可能的重复项：
MYSQL 选择共同的朋友
我有一个友谊表，友谊只存储在一行中。因此，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（连接，内部连接）将帮助我在人员#1和人员#3之间找到共同（共同）的朋友？此示例中的输入为 {1，3}，输出应{2}，因为人员 #2 是机器人 #1 和 #3 的好友。
好吧，到目前为止唯一可能有效的查询是西蒙的...但这真的是矫枉过正 - 如此复杂的令人讨厌的查询（2 个子查询和 2 个联合！:-)如果你有 1000+ 个用户，查询会很慢 - 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！
我建议再次重新考虑设计，并允许 2 行重复用于友谊：
id Person1 Person2 status
1 1 2 friend
2 2 1 friend
3 1 3 friend
4 3 1 friend
您可能认为这效率低下，但以下简化将允许将查询重写为简单连接：
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'
这将很快如地狱！（不要忘记为 Person1，2 添加索引。我建议在其他非常讨厌的数据结构中进行类似的简化（将子查询重写为连接），它将查询从永恒加速到闪电即时！
因此，可能看起来很大的开销（一个友谊的 2 行）实际上是一个很大的优化：-）
此外，它将使"查找X的所有朋友"之类的查询变得更加容易。而且不需要花费更多的赏金：-）
此查询的工作原理是假设友谊表中没有自我友好和重复项，如果这些条件不满足使其工作所需的小调整。
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
还有一个答案。
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'
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)
我认为这是通过这个简单地实现
的SELECT * FROM friends
WHERE
(Person1 = '1' or Person2 = '1') &&
(Person1 = '2' or Person2 = '2') &&
status = 'friend'
鉴于您正在尝试在人 1 和 2 之间找到相互关系
我问编号较低的用户是否总是Person1，但我最终编写了一个不在乎这是否属实的查询。
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
查找用户好友的子查询可以替换为@Nirmal thInk 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。
如果各种回复或评论之一已经建议了这一点，我深表歉意，但是怎么样：
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;
内部查询仅获取第一人称的 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，它会抓取第二个位置......如果第二个位置的人 ID = 1，那么它抓住第一个位置。
完成此操作后，我们知道谁是人 1 的单一朋友列表......做。 现在，再次加入友谊表，但仅适用于那些 FIRST 有资格成为人 1 的朋友之一的人...... 一旦符合条件，那么请确保第二张桌子上的另一个人是你正在寻找共同点的人 3。
确保两个 person1 上的索引和人员 2 上的另一个索引，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 )
另一种选择是将人员"3"预先标记为结果集中的公共人员，因此我们以后不需要显式限定 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，对友谊执行双重左联接，以显式测试可以在 X/Y 或 Y/X 组合中找到一个人的两个组合。 所以最后一个 where 子句只是说只要在 LEFT JOIN 条件下找到记录，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
此查询返回"22"作为结果，因为它在 1 和 3 中都很常见您可能需要过滤掉不同的PERSON1/PERSON2如果我能优化这个查询，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'))
This should answer your current question although I would advise against doing it like this. In this situation I would always opt for storing two copies of the relationship, one in each direction.
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'
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