如果你是"t1".persona_1_id = 2，预期的结果应该返回persona_id = 4。
like
---
id persona_1_id persona_2_id liked
1 2 1 FALSE
2 3 1 TRUE
3 4 2 TRUE -- 4 likes 2
4 2 4 TRUE -- 2 likes 4
-- (2 and 4 like each other)
chat_persona
---
id chat_id persona_id -- but same chat has not been created between 2 and 4
1 1 3
2 1 2
3 2 4
4 2 1
5 3 5
6 3 1
-- so persona_id = 4 is the answer
我试图返回彼此相似的用户，他们之间没有聊天。
“喜欢彼此”的作品，但我正在另外过滤“聊天不存在”：
SELECT DISTINCT
"t1".id, "t1".read_at as read_at, "t1".created_at as created_at,
"persona".id as persona_id, "persona".profile_id as profile_id, "persona".name as persona_name, "chat_persona".chat_id as chat_id, "chat_persona".id as chat_persona_id
FROM "like" as "t1"
JOIN "persona" ON "t1".persona_2_id = "persona".id
JOIN "chat_persona" on "t1".persona_2_id = "chat_persona".persona_id
WHERE
"t1".persona_1_id = 2
AND EXISTS (
SELECT 1
FROM "like" as "t2"
WHERE
"t1".persona_1_id = "t2".persona_2_id
AND "t1".persona_2_id = "t2".persona_1_id
AND "t2".liked = true
)
AND "t1".liked = true
AND "chat_persona".id IS NULL -- throws out the correct rows if ANY person chatted with them already... make sense
而不是AND "chat_persona".id IS NULL，也试过：
AND NOT EXISTS (
SELECT 1
FROM "chat_persona" as "t2"
WHERE
"t1".persona_1_id = "t2".persona_id
AND "t1".persona_2_id = "t2".persona_id
) -- doesn't throw out any rows
最终答案：
SELECT DISTINCT
"l1".id, "l1".read_at as read_at, "l1".created_at as created_at,
"persona".id as persona_id, "persona".profile_id as profile_id, "persona".name as persona_name
FROM "like" l1
JOIN "persona" ON "l1".persona_2_id = "persona".id
WHERE
"l1".persona_1_id = 2
AND "l1".liked = true
AND EXISTS (
SELECT 1
FROM "like" l2
WHERE
"l1".persona_1_id = "l2".persona_2_id
AND "l1".persona_2_id = "l2".persona_1_id
AND "l2".liked = true
)
AND NOT EXISTS (
SELECT 1
FROM "chat_persona" c
WHERE c.persona_id IN ("l1".persona_1_id, "l1".persona_2_id)
GROUP BY c.chat_id
HAVING count(*) = 2
)
我在想not exists，有一个子查询，检查两者是否在同一个chat：
select l.persona_1_id, l.persona_2_id
from l
where not exists (select 1
from chats c
where c.persona_id in (l.persona_1_id, l.persona_2_id)
group by c.chat_id
having count(*) = 2 -- both are in chat
);
我的机器上没有postgres，在SQL服务器上测试过，它给出了你没有聊天的行，并给出4作为人员ID。
您可能必须以这种方式修改查询。我修改了几个连接并且更改存在以不存在并更改了内部查询。你可以在SQL server中看到解决方案，概念应该是一样的。
SELECT DISTINCT
"t1".id, "t1".read_at as read_at, "t1".created_at as created_at,
"persona".id as persona_id, "persona".profile_id as profile_id, "persona".name as persona_name, "chat_persona".chat_id as chat_id, "chat_persona".id as chat_persona_id
FROM "like" as "t1"
JOIN "persona" ON "t1".persona_2_id = "persona".id
-- JOIN "chat_persona" on "t1".persona_2_id = --
--"chat_persona".persona_id --removed this join as you are looking
--for the records that don't have chat
WHERE
"t1".liked = true --changed it here
AND not EXISTS (
SELECT 1
FROM "chat_persona" as "t2" --changed it here
WHERE
"t2".Chat_ID = "t1".persona_1_id
AND "t2".Persona_ID = "t1".persona_2_id
)
SQL服务器查询：
select 1 as ID, 2 as persona_1_id, 1 as persona_2_id, 'FALSE' as Liked into #templike union all
select 2 as ID, 3 as persona_1_id, 1 as persona_2_id, 'TRUE' as Liked union all
select 3 as ID, 4 as persona_1_id, 2 as persona_2_id, 'TRUE' as Liked union all
select 4 as ID, 2 as persona_1_id, 4 as persona_2_id, 'TRUE' as Liked
select 1 as ID, 1 as Chat_ID, 3 as Persona_ID into #chat_persona union all
select 2 as ID, 1 as Chat_ID, 2 as Persona_ID union all
select 3 as ID, 2 as Chat_ID, 4 as Persona_ID union all
select 4 as ID, 2 as Chat_ID, 1 as Persona_ID union all
select 5 as ID, 3 as Chat_ID, 5 as Persona_ID union all
select 6 as ID, 3 as Chat_ID, 1 as Persona_ID
select * from #templike t
where Liked = 'TRUE' and not exists (select 1 from #chat_persona cp where cp.Chat_ID = t.persona_1_id and cp.Persona_ID = t.persona_2_id)
输出：
ID persona_1_id persona_2_id Liked
3 4 2 TRUE