可能重复：
MYSQL 选择共同的朋友
我有一张友谊表，友谊只存储在一行中。所以没有重复的条目。
id Person1 Person2 status
1 1 2 friend
2 1 3 friend
3 2 3 friend
4 3 4 friend
什么 MySQL 查询（join、inner join）可以帮助我找到第 1 个人和第 3 个人之间的共同（共同）朋友？此示例中的输入是 {1,3}，输出应该是 {2}，因为 Person #2 是 bot #1 和 #3 的朋友。
好吧，到目前为止，唯一可行的查询是 Simon 的……但这确实有点矫枉过正——如此复杂的令人讨厌的查询（2 个子查询和 2 个联合！）对于如此简单的事情，您需要悬赏？:-) 如果您有 1000 多个用户，那么查询会非常慢 - 请记住，它是二次的，并且由于子查询中的联合，几乎不会使用任何索引！
我建议重新考虑设计并允许 2 行重复以建立友谊：
id Person1 Person2 status
1 1 2 friend
2 2 1 friend
3 1 3 friend
4 3 1 friend
您可能认为这是低效的，但以下简化将允许将查询重写为简单的连接：
select f1.Person2 as common_friend
from friends as f1 join friends as f2
using (Person2)
where f1.Person1 = '$id1' and f2.Person1 = '$id2'
and f1.status = 'friend' and f2.status = 'friend'
这会很快！（不要忘记为 Person1,2 添加索引。）我已经建议在其他非常讨厌的数据结构中进行类似的简化（将子查询重写为联接），它已经加快了从永恒到闪电战的查询！
所以可能看起来很大的开销（一个友谊2行）实际上是一个很大的优化:-)
此外，它还将使诸如“查找 X 的所有朋友”之类的查询更加容易。并且不需要花费更多的赏金:-)
再来一个答案。
select
(case when f1.person1 = 1 then f1.person2 else f1.person1 end) as fid
from friends f1
where f1.person1 = 1 or f1.person2 = 1
and f1.status = 'friend'
intersect
select
(case when f1.person1 = 3 then f1.person2 else f1.person1 end) as fid
from friends f1
where f1.person1 = 3 or f1.person2 = 3
and f1.status = 'friend'
如果这些条件不满足使其工作所需的少量调整，则此查询的工作假设是友谊表中没有自我交友且没有重复。
SELECT fid FROM
(
--FIRST PERSON (X) FRIENDLIST
SELECT
(CASE WHEN Person1 = X THEN Person2 ELSE Person1 END) AS fid
FROM Friendships WHERE (Person1 = X OR Person2 = X) AND status = "friend"
UNION ALL --DO NOT REMOVE DUPLICATES WITH ALL JOIN
--SECOND PERSON (Y) FRIENDLIST
SELECT
(CASE WHEN Person1 = Y THEN Person2 ELSE Person1 END) AS fid
FROM Friendships WHERE (Person1 = Y OR Person2 = Y) AND status = "friend"
) FLIST
GROUP BY fid
HAVING COUNT(*) = 2
set search_path='tmp';
DROP TABLE friendship CASCADE;
CREATE TABLE friendship
( id integer not null PRIMARY KEY
, person1 integer not null
, person2 integer not null
, status varchar
, CONSTRAINT pk1 UNIQUE (status,person1,person2)
, CONSTRAINT pk2 UNIQUE (status,person2,person1)
, CONSTRAINT neq CHECK (person1 <> person2)
);
INSERT INTO friendship(id,person1,person2,status) VALUES
(1,1,2,'friend' ) ,(2,1,3,'friend' ) ,(3,2,3,'friend' ) ,(4,3,4,'friend' )
;
-- -----------------------------------------
-- For implementations that don't have CTEs,
-- a view can be used to emulate a CTE.
