SQL选择您可能认识的人

Question

The question you're asking appears subjective and is likely to be closed.

当我填写标题字段时，看到上述 可怕的警告 时，我并不感到惊讶。

我阅读了几乎所有谈论的话题，friends of friends或者mutual friends不确定我是否找到了我想做的正确解决方案。

对不起，我不擅长英语或SQL。

在不擅长两种语言的情况下，如何找到正确的答案？

我决定要问。我不会为down-votes或任何duplication warnings失望。

当我想要答案时，我将尽可能真诚地写下来，以帮助您解决其他任何类似的问题。

我有一张与朋友关系的桌子。

FRIEND (TABLE)
-----------------------------------
PLAYER_ID(PK,FK)   FRIEND_ID(PK,FK)
-----------------------------------
1                  2                 // 1 knows 2
2                  1                 // 2 knows 1
1                  3                 // 1 knows 3
2                  3                 // 2 knows 3
2                  4                 // 2 knows 4
2                  5                 // 2 knows 5 // updated
3                  5                 // 3 knows 5 // updated
1                  100
1                  200
1                  300
100                400
200                400
300                400

两者composite primary keys也是PLAYER表中的外键。

我问并从这样的好人那里得到答案，因为他们“彼此认识”。

表中的熟人的SQL视图。

我有这样的看法。

ACQUAINTANCE (VIEW)
-----------------------------------
PLAYER_ID(PK,FK)   FRIEND_ID(PK,FK)
-----------------------------------
1                  2                 // 1 knows 2
2                  1                 // 2 knows 1

您可能会注意到，这种关系的业务逻辑具有以下两个目的。

1. 一个玩家可以说他或她认识其他人。
2. 当两个人都说彼此认识时，可以说是相识。

而且，现在，我想知道有什么好的方法可以

1. 选择其他PLAYER_ID
2. 使用给定的PLAYER（PLAYER_ID）（例如1）
3. 每个人都是“给定玩家直接朋友的朋友”之一
4. 每个都不是玩家本人（不包括1-> 2-> 1）
5. 每个都不是玩家的直接朋友（不包括1-> 2-> 3 x 1-> 3中的3）
6. 如果可能，请按共同朋友的数量订购。

提前致谢。

如果此主题确实重复且不必要，我将关闭线程。

更新 - - - - - - - - - - - - - - - - - - - - - - - - - -------------

对于Rapha毛l Althaus的评论whose name is same with my future daughter（是男孩的名字吗？），

3是friends of friends of 1因为

1 knows 2
2 knows 3

但被排除在外是因为

1 already knows 3

基本上，我想以服务为given player中

people he or she may know
which is not himself or herself // this is nothing but obvious
which each is not already known to himself

与上表

by 1 -> 2 -> 4 and 1 -> 3 -> 5

4 and 5 can be suggested for 1 as 'people you may know'

order by number of mutual friends will be perfect
but I don't think I can understand even if someone show me how. sorry.

谢谢你。

更新 - - - - - - - - - - - - - - - - - - - - - - - - - --------------------

我认为，FROM HERE WITH VARIOUS PEOPLE即使这不是正确的答案，我也必须根据自己的经验逐步尝试。如果我做错了任何事情，请告诉我。

首先，让我自己加入FRIEND表本身。

SELECT *
FROM FRIEND F1 INNER JOIN FRIEND F2 ON F1.FRIEND_ID = F2.PLAYER_ID

打印

+-----------+-----------+-----------+-----------+
| PLAYER_ID | FRIEND_ID | PLAYER_ID | FRIEND_ID |
+-----------+-----------+-----------+-----------+
|         1 |         2 |         2 |         1 |
|         1 |         2 |         2 |         3 |
|         1 |         2 |         2 |         4 |
|         1 |         2 |         2 |         5 |
|         1 |         3 |         3 |         5 |
|         2 |         1 |         1 |         2 |
|         2 |         1 |         1 |         3 |
|         2 |         3 |         3 |         5 |
+-----------+-----------+-----------+-----------+

仅F2.FRIEND_ID

SELECT F2.FRIEND_ID
FROM FRIEND F1 INNER JOIN FRIEND F2 ON F1.FRIEND_ID = F2.PLAYER_ID

打印

+-----------+
| FRIEND_ID |
+-----------+
|         1 |
|         3 |
|         4 |
|         5 |
|         5 |
|         2 |
|         3 |
|         5 |
+-----------+

1只

SELECT F2.FRIEND_ID
FROM FRIEND F1 INNER JOIN FRIEND F2 ON F1.FRIEND_ID = F2.PLAYER_ID
WHERE F1.PLAYER_ID = 1;

打印

+-----------+
| FRIEND_ID |
+-----------+
|         1 |
|         3 |
|         4 |
|         5 |
|         5 |
+-----------+

不是1

SELECT F2.FRIEND_ID
FROM FRIEND F1 INNER JOIN FRIEND F2 ON F1.FRIEND_ID = F2.PLAYER_ID
WHERE F1.PLAYER_ID = 1 
AND F2.FRIEND_ID != 1;

打印

+-----------+
| FRIEND_ID |
+-----------+
|         3 |
|         4 |
|         5 |
|         5 |
+-----------+

不是1的直接已知

SELECT F2.FRIEND_ID
FROM FRIEND F1 INNER JOIN FRIEND F2 ON F1.FRIEND_ID = F2.PLAYER_ID
WHERE F1.PLAYER_ID = 1
AND F2.FRIEND_ID != 1
AND F2.FRIEND_ID NOT IN (SELECT FRIEND_ID FROM FRIEND WHERE PLAYER_ID = 1);

印刷

+-----------+
| FRIEND_ID |
+-----------+
|         4 |
|         5 |
|         5 |
+-----------+

我想我要去那里。

更新 - - - - - - - - - - - - - - - - - - - - - - - - - ----------------

添加了以下路径

1 -> 100 -> 400
1 -> 200 -> 400
1 -> 300 -> 400

和最后一个查询打印（再次）

+-----------+
| FRIEND_ID |
+-----------+
|         4 |
|         5 |
|         5 |
|       400 |
|       400 |
|       400 |
+-----------+

最后，我得到了候选人：4、5、400

把distinct肯定工作的首要目标

SELECT DISTINCT F2.FRIEND_ID
FROM FRIEND F1 INNER JOIN FRIEND F2 ON F1.FRIEND_ID = F2.PLAYER_ID
WHERE F1.PLAYER_ID = 1
AND F2.FRIEND_ID != 1
AND F2.FRIEND_ID NOT IN (SELECT FRIEND_ID FROM FRIEND WHERE PLAYER_ID = 1);

印刷

+-----------+
| FRIEND_ID |
+-----------+
|         4 |
|         5 |
|       400 |
+-----------+

而且，现在需要按相互计数的顺序进行订购。

这是每个候选人的共同朋友数。

+-----------+
| FRIEND_ID |
+-----------+
|         4 | 1 (1 -> 2 -> 4)
|         5 | 2 (1 -> 2 -> 5, 1 -> 3 -> 5)
|       400 | 3 (1 -> 100 -> 400, 1 -> 200 -> 400, 1 -> 300 -> 400)
+-----------+

我如何才能通过这么多共同的朋友来计算和排序？

SELECT F2.FRIEND_ID, COUNT(*)
FROM FRIEND F1 INNER JOIN FRIEND F2 ON F1.FRIEND_ID = F2.PLAYER_ID
WHERE F1.PLAYER_ID = 1
AND F2.FRIEND_ID != 1
AND F2.FRIEND_ID NOT IN (SELECT FRIEND_ID FROM FRIEND WHERE PLAYER_ID = 1)
GROUP BY F2.FRIEND_ID;

印刷

+-----------+----------+
| FRIEND_ID | COUNT(*) |
+-----------+----------+
|         4 |        1 |
|         5 |        2 |
|       400 |        3 |
+-----------+----------+

我得到了它！

SELECT F2.FRIEND_ID, COUNT(*) AS MFC
FROM FRIEND F1 INNER JOIN FRIEND F2 ON F1.FRIEND_ID = F2.PLAYER_ID
WHERE F1.PLAYER_ID = 1
AND F2.FRIEND_ID != 1
AND F2.FRIEND_ID NOT IN (SELECT FRIEND_ID FROM FRIEND WHERE PLAYER_ID = 1)
GROUP BY F2.FRIEND_ID
ORDER BY MFC DESC;

印刷

+-----------+-----+
| FRIEND_ID | MFC |
+-----------+-----+
|       400 |   3 |
|         5 |   2 |
|         4 |   1 |
+-----------+-----+

有人可以确认吗？该查询是否最佳？将其作为视图时是否可能存在性能问题？

谢谢你。

更新 - - - - - - - - - - - - - - - - - - - - - - - - -

我创建了一个视图

CREATE VIEW FOLLOWABLE AS
    SELECT F1.PlAYER_ID, F2.FRIEND_ID AS FOLLOWABLE_ID, COUNT(*) AS MFC
    FROM FRIEND F1 INNER JOIN FRIEND F2 ON F1.FRIEND_ID = F2.PLAYER_ID
    WHERE F2.FRIEND_ID != F1.PLAYER_ID
    AND F2.FRIEND_ID NOT IN (SELECT FRIEND_ID FROM FRIEND WHERE PLAYER_ID = F1.PLAYER_ID)
    GROUP BY F2.FRIEND_ID
    ORDER BY MFC DESC;

并经过测试。

mysql> select * from FOLLOWABLE;
+-----------+---------------+-----+
| PlAYER_ID | FOLLOWABLE_ID | MFC |
+-----------+---------------+-----+
|         1 |           400 |   3 |
|         1 |             5 |   2 |
|         2 |           100 |   1 |
|         2 |           200 |   1 |
|         2 |           300 |   1 |
|         1 |             4 |   1 |
+-----------+---------------+-----+
6 rows in set (0.01 sec)

mysql> select * from FOLLOWABLE WHERE PLAYER_ID = 1;
+-----------+---------------+-----+
| PlAYER_ID | FOLLOWABLE_ID | MFC |
+-----------+---------------+-----+
|         1 |           400 |   3 |
|         1 |             5 |   2 |
|         1 |             4 |   1 |
+-----------+---------------+-----+
3 rows in set (0.00 sec)

Answer 1

使用此 编辑

SELECT `friend_id` AS `possible_friend_id`
FROM `friends`
WHERE `player_id` IN (        --selecting those who are known
    SELECT `friend_id`        --by freinds of #1
    FROM `friends`
    WHERE `player_id` = 1) 
AND `friend_id` NOT IN (      --but not those who are known by #1
    SELECT `friend_id`
    FROM `friends`
    WHERE `player_id` = 1)
AND NOT `friend_id` = 1       --and are not #1 himself
                              --if one is known by multiple people
                              --he'll be multiple time in the list
GROUP BY `possible_friend_id` --so we group
ORDER BY COUNT(*) DESC        --and order by amount of repeatings