SQLite：具有 30 列以上的四个表的完全外连接

Question

这个问题是这个post的扩展
我想用 SQLite 连接四个不同的表，它们只有两列共有。但是，假设有 30 多列，即不仅仅是列 a - h。请看下面的例子

表 1：

a   b   lon lat
---------------
22  33  11  22

表 2：

c   d   lon lat
---------------
1   2   44  45

Table3

e       f       lon lat
-----------------------
NULL    NULL    100 101

Table4

g       h       lon lat
-----------------------
NULL    NULL    200 201

目前的解决方案如下

SELECT a,b,NULL AS c, NULL AS d,NULL AS e, NULL AS f, NULL AS g, NULL AS h,
lon,lat
FROM table1
UNION ALL
SELECT NULL, NULL,c,d,NULL AS e, NULL AS f, NULL AS g, NULL AS h, lon,lat
FROM table2
UNION ALL
SELECT NULL, NULL,NULL,NULL,e,f, NULL AS g, NULL AS h, lon,lat
FROM table3
UNION ALL
SELECT NULL, NULL,NULL,NULL,NULL,NULL,g,h, lon,lat
from table4

结果：

+------+------+------+------+------+------+------+------+-----+-----+
|  a   |  b   |  c   |  d   |  e   |  f   |  g   |  h   | lon | lat |
+------+------+------+------+------+------+------+------+-----+-----+
| 22   | 33   | NULL | NULL | NULL | NULL | NULL | NULL |  11 |  22 |
| NULL | NULL | 1    | 2    | NULL | NULL | NULL | NULL |  44 |  45 |
| NULL | NULL | NULL | NULL | NULL | NULL | NULL | NULL | 100 | 101 |
| NULL | NULL | NULL | NULL | NULL | NULL | NULL | NULL | 200 | 201 |
+------+------+------+------+------+------+------+------+-----+-----+

DEMO

• 问题：如果我不仅有 a 到 h 列，还有 a 到 z 列，即 table1、table2、table3 和 table4 中的许多列 -> 这将非常耗时且结构不会清楚我是否必须在我的 sql 语句中到处写 NULL 作为 [字母]
• @zarruq 帮了大忙，他建议在这种情况下，我可以使用 `UNION ALL` 然后使用 `PIVOT` 将列转换为行
• 但是，我不知道该怎么做。而且，我不知道 100% 他的意思。
• 编辑：SQLite 不支持数据透视：还有其他建议吗？

Answer 1

当您有 30 多列时，我认为没有一种特别巧妙的方法可以做到这一点。以下是我能做的最好的事情，使用嵌套 CTE 实现 full outer joins，然后使用 coalesce 选择第一个非空纬度/经度。

仍然需要枚举顶部 SELECT 语句中的所有 30 多个字段，但至少不需要 NULL AS ... 的海量列表：

SELECT 
  a, b, c, d, e, f, g, h,
  coalesce(lat1, lat2, lat3, lat4) AS lat,
  coalesce(lon1, lon2, lon3, lon4) AS lon
  FROM (
    WITH t1_x_t2 AS (
        SELECT t1.*, t2.*, 
        t1.lat AS lat1, t2.lat AS lat2, t1.lon AS lon1, t2.lon AS lon2
        FROM table1 t1 LEFT OUTER JOIN table2 t2 ON 0
      UNION ALL
        SELECT t1.*, t2.*, 
        t1.lat AS lat1, t2.lat AS lat2, t1.lon AS lon1, t2.lon AS lon2
        FROM table2 t2 LEFT OUTER JOIN table1 t1 ON 0
    ), t3_x_t4 AS (
        SELECT t3.*, t4.*, 
        t3.lat AS lat3, t4.lat AS lat4, t3.lon AS lon3, t4.lon AS lon4
        FROM table3 t3 LEFT OUTER JOIN table4 t4 ON 0
      UNION ALL
        SELECT t3.*, t4.*, 
        t3.lat AS lat3, t4.lat AS lat4, t3.lon AS lon3, t4.lon AS lon4
        FROM table4 t4 LEFT OUTER JOIN table3 t3 ON 0
    )
    SELECT t1_x_t2.*, t3_x_t4.* FROM t1_x_t2 LEFT OUTER JOIN t3_x_t4 ON 0
    UNION ALL
    SELECT t1_x_t2.*, t3_x_t4.* FROM t3_x_t4 LEFT OUTER JOIN t1_x_t2 ON 0
)

Answer 2

使用pivot()（如果您的dbms 支持）导出结果的一种方法可能如下。

select a,b,c,d,e,f,g,h,lon,lat
from (
select 'a' as columnName1, a as val1,'b' as ColumnName2, b as val2, lon,lat
from table1
union all
select 'c',c,'d', d, lon,lat
from table2
union all
select 'e',e,'f', f, lon,lat
from table3
union all
select 'g',g,'h', h, lon,lat
from table4
) t1
PIVOT (MAX(val1)
      FOR columnName1 IN (a,c,e,g)) as Pivot1
PIVOT (MAX(val2)
      FOR columnName2 IN (b,d,f,h)) AS Pivot2;

结果：

+------+------+------+------+------+------+------+------+-----+-----+
|  a   |  b   |  c   |  d   |  e   |  f   |  g   |  h   | lon | lat |
+------+------+------+------+------+------+------+------+-----+-----+
| 22   | 33   | NULL | NULL | NULL | NULL | NULL | NULL |  11 |  22 |
| NULL | NULL | 1    | 2    | NULL | NULL | NULL | NULL |  44 |  45 |
| NULL | NULL | NULL | NULL | NULL | NULL | NULL | NULL | 100 | 101 |
| NULL | NULL | NULL | NULL | NULL | NULL | NULL | NULL | 200 | 201 |
+------+------+------+------+------+------+------+------+-----+-----+

DEMO

附言请注意，上面的查询是针对sql-Server，所以你可能需要为你的dbms tweek 它。