mysql - SQL 选择你可能认识的人

Question

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

当我在填写标题字段时看到上述可怕的警告时，我并不感到惊讶。

我阅读了几乎所有讨论的主题friends of friends，mutual friends但我不确定我是否找到了我想做的正确解决方案。

对不起，我不擅长英语和 SQL。

如何在两种语言都不擅长的情况下找到正确的答案？

我决定我必须问。我不会因为down-votes 或任何duplication warnings 而让自己失望。

由于我想要答案，我会尽可能真诚地写下来，以便任何进一步的类似问题可以得到帮助。

我有一张朋友关系的桌子。

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。

我问这些好人并得到了“人们互相认识”的回答。

表中熟人的 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有以下两个目的。

一名玩家可以说他或她认识其他人。
当两个人都说他们认识时，他们可以说是熟人。

而且，现在，我想知道有什么好的方法

选择其他 PLAYER_ID
使用给定的 PLAYER(PLAYER_ID)（比如 1）
其中每个都是“给定玩家的直接朋友的朋友”之一
其中每个不是PLAYER本人（不包括1 -> 2 -> 1）
哪个不是PLAYER的直接好友（不包括3 from 1 -> 2 -> 3 by 1 -> 3）
如果可能的话，按共同朋友的数量排序。

我认为贾斯汀·尼斯纳在“你可能认识的人”sql 查询中的回答是我必须遵循的最接近的路径。

提前致谢。

如果这个主题真的重复并且没有必要，我会关闭线程。

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

对于 Raphaël 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，因为

1 already knows 3

基本上我想given player为

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 - - - - - - - - - - - - - - - - - - - - - - - - - --------------------

我想我必须从我所学到的东西中一步一步地尝试，FROM HERE WITH VARIOUS PEOPLE即使这不是正确的答案。如果我做错了什么，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

印刷

+-----------+-----------+-----------+-----------+
| 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 |
+-----------+-----------+-----------+-----------+

仅限 F2.FRIEND_ID

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

印刷

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

仅限 1

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

印刷

+-----------+
| 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;

印刷

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

不是 1 的直接已知信息

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 |
+-----------+

我想我快到了。

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

添加了以下路径

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

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

确保distinct为主要目标工作

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，需要按相互计数排序。

这是每个候选人的共同朋友的数量。

+-----------+
| 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)
+-----------+

我如何计算和订购这些共同朋友的数量？

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 |
+-----------+----------+

我知道了！

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？该查询是最优的吗？将其作为视图时是否存在任何可能的性能问题？

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)

score 8 · Accepted Answer

使用这个编辑

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

mysql - SQL 选择你可能认识的人

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