Question

我想根据共同好友的数量来推荐用户，就像这样。

Suggested Friends:

Amy Adams (42 mutual friends)
Brian Bautista (21 mutual friends)
Chris Cross (6 mutual friends)

注意：我已经阅读了几个类似的帖子和答案，但是解决方案非常依赖于表结构，而且我无法解决我们的 Friendships 表的设置方式。见下文。

USERS TABLE
id
name

FRIENDSHIPS TABLE
id
initiatingUserId (FK to Users Table)
targetUserId (FK to Users Table)
status ('active', 'denied', 'pending')

如您所见，每个好友都占一行。发送好友请求的用户为发起用户，接受好友请求的用户为目标用户。这两列都有索引。

如何有效地获取当前用户还不是好友的用户列表，以及他们共同的好友数量？

这是一个示例数据集，根据@Strawberry 的要求...

CREATE TABLE `users` (
  `id` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `firstName` varchar(255) COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `lastName` varchar(255) COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  PRIMARY KEY (`id`)
);

CREATE TABLE `friends` (
  `id` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `initiatingUserId` int(10) unsigned NOT NULL,
  `targetUserId` int(10) unsigned NOT NULL,
  `status` varchar(255) COLLATE utf8mb4_unicode_ci NOT NULL DEFAULT 'pending',
  PRIMARY KEY (`id`),
  UNIQUE KEY `compositeInitiatingUserIdTargetUserIdIndex` (`initiatingUserId`,`targetUserId`),
  KEY `friends_initiating_user_id` (`initiatingUserId`),
  KEY `friends_target_user_id` (`targetUserId`),
  CONSTRAINT `friends_ibfk_1` FOREIGN KEY (`initiatingUserId`) REFERENCES `users` (`id`) ON DELETE CASCADE ON UPDATE CASCADE,
  CONSTRAINT `friends_ibfk_2` FOREIGN KEY (`targetUserId`) REFERENCES `users` (`id`) ON DELETE CASCADE ON UPDATE CASCADE
);

INSERT INTO `users` (`id`, `firstName`, `lastName`) VALUES 
(1, 'A', 'A'),
(2, 'B', 'B'),
(3, 'C', 'C'),
(4, 'D', 'D'),
(5, 'E', 'E'),
(6, 'F', 'F'),
(7, 'G', 'G');

INSERT INTO `friends` (`id`, `initiatingUserId`, `targetUserId`, `status`) VALUES 
( 1, 1, 2, 'active'),
( 2, 1, 3, 'active'),
( 3, 4, 1, 'active'),
( 4, 2, 3, 'active'),
( 5, 2, 4, 'active'),
( 6, 5, 2, 'active'),
( 7, 2, 6, 'active'),
( 8, 2, 7, 'active'),
( 9, 3, 5, 'active'),
(10, 6, 3, 'active'),
(11, 4, 5, 'active'),
(12, 6, 7, 'active');

...和SQL Fiddle of same

假设我们为用户 ID 1 提取建议用户，所需的输出是：

+----------------+-------------+
| strangerUserId | mutualCount |
+----------------+-------------+
| 5              | 2           |
+----------------+-------------+
| 6              | 2           |
+----------------+-------------+
| 4              | 1           |
+----------------+-------------+
| 7              | 1           |
+----------------+-------------+

我确实有一个解决方案，但它假设我们已经拥有用户朋友的 ID，我不确定它的运行速度：

select
count(*) as 'mutualCount',
case when f.initiatingUserId in (2, 3) then f.targetUserId else f.initiatingUserId end as strangerUserId
from friends f
    join users initiating on f.initiatingUserId=initiating.id
    join users target on targetUserId=target.id
and (
(
    initiatingUserId in (2, 3)
    and targetUserId not in (1, 2, 3)
)
or 
(
    targetUserId in (2, 3)
    and initiatingUserId not in (1, 2, 3)
)
)
group by strangerUserId
order by mutualCount desc;

Answer 1

我想出了这个快速（仅限 MySql）解决方案。

<块引用>

注意：此查询未针对您现有的索引进行优化，总体而言，在较大的数据集上可能会出现性能问题。如果您需要添加一些额外的条件，也很难修改查询。我会使用查询的各个部分先拉朋友列表，然后是非朋友列表，然后运行单独的查询来确定共同朋友的数量。

因此，这里是拉取 userId '1' 的共同好友计数的非好友列表的查询。您必须使用动态变量替换查询中的“1”硬编码 userId 值：

SELECT COUNT(fnf.friendOfNonFriendId) AS mutualFriednCount, fnf.nonFriendId
 FROM
    -- Subquery to pull the list of non-friends and their friends
    (SELECT
        IF(f.initiatingUserId = nf.nonFriendId, f.targetUserId, f.initiatingUserId) AS friendOfNonFriendId,
        nf.nonFriendId
    FROM friends f
    JOIN
        (SELECT u.id AS nonFriendId
        FROM users u
        LEFT JOIN (
            SELECT IF(f.initiatingUserId = 1, f.targetUserId, f.initiatingUserId) AS friendId
            FROM friends f
            WHERE f.status = 'active' AND (f.initiatingUserId = 1 OR f.targetUserId = 1)) f
                                           ON f.friendId = u.id
        WHERE f.friendId IS NULL AND u.id != 1) nf ON nf.nonFriendId = f.initiatingUserId
          OR  nf.nonFriendId = f.targetUserId) fnf
 -- Joining the subquery with the list of friends of a user with ID 1 to filter the above subquery to only mutual friends 
 JOIN 
    (SELECT IF(f.initiatingUserId = 1, f.targetUserId, f.initiatingUserId) AS friendId
    FROM friends f
    WHERE f.status = 'active' AND (f.initiatingUserId = 1 OR f.targetUserId = 1)) f
                                   ON f.friendId = fnf.friendOfNonFriendId
 GROUP BY fnf.nonFriendId

根据共同好友数查询推荐好友？

1 个答案: