使用full outer join将多个表连接在一起，它接近正确的结果但由于join子句有一些重复的行。我有几个包含id，date，value列的表。我希望得到一个表，每个id都有一行，日期对包含每个表的所有值。

Here is a SQLFiddle if you want to play with it.

这是我到目前为止所得到的：


SELECT
  COALESCE(T1.ID, T2.ID, T3.ID, t4.id) AS ID,
  COALESCE(T1.event_dt, T2.event_dt, T3.event_dt, t4.event_dt) AS DATE,
  T1.AMT1, T2.AMT2, T3.AMT3, t4.AMT4
FROM T1
FULL OUTER JOIN T2
ON
  T2.id = T1.id
  AND T2.event_dt = T1.event_dt
FULL OUTER JOIN T3
ON
  T3.id = T1.id
  AND T3.event_dt = T1.event_dt
FULL OUTER JOIN T4
ON
  T4.id = T1.id
  AND T4.event_dt = T1.event_dt
ORDER BY ID, DATE

这几乎可以工作，但是我得到了一些重复的行，例如T4有一个ID，event_dt对不在T1中（因为这就是我正在加入的东西）。例如，我会得到类似的东西：


1   April, 06 2012 00:00:00+0000    (null)  2   (null)  (null)
1   April, 06 2012 00:00:00+0000    (null)  (null)  (null)  4
1   April, 06 2012 00:00:00+0000    (null)  (null)  3   (null)



When I'm looking to get:
1   April, 06 2012 00:00:00+0000    (null)   2   3   4

有没有办法将这些行压平/合并在一起，还是有更好的方法来解决这个问题？

Answer 1

您始终可以在amount列周围使用聚合：

SELECT
  COALESCE(T1.ID, T2.ID, T3.ID, t4.id) AS ID,
  COALESCE(T1.event_dt, T2.event_dt, T3.event_dt, t4.event_dt) AS DATE,
  max(coalesce(T1.AMT1, 0)) AMT1,  -- use coalesce to replace the null with zero
  max(coalesce(T2.AMT2, 0)) AMT2, 
  max(coalesce(T3.AMT3, 0)) AMT3, 
  max(coalesce(t4.AMT4, 0)) AMT4
FROM T1
FULL OUTER JOIN T2
  ON T2.id = T1.id
  AND T2.event_dt = T1.event_dt
FULL OUTER JOIN T3
  ON T3.id = T1.id
  AND T3.event_dt = T1.event_dt
FULL OUTER JOIN T4
  ON T4.id = T1.id
  AND T4.event_dt = T1.event_dt
group by  COALESCE(T1.ID, T2.ID, T3.ID, t4.id), 
  COALESCE(T1.event_dt, T2.event_dt, T3.event_dt, t4.event_dt)
ORDER BY ID, DATE;

请参阅Demo

Answer 2

我认为你加入citeria根本不是你真正想要的东西。这个应该做的伎俩：

SELECT
  COALESCE(T1.ID, T2.ID, T3.ID, t4.id) AS ID,
  COALESCE(T1.event_dt, T2.event_dt, T3.event_dt, t4.event_dt) AS DATE,
  T1.AMT1, T2.AMT2, T3.AMT3, t4.AMT4
FROM T1
FULL OUTER JOIN T2
ON
  T2.id = T1.id
  AND T2.event_dt = T1.event_dt
FULL OUTER JOIN T3
ON
  T3.id = coalesce(T1.id, T2.id)
  AND T3.event_dt = coalesce(T1.event_dt, T2.event_dt)
FULL OUTER JOIN T4
ON
  T4.id = coalesce(T1.id, T2.id, T3.id)
  AND T4.event_dt = coalesce(T1.event_dt, T2.event_dt, T3.event_dt)
ORDER BY ID, DATE

SQL-Fiddle here为您提供2012-04-06所需的输出。

Answer 3

陷阱为NULL，用零重新移动它们，然后在每列中找到MAX值。

SELECT
  COALESCE(T1.ID, T2.ID, T3.ID, t4.id) AS ID,
  COALESCE(T1.event_dt, T2.event_dt, T3.event_dt, t4.event_dt) AS DATE,
  max( coalesce(T1.AMT1,0)) as amt1
, max( coalesce(T2.AMT2,0)) as amt2
, max( coalesce(T3.AMT3,0)) as amt3
, max( coalesce(t4.AMT4,0)) as amt4
FROM T1
FULL OUTER JOIN T2
ON
  T2.id = T1.id
  AND T2.event_dt = T1.event_dt
FULL OUTER JOIN T3
ON
  T3.id = T1.id
  AND T3.event_dt = T1.event_dt
FULL OUTER JOIN T4
ON
  T4.id = T1.id
  AND T4.event_dt = T1.event_dt
group by   COALESCE(T1.ID, T2.ID, T3.ID, t4.id),
  COALESCE(T1.event_dt, T2.event_dt, T3.event_dt, t4.event_dt)
ORDER BY ID, DATE

这是Fiddle

Answer 4

（假设OP想要一个完全对称的外部4连接）

WITH four AS (
        SELECT id, event_dt FROM t1
        UNION
        SELECT id, event_dt FROM t2
        UNION
        SELECT id, event_dt FROM t3
        UNION
        SELECT id, event_dt FROM t4
        )
SELECT f.id, f.event_dt
        , t1.amt1
        , t2.amt2
        , t3.amt3
        , t4.amt4
FROM four f
LEFT JOIN t1 ON t1.id = f.id AND t1.event_dt = f.event_dt
LEFT JOIN t2 ON t2.id = f.id AND t2.event_dt = f.event_dt
LEFT JOIN t3 ON t3.id = f.id AND t3.event_dt = f.event_dt
LEFT JOIN t4 ON t4.id = f.id AND t4.event_dt = f.event_dt
ORDER BY id, event_dt
        ;

结果：

 id |  event_dt  | amt1 | amt2 | amt3 | amt4 
----+------------+------+------+------+------
  1 | 2012-04-01 |    1 |      |      |     
  1 | 2012-04-02 |    1 |      |    3 |     
  1 | 2012-04-03 |    1 |      |    3 |     
  1 | 2012-04-06 |      |    2 |    3 |    4
  1 | 2012-04-07 |      |    2 |      |     
  2 | 2012-04-01 |   40 |      |      |     
  2 | 2012-04-02 |      |      |    3 |     
  2 | 2012-04-03 |      |      |    3 |     
  2 | 2012-04-04 |   40 |      |      |     
(9 rows)

BTW：在UNION四之后，LEFT JOIN将与FULL JOIN s相同（联合四已经拥有所有可能的{id，event_dt}对）

SQL全外连接或替代解决方案

4 个答案: