Question

我对PostgrSQL和SQL很新。我正在尝试使用FULL OUTER JOIN函数根据相应的11列连接11个表。您可以在下面找到我的代码：

SELECT * FROM _0_general_view
FULL OUTER JOIN _1_foundation_view
ON _0_general_view.concatenate_0_general = _1_foundation_view.concatenate_1_foundation
FULL OUTER JOIN _1_plinth_view
ON _0_general_view.concatenate_0_general = _1_plinth_view.concatenate_1_plinth
FULL OUTER JOIN _1_rc_beams_columns_view
ON _0_general_view.concatenate_0_general = _1_rc_beams_columns_view.concatenate_1_rc_beams_columns
FULL OUTER JOIN _1_vertical_members_view
ON _0_general_view.concatenate_0_general = _1_vertical_members_view.concatenate_1_vertical_members
FULL OUTER JOIN _2_floor_view
ON _0_general_view.concatenate_0_general = _2_floor_view.concatenate_2_floor
FULL OUTER JOIN _2_horizontal_bands_view
ON _0_general_view.concatenate_0_general = _2_horizontal_bands_view.concatenate_2_horizontal_bands
FULL OUTER JOIN _2_rc_beams_columns_view
ON _0_general_view.concatenate_0_general = _2_rc_beams_columns_view.concatenate_2_rc_beams_columns
FULL OUTER JOIN _2_vertical_members_view
ON _0_general_view.concatenate_0_general = _2_vertical_members_view.concatenate_2_vertical_members
FULL OUTER JOIN _2_walls_view
ON _0_general_view.concatenate_0_general = _2_walls_view.concatenate_2_walls
FULL OUTER JOIN _3_roof_view
ON _0_general_view.concatenate_0_general = _3_roof_view.concatenate_3_roof
;

在上面的代码中，我假设我的表格结构如下：

表1 （_0_general_view）

╔══════════╦═════════╦═════════╗
║   col_1  ║  col_2  ║  col_3  ║
╠══════════╬═════════╬═════════╣
║     a    ║ 3.5046  ║   Jan   ║
║     b    ║ 3.7383  ║   Mar   ║
║     c    ║ 3.9719  ║   Jul   ║
║     d    ║ 6.1915  ║   Feb   ║
╚══════════╩═════════╩═════════╝

表2 （_1_plinth_view）

╔══════════╦═════════╦═════════╗
║   col_4  ║  col_5  ║  col_6  ║
╠══════════╬═════════╬═════════╣
║     a    ║ 2.8846  ║   Dec   ║
║     d    ║ 5.2244  ║   Aug   ║
╚══════════╩═════════╩═════════╝

表n （_3_xxxx_view）

╔══════════╦═════════╦═════════╗
║   col_7  ║  col_8  ║  col_9  ║
╠══════════╬═════════╬═════════╣
║     b    ║ 1.2365  ║   May   ║
║     c    ║ 2.5432  ║   Sep   ║
║     d    ║ 8.1515  ║   Oct   ║
╚══════════╩═════════╩═════════╝

换句话说，我假设col_1（在第一个表/视图中名为concatenate_0_general，名为_0_general_view）包含我需要加入的所有记录，但是如果这不是真的怎么办？？

我无法找到一种方法来考虑我的11个表中的所有可能组合。是否有可能以某种方式？

如果是这样，考虑到每个表平均有150列，并且每个表中的行数可能大约为10000，您认为这将是一个非常漫长的过程。

编辑：可能我们将不得不以更少的记录进行检索，让我们说300-400左右，因为我们将添加一个WHERE子句来按日期过滤它们。

谢谢，Stefano

Answer 1

如果join列的名称相同，则可以使用using：

select . . .
from a join
     b
     on (id) join
     c
     on (id) . . .;

在您的情况下，union all和聚合是最好的方法 - 假设您每个“id”最终会有一行：

select id, max(col1) as col1, max(col2) as col2, max(col3) as col3,
       max(col4) as col4, max(col5) as col5, max(col6) as col6
from ((select concatenate_0_general as id, col1, col2, NULL as col3, NULL as col4, NULL as col5, NULL as col6
       from _0_general_view
      ) union all
      (select concatenate_1_foundation, NULL, NULL, col3, col4, NULL, NULL
       from _1_foundation_view
      ) union all
      (select concatenate_1_plinth, NULL, NULL, NULL, NULL, col5, col6
       from _1_plinth_view
      )
     ) x
group by id;

如果这些都不起作用，那么您可以构建完整的ID列表并使用left join：

select . . .
from ((select concatenate_0_general as id from _0_general_view
      ) union -- on purpose to remove duplicates
      (select concatenate_1_foundation from _1_foundation_view
      ) union
      (select concatenate_1_plinth from _1_plinth_view
      ) union
      . . .
     ) ids left join
     _0_general_view gv
     on gv.concatenate_0_general = ids.id left join
     . . .

Answer 2

假设第一列是唯一的（否则您的结果集将会爆炸，无论是左连接还是全连接），您可以先创建所有现有值的列表，然后将所有表连接到左边：

select _0_general_view.*,
   _1_foundation_view.*,
   _1_plinth_view.*
...
from
 (
   select concatenate_0_general as col from _0_general_view
   union 
   select concatenate_1_foundation from _1_foundation_view
   union
   select concatenate_1_plinth from _1_plinth_view
   ...
 ) as dt
left join _0_general_view 
  on col = concatenate_0_general
left join _1_foundation_view
  on col = concatenate_1_foundation
left join _1_plinth_view
  on col = concatenate_1_plinth
...

PostgreSQL，多个表上的FULL OUTER JOIN（所有“组合”）

2 个答案: