postgresql:与数组的自连接

我的问题是关于为下面的用例生成PostgresSQL查询

方法#1

insert into table1 values 
(1,'a','uuid-1'),
(2,'a','uuid-2'),
(3,'a','uuid-3'),
(4,'a','uuid-4'),
(5,'a','uuid-5'),
(6,'b','uuid-1'),
(7,'b','uuid-2'),
(8,'b','uuid-3'),
(9,'b','uuid-6'),
(10,'c','uuid-1'),
(11,'c','uuid-2'),
(12,'c','uuid-3'),
(13,'c','uuid-6'),
(14,'c','uuid-7'),
(15,'d','uuid-6'),
(16,'d','uuid-2')

我有一个如下表，其中我跨不同类型（a、b、c、d）生成相同的uuid，就像映射不同类型一样

+----+------+-------------+
| id | type | master_guid |
+----+------+-------------+
|  1 | a    | uuid-1      |
|  2 | a    | uuid-2      |
|  3 | a    | uuid-3      |
|  4 | a    | uuid-4      |
|  5 | a    | uuid-5      |
|  6 | b    | uuid-1      |
|  7 | b    | uuid-2      |
|  8 | b    | uuid-3      |
|  9 | b    | uuid-6      |
| 10 | c    | uuid-1      |
| 11 | c    | uuid-2      |
| 12 | c    | uuid-3      |
| 13 | c    | uuid-6      |
| 14 | c    | uuid-7      |
| 15 | d    | uuid-6      |
| 16 | d    | uuid-2      |
+----+------+-------------+

方法#2

+----+-------------+
| id | master_guid |
+----+-------------+
|  1 | uuid-1      |
|  2 | uuid-2      |
|  3 | uuid-3      |
|  4 | uuid-4      |
|  5 | uuid-5      |
|  6 | uuid-1      |
|  7 | uuid-2      |
|  8 | uuid-3      |
|  9 | uuid-6      |
| 10 | uuid-1      |
| 11 | uuid-2      |
| 12 | uuid-3      |
| 13 | uuid-6      |
| 14 | uuid-7      |
| 15 | uuid-6      |
| 16 | uuid-2      |
+----+-------------+

insert into table1 values 
(1,'a'),
(2,'a'),
(3,'a'),
(4,'a'),
(5,'a'),
(6,'b'),
(7,'b'),
(8,'b'),
(9,'b'),
(10,'c'),
(11,'c'),
(12,'c'),
(13,'c'),
(14,'c'),
(15,'d'),
(16,'d')

insert into table2 values 
(1,'uuid-1'),
(2,'uuid-2'),
(3,'uuid-3'),
(4,'uuid-4'),
(5,'uuid-5'),
(6,'uuid-1'),
(7,'uuid-2'),
(8,'uuid-3'),
(9,'uuid-6'),
(10,'uuid-1'),
(11,'uuid-2'),
(12,'uuid-3'),
(13,'uuid-6'),
(14,'uuid-7'),
(15,'uuid-6'),
(16,'uuid-2')

我创建了两个表，分别输入id和主控guid，如下所示

表1:

+----+------+
| id | type |
+----+------+
|  1 | a    |
|  2 | a    |
|  3 | a    |
|  4 | a    |
|  5 | a    |
|  6 | b    |
|  7 | b    |
|  8 | b    |
|  9 | b    |
| 10 | c    |
| 11 | c    |
| 12 | c    |
| 13 | c    |
| 14 | c    |
| 15 | d    |
| 16 | d    |
+----+------+

表2

+----+-------------+
| id | master_guid |
+----+-------------+
|  1 | uuid-1      |
|  2 | uuid-2      |
|  3 | uuid-3      |
|  4 | uuid-4      |
|  5 | uuid-5      |
|  6 | uuid-1      |
|  7 | uuid-2      |
|  8 | uuid-3      |
|  9 | uuid-6      |
| 10 | uuid-1      |
| 11 | uuid-2      |
| 12 | uuid-3      |
| 13 | uuid-6      |
| 14 | uuid-7      |
| 15 | uuid-6      |
| 16 | uuid-2      |
+----+-------------+

insert into table1 values 
(1,'a'),
(2,'a'),
(3,'a'),
(4,'a'),
(5,'a'),
(6,'b'),
(7,'b'),
(8,'b'),
(9,'b'),
(10,'c'),
(11,'c'),
(12,'c'),
(13,'c'),
(14,'c'),
(15,'d'),
(16,'d')

insert into table2 values 
(1,'uuid-1'),
(2,'uuid-2'),
(3,'uuid-3'),
(4,'uuid-4'),
(5,'uuid-5'),
(6,'uuid-1'),
(7,'uuid-2'),
(8,'uuid-3'),
(9,'uuid-6'),
(10,'uuid-1'),
(11,'uuid-2'),
(12,'uuid-3'),
(13,'uuid-6'),
(14,'uuid-7'),
(15,'uuid-6'),
(16,'uuid-2')

我希望通过这两种方法获得如下输出：

+----+------+--------+------------+
| id | type |  uuid  | mapped_ids |
+----+------+--------+------------+
|  1 | a    | uuid-1 | [6,10]     |
|  2 | a    | uuid-2 | [7,11]     |
|  3 | a    | uuid-3 | [8,12]     |
|  4 | a    | uuid-4 | null       |
|  5 | a    | uuid-5 | null       |
+----+------+--------+------------+

我曾尝试在ID上使用array_agg进行自连接，并基于uuid进行分组，但未能获得所需的输出

使用以下查询填充数据：

方法#1

insert into table1 values 
(1,'a','uuid-1'),
(2,'a','uuid-2'),
(3,'a','uuid-3'),
(4,'a','uuid-4'),
(5,'a','uuid-5'),
(6,'b','uuid-1'),
(7,'b','uuid-2'),
(8,'b','uuid-3'),
(9,'b','uuid-6'),
(10,'c','uuid-1'),
(11,'c','uuid-2'),
(12,'c','uuid-3'),
(13,'c','uuid-6'),
(14,'c','uuid-7'),
(15,'d','uuid-6'),
(16,'d','uuid-2')

方法#2

+----+-------------+
| id | master_guid |
+----+-------------+
|  1 | uuid-1      |
|  2 | uuid-2      |
|  3 | uuid-3      |
|  4 | uuid-4      |
|  5 | uuid-5      |
|  6 | uuid-1      |
|  7 | uuid-2      |
|  8 | uuid-3      |
|  9 | uuid-6      |
| 10 | uuid-1      |
| 11 | uuid-2      |
| 12 | uuid-3      |
| 13 | uuid-6      |
| 14 | uuid-7      |
| 15 | uuid-6      |
| 16 | uuid-2      |
+----+-------------+

insert into table1 values 
(1,'a'),
(2,'a'),
(3,'a'),
(4,'a'),
(5,'a'),
(6,'b'),
(7,'b'),
(8,'b'),
(9,'b'),
(10,'c'),
(11,'c'),
(12,'c'),
(13,'c'),
(14,'c'),
(15,'d'),
(16,'d')

insert into table2 values 
(1,'uuid-1'),
(2,'uuid-2'),
(3,'uuid-3'),
(4,'uuid-4'),
(5,'uuid-5'),
(6,'uuid-1'),
(7,'uuid-2'),
(8,'uuid-3'),
(9,'uuid-6'),
(10,'uuid-1'),
(11,'uuid-2'),
(12,'uuid-3'),
(13,'uuid-6'),
(14,'uuid-7'),
(15,'uuid-6'),
(16,'uuid-2')

