我有关系
+-----+----+
| seq | id |
+-----+----+
| 1 | A1 |
| 2 | B1 |
| 3 | C1 |
| 4 | D1 |
+-----+----+
并希望通过以下方式将其加入PostgreSQL
+----+-------+
| id | alter |
+----+-------+
| B1 | B2 |
| D1 | D2 |
+----+-------+
因此我得到了所有可能的替换组合(即或多或少替换的笛卡尔积)。因此，组1没有更新，组2仅是B2，组3仅是D2，组4均为B2和D2。
末端应该看起来像这样，但是应该向更多的人开放(例如为D1额外添加一个D3)
+-------+-----+----+
| group | seq | id |
+-------+-----+----+
| 1 | 1 | A1 |
| 1 | 2 | B1 |
| 1 | 3 | C1 |
| 1 | 4 | D1 |
| 2 | 1 | A1 |
| 2 | 2 | B2 |
| 2 | 3 | C1 |
| 2 | 4 | D1 |
| 3 | 1 | A1 |
| 3 | 2 | B1 |
| 3 | 3 | C1 |
| 3 | 4 | D2 |
| 4 | 1 | A1 |
| 4 | 2 | B2 |
| 4 | 3 | C1 |
| 4 | 4 | D2 |
+-------+-----+----+
编辑:
另一个可能的替换表可能是
+----+-------+
| id | alter |
+----+-------+
| B1 | B2 |
| D1 | D2 |
| D1 | D3 |
+----+-------+
可能会导致6组(我希望我没有忘记一个案子)
+-------+-----+----+
| group | seq | id |
+-------+-----+----+
| 1 | 1 | A1 |
| 1 | 2 | B1 |
| 1 | 3 | C1 |
| 1 | 4 | D1 |
| 2 | 1 | A1 |
| 2 | 2 | B2 |
| 2 | 3 | C1 |
| 2 | 4 | D1 |
| 3 | 1 | A1 |
| 3 | 2 | B2 |
| 3 | 3 | C1 |
| 3 | 4 | D2 |
| 4 | 1 | A1 |
| 4 | 2 | B2 |
| 4 | 3 | C1 |
| 4 | 4 | D3 |
| 5 | 1 | A1 |
| 5 | 2 | B1 |
| 5 | 3 | C1 |
| 5 | 4 | D2 |
| 6 | 1 | A1 |
| 6 | 2 | B1 |
| 6 | 3 | C1 |
| 6 | 4 | D3 |
+-------+-----+----+
如果您有三个替代品，例如
+----+-------+
| id | alter |
+----+-------+
| B1 | B2 |
| C1 | C2 |
| D1 | D3 |
+----+-------+
这将导致8组。
到目前为止，我所做的尝试并没有真正的帮助:
WITH a as (SELECT * FROM (values (1,'A1'),(2,'B1'), (3,'C1'), (4,'D1') ) as a1(seq, id) )
, b as (SELECT * FROM (values ('B1','B2'), ('D1','D2')) as b1(id,alter) )
---------
SELECT row_number() OVER (PARTITION BY a.id) as g, * FROM
a
CROSS JOIN b as b1
CROSS JOIN b as b2
LEFT JOIN b as b3 ON a.id=b3.id
ORDER by g,seq;
我很高兴为标题提供更好的建议。
最佳答案
修改问题后的答案已更新
这个问题中最棘手的部分是生成替换的功率集。但是，幸运的是，postgres支持递归查询，并且可以递归计算幂集。因此，我们可以为该问题建立一个通用的解决方案，无论替换集的大小如何，该解决方案都将起作用。
让我们将第一个表称为source，将第二个表称为replacements，我将避免使用其他名称来代替讨厌的名称alter:
CREATE TABLE source (seq, id) as (
VALUES (1, 'A1'), (2, 'B1'), (3, 'C1'), (4, 'D1')
);
CREATE TABLE replacements (id, sub) as (
VALUES ('B1', 'B2'), ('D1', 'D2')
);
需要生成要替换的ID的第一个功率集。空集可能会被省略，因为它无论如何都无法与联接一起使用，最后
source表可以被union'到中间结果以提供相同的输出。在递归步骤中，JOIN条件
rec.id > repl.id确保每个id对于每个生成的子集仅出现一次。最后一步:
交叉联接将源散开N次，其中N是替换的非空组合数(有变化)
组名是使用
seq上经过过滤的运行时总和生成的。如果替换ID等于源ID，则子集是未嵌套的＆合并ID。
WITH RECURSIVE rec AS (
SELECT ARRAY[(id, sub)] subset, id FROM replacements
UNION ALL
SELECT subset || (repl.id, sub), repl.id
FROM replacements repl
JOIN rec ON rec.id > repl.id
)
SELECT NULL subset, 0 set_name, seq, id FROM source
UNION ALL
SELECT subset
, SUM(seq) FILTER (WHERE seq = 1) OVER (ORDER BY subset, seq) set_name
, seq
, COALESCE(sub, source.id) id
