我有一张友谊表，友谊只存储在一行。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（加入，内部联接）将帮助我找到人＃1和人＃3之间的共同（共同）朋友？此示例中的输入为{1,3}，输出应为{2}，因为Person＃2是bot＃1和＃3的朋友。
答案 0 :(得分：8)
嗯，到目前为止唯一可能工作的查询是Simon的......但这真是一种过度杀伤 - 这种复杂讨厌的查询（2个子查询有2个联盟！）这么简单，你需要放置一笔赏金吗？ :-)如果你有1000多个用户，那么查询将会很慢 - 因为地狱 - 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个重复的行来建立友谊：
id Person1 Person2 status
1 1 2 friend
2 2 1 friend
3 1 3 friend
4 3 1 friend
您可能认为效率低下，但简化后会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'
这将是快死的！ （不要忘记为Person1,2添加索引。）I've advised a similar simplification (rewriting subqueries to joins) in other very nasty data structure and it has speeded up the query从永恒到闪电战瞬间！
所以看起来可能是一个很大的开销（一个友谊的两行）实际上是一个很大的优化： - ）
此外，它将使“查找所有X的朋友”等查询变得更加容易。并且不需要花更多的奖励： - ）
答案 1 :(得分：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)
答案 2 :(得分：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'
答案 3 :(得分：1)
如果这些条件不符合使其工作所需的小调整，则此查询的假设是友谊表中没有自助友好且没有重复项。
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
答案 4 :(得分：0)
我询问编号较低的用户是否总是Person1，但我最后写了一个不关心这是否属实的查询。
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
用于查找用户朋友的子查询可以替换为使用的@Nirmal-thInk beYond：
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得更好。
答案 5 :(得分：0)
如果各种回复或评论之一已经提出过这一点，我很抱歉，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;
答案 6 :(得分：0)
内部查询专门只获取第一个人的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ID = 1，则抓住第二个...如果第二个位置的人ID = 1，则抓住第一个。
通过这样做，我们知道朋友的单一名单是谁1 ...完成。现在，再次加入友谊表，但仅限于那些第一人被认定为人1的朋友之一...一旦合格，那么确保第二张桌上的另一个人是那个人3你正在寻找共性。
确保在person2上的索引和person2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 )
另一个选项是将“3”作为常用标记预先标记到结果集中，这样我们就不需要在以后明确限定3。此外，通过使用MySQL变量，易于编写脚本并实现为参数。在内部查询之后，对友谊进行DOUBLE左连接以明确地测试可以在X / Y或Y / X组合中找到人的两种组合。所以最后的where子句就是说只要在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
答案 7 :(得分：0)
此查询返回'22'作为结果，因为它对于1和3都是常见的 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'))
答案 8 :(得分：-1)
我认为这可以通过这个
来实现SELECT * FROM friends
WHERE
(Person1 = '1' or Person2 = '1') &&
(Person1 = '2' or Person2 = '2') &&
status = 'friend'
鉴于你正试图找到1号和2号人之间的相互作用
答案 9 :(得分：-1)
这应该回答你当前的问题，虽然我建议不要这样做。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'
答案 10 :(得分：-1)
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