-- -----------------------------------------
CREATE VIEW flip AS (
SELECT person1 AS one
, person2 AS two
FROM friendship WHERE status = 'friend'
UNION
SELECT person2 AS one
, person1 AS two
FROM friendship WHERE status = 'friend'
);
SELECT DISTINCT
f1.two AS common
FROM flip f1
JOIN flip f2 ON f1.two = f2.two
WHERE f1.one = 1
AND f2.one = 3
;
DROP VIEW flip;
-- ------------------------------
-- The same query with a real CTE
-- ------------------------------
with flip AS (
SELECT person1 AS one
, person2 AS two
FROM friendship WHERE status = 'friend'
UNION
SELECT person2 AS one
, person1 AS two
FROM friendship WHERE status = 'friend'
)
SELECT DISTINCT
f1.two AS common
FROM flip f1
JOIN flip f2 ON f1.two = f2.two
WHERE f1.one = 1
AND f2.one = 3
;
结果：
SET
DROP TABLE
NOTICE: CREATE TABLE / PRIMARY KEY will create implicit index "friendship_pkey" for table "friendship"
NOTICE: CREATE TABLE / UNIQUE will create implicit index "pk1" for table "friendship"
NOTICE: CREATE TABLE / UNIQUE will create implicit index "pk2" for table "friendship"
CREATE TABLE
INSERT 0 4
CREATE VIEW
common
--------
2
(1 row)
DROP VIEW
common
--------
2
(1 row)
内部查询仅获取第一个人的 FRIEND ID，并将其标准化为单个列“FriendID”。如果找到的记录在第一个位置有人员 ID = 1，它会抓取第二个......如果在第二个位置有人员 ID = 1，那么它会抓取第一个。
完成后，我们知道谁是第 1 个人的唯一朋友列表……完成。现在，再次加入友谊表，但仅限于那些 FIRST 有资格成为第 1 个人的朋友之一的人......一旦有资格，然后确保第二张桌子上的另一个人是第 3 个人你在寻找共同点。
确保在 person1 和 person2 上都有一个索引，以利用 OR 条件。
select
JustPerson1Friends.FriendID
from
( select
if( f.Person1 = 1, f.Person2, f.Person1 ) as FriendID
from
Friendships f
where
( f.Person1 = 1
OR f.Person2 = 1 )
AND f.status = "friend" ) JustPerson1Friends
JOIN Friendships f2
on ( JustPerson1Friends.FriendID = f2.Person1
OR JustPerson1Friends.FriendID = f2.Person2 )
AND f2.status = "friend"
AND ( f2.Person1 = 3 OR f2.person2 = 3 )
另一种选择将人“3”作为常见的人预先标记到结果集中，因此我们以后不需要明确限定 3。此外，通过使用 MySQL 变量，易于编写脚本并作为参数实现。在内部查询之后，对友谊进行 DOUBLE left-join 以明确测试可以在 X/Y 或 Y/X 组合中找到一个人的 BOTH 组合。所以最后的 where 子句只是说只要在 EITHER LEFT-JOIN 条件中找到一条记录，它就是一个共同的朋友并包含在结果集中。
select
JustPerson1Friends.FriendID
from
( select
@WantPerson2 as FindInCommonWith,
if( f.Person1 = @WantPerson1, f.Person2, f.Person1 ) as FriendID
from
( select @WantPerson1 := 1,
@WantPerson2 := 3 ) sqlvars
Friendships f,
(
where
( f.Person1 = @WantPerson1
OR f.Person2 = @WantPerson2 )
AND f.status = "friend" ) JustPerson1Friends
LEFT JOIN Friendships f2
on JustPerson1Friends.FindInCommonWith = f2.Person1
AND JustPerson1Friends.FriendID = f2.Person2
AND f2.status = "friend"
LEFT JOIN Friendships f3
on JustPerson1Friends.FindInCommonWith = f2.Person2
AND JustPerson1Friends.FriendID = f2.Person1
AND f2.status = "friend"
where
f2.Person1 > 0
OR f3.Person1 > 0
如果各种回复或评论之一已经提出了这一点，我深表歉意，但是如何：
select Person2 mutual_friend from
(select Person1, Person2 from friends
where Person1 in (1,3) union
select Person2, Person1 from friends
where Person2 in (1,3)
) t
group by Person2 having count(*) > 1;
此查询返回 '22' 作为结果，因为它对于 1 和 3 都很常见 您可能必须过滤掉不同的 PERSON1/PERSON2 如果我可以优化此查询，我会更新它
SELECT DISTINCT (REPLACE(TRANSLATE((WM_CONCAT(DISTINCT F.PERSON1) || ',' ||
WM_CONCAT(DISTINCT F.PERSON2)),
'1,3',
' '),
' ',
'')) AS COMMON_FRIEND
FROM FRIENDSHIP F
WHERE UPPER(F.STATUS) = 'FRIEND'
AND ((SELECT DISTINCT WM_CONCAT(F1.PERSON1)
FROM FRIENDSHIP F1
WHERE F1.PERSON2 = '3') LIKE ('%' || F.PERSON1 || '%') OR
(SELECT DISTINCT WM_CONCAT(F1.PERSON2)
FROM FRIENDSHIP F1
WHERE F1.PERSON1 = '3') LIKE ('%' || F.PERSON2 || '%'))
AND ((SELECT DISTINCT WM_CONCAT(F1.PERSON1)
FROM FRIENDSHIP F1
WHERE F1.PERSON2 = '1') LIKE ('%' || F.PERSON1 || '%') OR
(SELECT DISTINCT WM_CONCAT(F1.PERSON2)
FROM FRIENDSHIP F1
WHERE F1.PERSON1 = '1') LIKE ('%' || F.PERSON2 || '%'))
AND NOT ((F.PERSON1 = '1' AND F.PERSON2 = '3') OR
(F.PERSON1 = '3' AND F.PERSON2 = '1'))
我认为这很简单
SELECT * FROM friends
WHERE
(Person1 = '1' or Person2 = '1') &&
(Person1 = '2' or Person2 = '2') &&
status = 'friend'
鉴于您正试图在人 1 和 2 之间找到相互
我问过编号较小的用户是否总是Person1，但我最后写了一个不关心是否属实的查询。
set @firstParty = 1, @secondParty = 3
select friends_of_first.friend
from (
select Person2 as friend from friends where Person1 = @firstParty
union
select Person1 as friend from friends where Person2 = @firstParty
) as friends_of_first
join (
select Person2 as friend from friends where Person1 = @secondParty
union
select Person1 as friend from friends where Person2 = @secondParty
) as friends_of_second
on friends_of_first.friend = friends_of_second.friend
查找用户朋友的子查询可以替换为@NirmalthInk beYond 使用的子查询：
select case when f1.person1 = @firstParty then f1.person2 else f1.person1 end
from friend f1 where f1.person1 = @firstParty or f1.person2 = @firstParty
我很想知道哪种替代方案表现更好。
id Person1 Person2 status
1 1 2 friend
2 1 3 friend
3 2 3 friend
4 3 4 friend
SELECT
DISTINCT
F1.Person
FROM
--Friends of 1
(
SELECT F.Person1 Person FROM People F WHERE F.Person2 = 1 AND F.status = 'friend'
UNION
SELECT F.Person2 Person FROM People F WHERE F.Person1 = 1 AND F.status = 'friend'
) F1
INNER JOIN
(
--Friends of 3
SELECT F.Person1 Person FROM People F WHERE F.Person2 = 3 AND F.status = 'friend'
UNION
SELECT F.Person2 Person FROM People F WHERE F.Person1 = 3 AND F.status = 'friend'
) F2 ON
F2.Person = F1.Person
输出：
Person
2
这应该回答您当前的问题，尽管我建议不要这样做。在这种情况下，我总是选择存储两个关系副本，每个方向一个。
SELECT IF(f1.person1 IN ($id1, $id3), f1.person2, f1.person1) AS mutual_friend
FROM friends f1
INNER JOIN friends f2
ON (f1.person1 = $id1 AND f2.person1 = $id3 AND f1.person2 = f2.person2)
OR (f1.person1 = $id1 AND f2.person2 = $id3 AND f1.person2 = f2.person1)
OR (f1.person2 = $id1 AND f2.person1 = $id3 AND f1.person1 = f2.person2)
OR (f1.person2 = $id1 AND f2.person2 = $id3 AND f1.person1 = f2.person1)
WHERE f1.status = 'friend' AND f2.status = 'friend'