使用允许您为每个组聚合

id

s（在您的情况下，这些组是您的

uuid

s）

结果:

| id | type |   uuid | mapped_ids |
|----|------|--------|------------|
|  1 |    a | uuid-1 |     10,6,1 |
|  2 |    a | uuid-2 |  16,2,7,11 |
|  3 |    a | uuid-3 |     8,3,12 |
|  4 |    a | uuid-4 |          4 |
|  5 |    a | uuid-5 |          5 |
|  6 |    b | uuid-1 |     10,6,1 |
|  7 |    b | uuid-2 |  16,2,7,11 |
|  8 |    b | uuid-3 |     8,3,12 |
|  9 |    b | uuid-6 |    15,13,9 |
| 10 |    c | uuid-1 |     10,6,1 |
| 11 |    c | uuid-2 |  16,2,7,11 |
| 12 |    c | uuid-3 |     8,3,12 |
| 13 |    c | uuid-6 |    15,13,9 |
| 14 |    c | uuid-7 |         14 |
| 15 |    d | uuid-6 |    15,13,9 |
| 16 |    d | uuid-2 |  16,2,7,11 |

| id | type |   uuid | mapped_ids |
|----|------|--------|------------|
|  1 |    a | uuid-1 |       10,6 |
|  2 |    a | uuid-2 |    16,7,11 |
|  3 |    a | uuid-3 |       8,12 |
|  4 |    a | uuid-4 |            |
|  5 |    a | uuid-5 |            |
|  6 |    b | uuid-1 |       10,1 |
|  7 |    b | uuid-2 |    16,2,11 |
|  8 |    b | uuid-3 |       3,12 |
|  9 |    b | uuid-6 |      15,13 |
| 10 |    c | uuid-1 |        6,1 |
| 11 |    c | uuid-2 |     16,2,7 |
| 12 |    c | uuid-3 |        8,3 |
| 13 |    c | uuid-6 |       15,9 |
| 14 |    c | uuid-7 |            |
| 15 |    d | uuid-6 |       13,9 |
| 16 |    d | uuid-2 |     2,7,11 |

| id | type |   uuid | mapped_ids |
|----|------|--------|------------|
|  1 |    a | uuid-1 |       10,6 |
|  2 |    a | uuid-2 |    16,7,11 |
|  3 |    a | uuid-3 |       8,12 |
|  4 |    a | uuid-4 |     (null) |
|  5 |    a | uuid-5 |     (null) |
|  6 |    b | uuid-1 |       10,1 |
|  7 |    b | uuid-2 |    16,2,11 |
|  8 |    b | uuid-3 |       3,12 |
|  9 |    b | uuid-6 |      15,13 |
| 10 |    c | uuid-1 |        6,1 |
| 11 |    c | uuid-2 |     16,2,7 |
| 12 |    c | uuid-3 |        8,3 |
| 13 |    c | uuid-6 |       15,9 |
| 14 |    c | uuid-7 |     (null) |
| 15 |    d | uuid-6 |       13,9 |
| 16 |    d | uuid-2 |     2,7,11 |

这些数组当前还包含当前行的id（

id=1的
mapped_id
包含
1
）。这可以通过使用
array\u remove
删除此元素来纠正：

SELECT 
    id, type, master_guid as uuid,  
    array_remove(array_agg(id) OVER (PARTITION BY master_guid), id) as mapped_ids
FROM table1
ORDER BY id

