1

我正在尝试使用 a 将多个表连接在一起full outer join，它接近正确的结果，但由于连接子句存在一些重复的行。我有几个表，其中包含 id、date、value 列。我正在寻找一个表，其中每个 id、日期对都有一行，其中包含每个表中的所有值。

如果你想玩它，这里有一个 SQLFiddle。

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，但是当例如 T4 有一个不在 T1 中的 ID,event_dt 对时，我会得到一些重复的行（正如预期的那样，因为这就是我要加入的）。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展平/合并在一起，或者有没有更好的方法来解决这个问题？

4

4 回答 4

3

您始终可以在列周围使用聚合amount：

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;

看演示

于 2013-05-10T16:49:59.893 回答
3

我认为您加入-citeria 根本不是您真正想要的。这个应该可以解决问题：

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为您提供了 2012-04-06 所需的输出。

于 2013-05-10T17:07:28.680 回答
2

捕获 NULL，用零替换它们，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

这是小提琴

于 2013-05-10T16:55:01.927 回答
1

（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)

顺便说一句：在UNION四个之后，LEFT JOINs 将与此处的 s 执行相同的操作FULL JOIN（联合四已经拥有所有可能的 {id, event_dt} 对）

于 2013-05-10T18:01:06.060 回答