在 PostgreSQL 12 上按 id（唯一）分组的查询中同时获取房间成员、房间所有者和管理员

Question

我想获取房间的 member 列表，房间的所有者 member 以防他不存在于其他 table 和管理员 member 同时.目前我单独获取它们。

CREATE TABLE public.room_members (
  id               bigint NOT NULL,
  member_id        bigint,
  room_id       bigint,
  group_id  bigint
);


CREATE TABLE public.rooms (
  id               bigint NOT NULL,
  member_id        bigint,
  group_id  bigint,
   name varchar(128)
);


CREATE TABLE public.members (
  id               bigint NOT NULL,
  group_id  bigint,
  username varchar(128),
    is_admin bool default false
);

CREATE TABLE public.groups (
  id               bigint NOT NULL,
  name varchar(128)
);


-- My Group created
INSERT INTO "groups" (id, name) VALUES (1, 'My Group');

-- Create users for this group. We have 4 users/members
INSERT INTO "members" (id, group_id, username, is_admin) VALUES (1, 1, 'Pratha', true);
INSERT INTO "members" (id, group_id, username) VALUES (2, 1, 'John');
INSERT INTO "members" (id, group_id, username) VALUES (3, 1, 'Mike');
INSERT INTO "members" (id, group_id, username) VALUES (4, 1, 'April');

-- April creates a room and he is owner of this room
INSERT INTO "rooms" (id, group_id, member_id, name) VALUES (1, 1, 4, 'My Room'); -- 4 is April

-- April also adds Mike to the room members. But she does not add herself. As she is owner.
INSERT INTO "room_members" (id, group_id, room_id, member_id) VALUES (1, 1, 1, 3); -- 3 is Mike

我想要的是：

1. room_members 'My Room' 列表 （目前只有 Mike）
2. 我的房间的所有者，以防他没有将自己添加到 room_members table。因为他是那个房间的主人(也就是四月)
3. 另外，管理员成员 （这是 Pratha）

而且这应该是独一无二的。例如，如果用户将自己添加到 room_members 和 owner，那么它应该只获取成员一次。

到目前为止我尝试了什么？

select * from members
left outer join room_members cm on cm.member_id = members.id
left outer join rooms c on c.id = cm.room_id
where c.name = 'My Room' or members.id = 1

我这里也不能用group by。我也不需要所有领域。仅 room_members table 个字段。

看这里：https://rextester.com/XWDS42043

room_members 的预期输出：

+-------------+------------+------------+
|  member_id  |  group_id  |  username  |
+-------------+------------+------------+
|      1      |      1     |   Pratha   |
+-------------+------------+------------+
|      3      |      1     |    Mike    |
+-------------+------------+------------+
|      4      |      1     |    April   |
+-------------+------------+------------+

• Pratha：因为他是ADMIN
• Mike：因为他是My Room的成员。 MEMBER
• April：因为她创造了那个房间。 OWNER

room_members可以很多。我只添加了 Mike，但它可以有多个成员，包括管理员和所有者。

Answer 1

您可以通过 UNION 来解决这个问题：

-- list the admin(s) of the room group
select m.id, m.group_id, m.username 
from rooms r 
inner join members m on m.group_id = r.group_id and m.is_admin = true
where r.name = 'My Room'
union
-- list the members of the room
select m.id, m.group_id, m.username 
from rooms r 
inner join room_members rm on r.id = rm.room_id
inner join members m on rm.member_id = m.id
where r.name = 'My Room'
union
-- recover the room owner
select m.id, m.group_id, m.username 
from rooms r 
inner join members m on r.member_id = m.id
where r.name = 'My Room'

UNION 消除了跨查询的重复项，因此如果用户同时是成员 and/or 组管理员 and/or 房间的所有者，他们只会出现一次。

在您的 fiddle 中，此查询 returns:

id  group_id    username
1   4   1       April
2   3   1       Mike
3   1   1       Pratha