FROM rec
CROSS JOIN source
LEFT JOIN LATERAL (
SELECT id, sub
FROM unnest(subset) x(id TEXT, sub TEXT)
) x ON source.id = x.id;
测验
使用替换值
('B1', 'B2'), ('D1', 'D2')，查询返回4个组。 subset | set_name | seq | id
-----------------------+----------+-----+----
| 0 | 1 | A1
| 0 | 2 | B1
| 0 | 3 | C1
| 0 | 4 | D1
{"(B1,B2)"} | 1 | 1 | A1
{"(B1,B2)"} | 1 | 2 | B2
{"(B1,B2)"} | 1 | 3 | C1
{"(B1,B2)"} | 1 | 4 | D1
{"(D1,D2)"} | 2 | 1 | A1
{"(D1,D2)"} | 2 | 2 | B1
{"(D1,D2)"} | 2 | 3 | C1
{"(D1,D2)"} | 2 | 4 | D2
{"(D1,D2)","(B1,B2)"} | 3 | 1 | A1
{"(D1,D2)","(B1,B2)"} | 3 | 2 | B2
{"(D1,D2)","(B1,B2)"} | 3 | 3 | C1
{"(D1,D2)","(B1,B2)"} | 3 | 4 | D2
(16 rows)
使用替换值
('B1', 'B2'), ('D1', 'D2'), ('D1', 'D3')，该查询返回6组: subset | set_name | seq | id
-----------------------+----------+-----+----
| 0 | 1 | A1
| 0 | 2 | B1
| 0 | 3 | C1
| 0 | 4 | D1
{"(B1,B2)"} | 1 | 1 | A1
{"(B1,B2)"} | 1 | 2 | B2
{"(B1,B2)"} | 1 | 3 | C1
{"(B1,B2)"} | 1 | 4 | D1
{"(D1,D2)"} | 2 | 1 | A1
{"(D1,D2)"} | 2 | 2 | B1
{"(D1,D2)"} | 2 | 3 | C1
{"(D1,D2)"} | 2 | 4 | D2
{"(D1,D2)","(B1,B2)"} | 3 | 1 | A1
{"(D1,D2)","(B1,B2)"} | 3 | 2 | B2
{"(D1,D2)","(B1,B2)"} | 3 | 3 | C1
{"(D1,D2)","(B1,B2)"} | 3 | 4 | D2
{"(D1,D3)"} | 4 | 1 | A1
{"(D1,D3)"} | 4 | 2 | B1
{"(D1,D3)"} | 4 | 3 | C1
{"(D1,D3)"} | 4 | 4 | D3
{"(D1,D3)","(B1,B2)"} | 5 | 1 | A1
{"(D1,D3)","(B1,B2)"} | 5 | 2 | B2
{"(D1,D3)","(B1,B2)"} | 5 | 3 | C1
{"(D1,D3)","(B1,B2)"} | 5 | 4 | D3
(24 rows)
使用替换值
('B1', 'B2'), ('C1', 'C2'), ('D1', 'D2')，查询返回8组: subset | set_name | seq | id
---------------------------------+----------+-----+----
| 0 | 1 | A1
| 0 | 2 | B1
| 0 | 3 | C1
| 0 | 4 | D1
{"(B1,B2)"} | 1 | 1 | A1
{"(B1,B2)"} | 1 | 2 | B2
{"(B1,B2)"} | 1 | 3 | C1
{"(B1,B2)"} | 1 | 4 | D1
{"(C1,C2)"} | 2 | 1 | A1
{"(C1,C2)"} | 2 | 2 | B1
{"(C1,C2)"} | 2 | 3 | C2
{"(C1,C2)"} | 2 | 4 | D1
{"(C1,C2)","(B1,B2)"} | 3 | 1 | A1
{"(C1,C2)","(B1,B2)"} | 3 | 2 | B2
{"(C1,C2)","(B1,B2)"} | 3 | 3 | C2
{"(C1,C2)","(B1,B2)"} | 3 | 4 | D1
{"(D1,D2)"} | 4 | 1 | A1
{"(D1,D2)"} | 4 | 2 | B1
{"(D1,D2)"} | 4 | 3 | C1
{"(D1,D2)"} | 4 | 4 | D2
{"(D1,D2)","(B1,B2)"} | 5 | 1 | A1
{"(D1,D2)","(B1,B2)"} | 5 | 2 | B2
{"(D1,D2)","(B1,B2)"} | 5 | 3 | C1
{"(D1,D2)","(B1,B2)"} | 5 | 4 | D2
{"(D1,D2)","(C1,C2)"} | 6 | 1 | A1
{"(D1,D2)","(C1,C2)"} | 6 | 2 | B1
{"(D1,D2)","(C1,C2)"} | 6 | 3 | C2
{"(D1,D2)","(C1,C2)"} | 6 | 4 | D2
{"(D1,D2)","(C1,C2)","(B1,B2)"} | 7 | 1 | A1
{"(D1,D2)","(C1,C2)","(B1,B2)"} | 7 | 2 | B2
{"(D1,D2)","(C1,C2)","(B1,B2)"} | 7 | 3 | C2
{"(D1,D2)","(C1,C2)","(B1,B2)"} | 7 | 4 | D2
(32 rows)
关于SQL:使用所有可能的组合进行更新，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/61106218/