结果:

| id | type |   uuid | mapped_ids |
|----|------|--------|------------|
|  1 |    a | uuid-1 |     10,6,1 |
|  2 |    a | uuid-2 |  16,2,7,11 |
|  3 |    a | uuid-3 |     8,3,12 |
|  4 |    a | uuid-4 |          4 |
|  5 |    a | uuid-5 |          5 |
|  6 |    b | uuid-1 |     10,6,1 |
|  7 |    b | uuid-2 |  16,2,7,11 |
|  8 |    b | uuid-3 |     8,3,12 |
|  9 |    b | uuid-6 |    15,13,9 |
| 10 |    c | uuid-1 |     10,6,1 |
| 11 |    c | uuid-2 |  16,2,7,11 |
| 12 |    c | uuid-3 |     8,3,12 |
| 13 |    c | uuid-6 |    15,13,9 |
| 14 |    c | uuid-7 |         14 |
| 15 |    d | uuid-6 |    15,13,9 |
| 16 |    d | uuid-2 |  16,2,7,11 |

| id | type |   uuid | mapped_ids |
|----|------|--------|------------|
|  1 |    a | uuid-1 |       10,6 |
|  2 |    a | uuid-2 |    16,7,11 |
|  3 |    a | uuid-3 |       8,12 |
|  4 |    a | uuid-4 |            |
|  5 |    a | uuid-5 |            |
|  6 |    b | uuid-1 |       10,1 |
|  7 |    b | uuid-2 |    16,2,11 |
|  8 |    b | uuid-3 |       3,12 |
|  9 |    b | uuid-6 |      15,13 |
| 10 |    c | uuid-1 |        6,1 |
| 11 |    c | uuid-2 |     16,2,7 |
| 12 |    c | uuid-3 |        8,3 |
| 13 |    c | uuid-6 |       15,9 |
| 14 |    c | uuid-7 |            |
| 15 |    d | uuid-6 |       13,9 |
| 16 |    d | uuid-2 |     2,7,11 |

| id | type |   uuid | mapped_ids |
|----|------|--------|------------|
|  1 |    a | uuid-1 |       10,6 |
|  2 |    a | uuid-2 |    16,7,11 |
|  3 |    a | uuid-3 |       8,12 |
|  4 |    a | uuid-4 |     (null) |
|  5 |    a | uuid-5 |     (null) |
|  6 |    b | uuid-1 |       10,1 |
|  7 |    b | uuid-2 |    16,2,11 |
|  8 |    b | uuid-3 |       3,12 |
|  9 |    b | uuid-6 |      15,13 |
| 10 |    c | uuid-1 |        6,1 |
| 11 |    c | uuid-2 |     16,2,7 |
| 12 |    c | uuid-3 |        8,3 |
| 13 |    c | uuid-6 |       15,9 |
| 14 |    c | uuid-7 |     (null) |
| 15 |    d | uuid-6 |       13,9 |
| 16 |    d | uuid-2 |     2,7,11 |

现在，例如
id=4
包含一个空数组，而不是
NULL
值。这可以通过使用
NULLIF
功能来实现。如果两个参数相等，则给出
NULL
，否则给出第一个参数

SELECT 
    id, type, master_guid as uuid,  
    NULLIF(
        array_remove(array_agg(id) OVER (PARTITION BY master_guid), id), 
        '{}'::int[]
    ) as mapped_ids 
FROM table1
ORDER BY id

结果:

| id | type |   uuid | mapped_ids |
|----|------|--------|------------|
|  1 |    a | uuid-1 |     10,6,1 |
|  2 |    a | uuid-2 |  16,2,7,11 |
|  3 |    a | uuid-3 |     8,3,12 |
|  4 |    a | uuid-4 |          4 |
|  5 |    a | uuid-5 |          5 |
|  6 |    b | uuid-1 |     10,6,1 |
|  7 |    b | uuid-2 |  16,2,7,11 |
|  8 |    b | uuid-3 |     8,3,12 |
|  9 |    b | uuid-6 |    15,13,9 |
| 10 |    c | uuid-1 |     10,6,1 |
| 11 |    c | uuid-2 |  16,2,7,11 |
| 12 |    c | uuid-3 |     8,3,12 |
| 13 |    c | uuid-6 |    15,13,9 |
| 14 |    c | uuid-7 |         14 |
| 15 |    d | uuid-6 |    15,13,9 |
| 16 |    d | uuid-2 |  16,2,7,11 |

