Question

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

当我在标题字段填写时看到上面的可怕警告时，我并不感到惊讶。

我几乎阅读了所有讨论friends of friends或mutual friends的帖子，但我不确定我找到了正确的解决方案。

对不起，我不擅长英语，也不擅长SQL。

如何在不擅长这两种语言的同时找到正确的答案？

我决定要问。我不会让自己失望down-vote或任何duplication warning。

正如我想要的答案，我会尽可能真诚地写下来，因为任何进一步的类似问题都可以得到帮助。

我有朋友关系表。

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

两个composite primary keys也是来自PLAYER表的外键。

我问过并从这些善良的人那里得到了答案，因为“人们互相认识”。

SQL view for acquaintance from table

我有这样的观点。

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两个目的。

一个玩家可以说他或她认识别人。
当两个人都说他们彼此认识时，他们可以说是熟人。

而且，现在，我想知道

选择其他PLAYER_ID
给定的PLAYER（PLAYER_ID）（比如1）
每个人都是“给予PLAYER直接朋友”的朋友之一
每个人不是玩家本人（不包括1 - > 2 - > 1）
哪一个不是PLAYER的直接朋友（不包括3从1 - ＆gt; 2 - ＆gt; 3 by 1 - ＆gt; 3）
如果可能，按共同朋友的数量排序。

我认为Justin Niessner在"people you may know" sql query中的答案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而且没有必要，我将关闭该主题。

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

对于RaphaëlAlthaus的评论whose name is same with my future daughter（这是男孩的名字吗？），

3是friends of friends of 1的候选人，因为

1 knows 2
2 knows 3

但排除因为

1 already knows 3

基本上我想为given player

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

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。

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

我想我必须从我所学到的FROM HERE WITH VARIOUS PEOPLE中逐步尝试，即使这不是正确的答案。如果我做错了，请告诉我。

首先，让我自己加入FRIEND表。

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

仅限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 |
+-----------+

仅限1个

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的直接知识

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);

打印

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

我想我到了那儿。

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

添加以下路径

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确实适用于主要目标

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);

打印

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

而且，现在，需要按相互排序进行排序。

每个候选人都有共同朋友的数量。

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

如何通过这些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;

打印

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

我明白了！

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;

打印

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

任何人都可以确认一下吗？该查询是否最佳？将它作为视图时会出现任何可能的性能问题吗？

谢谢。

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

使用它的修改

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

SQL选择您可能认识的人

1 个答案: