两个用户之间的共同朋友

Question

我有两张桌子

friends表：

users表：

并希望得到两个用户的共同朋友的名字（nom）。即： 1是2的朋友; 3是1和2的朋友。

用户1和2有共同的朋友3.我想得到他的名字＆＃39; sammy＆＃39;在一个查询中。

我该怎么做？

Answer 1

排除date_confirmation检查的逻辑，您有f1_id和f2_id

SELECT f.id_des, u.nom
FROM friends f
LEFT JOIN users u ON u.id=f.id_des
WHERE 
f.id_exp=$f2_id AND
f.id_des IN (
   SELECT id_des
   FROM friends
   WHERE id_exp=$f1_id
)

这只会检查 ONE 方向。两者都邀请了同一个朋友。更复杂的部分是让另一个朋友邀请他们中的一个（圆形邀请！）

对于完整的循环答案，有4个条件。您需要在每个之间执行UNION。

1. 两人都邀请了同一个人（我给的例子）
2. 由同一个人邀请
3. 由F2邀请的F1邀请
4. F2邀请了谁邀请了

除了您使用id_exp和id_des以及$f1_id和$f2_id

之外，所有查询都与上述查询相同

这是第二个：

SELECT f.id_des, u.nom
FROM friends f
LEFT JOIN users u ON u.id=f.id_des
WHERE 
f.id_exp=$f2_id AND
f.id_des IN (
   SELECT id_des
   FROM friends
   WHERE id_exp=$f1_id
)
UNION DISTINCT
SELECT f.id_des, u.nom
FROM friends f
LEFT JOIN users u ON u.id=f.id_des
WHERE 
f.id_des=$f2_id AND
f.id_exp IN (
   SELECT id_exp
   FROM friends
   WHERE id_des=$f1_id
)
UNION DISTINCT
SELECT f.id_des, u.nom
FROM friends f
LEFT JOIN users u ON u.id=f.id_des
WHERE 
f.id_des=$f2_id AND
f.id_des IN (
   SELECT id_des
   FROM friends
   WHERE id_exp=$f1_id
)
UNION DISTINCT
SELECT f.id_des, u.nom
FROM friends f
LEFT JOIN users u ON u.id=f.id_des
WHERE 
f.id_exp=$f2_id AND
f.id_des IN (
   SELECT id_des
   FROM friends
   WHERE id_des=$f1_id
)

不容易测试，但这是概念。然后你必须消除重复： - ）

Answer 2

我试过这段代码：

SELECT f.id_des, u.nom FROM friends f LEFT JOIN users u ON u.id=f.id_des WHERE f.id_exp=445 AND f.id_des IN ( SELECT id_des FROM friends WHERE id_exp=400 ) UNION DISTINCT SELECT f.id_des, u.nom FROM friends f LEFT JOIN users u ON u.id=f.id_des WHERE f.id_des=445 AND f.id_exp IN ( SELECT id_exp FROM friends WHERE id_des=400 ) UNION DISTINCT SELECT f.id_des, u.nom FROM friends f LEFT JOIN users u ON u.id=f.id_des WHERE f.id_des=445 AND f.id_des IN ( SELECT id_des FROM friends WHERE id_exp=400 ) UNION DISTINCT SELECT f.id_des, u.nom FROM friends f LEFT JOIN users u ON u.id=f.id_des WHERE f.id_exp=445 AND f.id_des IN ( SELECT id_des FROM friends WHERE id_des=400 )

他告诉我：enter image description here 这是ma表朋友：enter image description here 查看朋友表的最后三行。

当我交换两个数字的位置时，它给了我另一个结果。 我认为代码不正确。 你怎么看？ 感谢。