| id | type |   uuid | mapped_ids |
|----|------|--------|------------|
|  1 |    a | uuid-1 |       10,6 |
|  2 |    a | uuid-2 |    16,7,11 |
|  3 |    a | uuid-3 |       8,12 |
|  4 |    a | uuid-4 |            |
|  5 |    a | uuid-5 |            |
|  6 |    b | uuid-1 |       10,1 |
|  7 |    b | uuid-2 |    16,2,11 |
|  8 |    b | uuid-3 |       3,12 |
|  9 |    b | uuid-6 |      15,13 |
| 10 |    c | uuid-1 |        6,1 |
| 11 |    c | uuid-2 |     16,2,7 |
| 12 |    c | uuid-3 |        8,3 |
| 13 |    c | uuid-6 |       15,9 |
| 14 |    c | uuid-7 |            |
| 15 |    d | uuid-6 |       13,9 |
| 16 |    d | uuid-2 |     2,7,11 |

| id | type |   uuid | mapped_ids |
|----|------|--------|------------|
|  1 |    a | uuid-1 |       10,6 |
|  2 |    a | uuid-2 |    16,7,11 |
|  3 |    a | uuid-3 |       8,12 |
|  4 |    a | uuid-4 |     (null) |
|  5 |    a | uuid-5 |     (null) |
|  6 |    b | uuid-1 |       10,1 |
|  7 |    b | uuid-2 |    16,2,11 |
|  8 |    b | uuid-3 |       3,12 |
|  9 |    b | uuid-6 |      15,13 |
| 10 |    c | uuid-1 |        6,1 |
| 11 |    c | uuid-2 |     16,2,7 |
| 12 |    c | uuid-3 |        8,3 |
| 13 |    c | uuid-6 |       15,9 |
| 14 |    c | uuid-7 |     (null) |
| 15 |    d | uuid-6 |       13,9 |
| 16 |    d | uuid-2 |     2,7,11 |

试试这个：

select
  t1.id, t1.type, t1.master_guid, array_agg (distinct t2.id)
from
  table1 t1
  left join table1 t2 on
    t1.master_guid = t2.master_guid and
    t1.id != t2.id
group by
  t1.id, t1.type, t1.master_guid

我没有得出与你列出的结果完全相同的结果，但我认为这很接近，也许你的预期有误，或者我的预期有点小错误。。。不管怎样，这都是一个潜在的起点

--编辑--

对于方法#2，我认为您只需要向表2添加一个内部联接即可获得GUID：

select
  t1.id, t1.type, t2.master_guid,
  array_agg (t2a.id)
from
  table1 t1
  join table2 t2 on t1.id = t2.id
  left join table2 t2a on
    t2.master_guid = t2a.master_guid and
    t2a.id != t1.id
where
  t1.type = 'a'
group by
  t1.id, t1.type, t2.master_guid

谢谢这似乎奏效了。我为type添加了where子句，可以对其进行筛选。但是，在相同的行上，如果我从表1中删除master_guid列，并创建具有id和master_guid列的新表2。那么查询会是什么样子？@Learner——您能在问题中添加新的表结构和一些示例数据吗？顺便说一句，您在构建问题框架和提供样本数据方面做得非常好。是的，这两种方法的效果都与预期一样。非常感谢！！解释得很好！只有一个查询，我试图使用type筛选出记录，在这种情况下，它不会返回所需的结果。我刚刚添加了where子句type='a'，仅仅添加where子句将不起作用，因为窗口函数是在过滤之后计算的。因此，它将找不到过滤类型的UUID。您只需在子查询中添加WHERE子句：请不要更改已回答问题的上下文。答案将不再适合这个问题。如果您有第二个问题，请打开一个新问题：）此外，我不清楚您想用“第二种方法”向我们展示什么。注意，下次会注意的。在第二种方法中，我进一步规范了它并检查了查询。