MySQL中的朋友关系

8

我正在MySQL中建立一种友谊关系，其中的朋友关系是相互的。如果A是B的朋友，则B是A的朋友。如果用户之一结束友谊，则关系下降。我想学习哪种方法更好。

我有一个正在运行的系统；

user
-----------
userid p.k
name 

friends
-------
userid
friendid
primary key (`userid`,`friendid`),
key `friendid` (`friendid`)

1 2
2 5
1 3


To get all of my friends;
SELECT u.name, f.friendid , IF(f.userid = $userid, f.friendid, f.userid) friendid 
FROM friends f 
    inner join user u  ON ( u.userid = IF(f.userid = $userid, f.friendid, f.userid)) 
WHERE ( f.userid = '$userid' or f.friendid = '$userid' )

此查询工作良好。也许我可以添加一个UNION。查询比下面的查询更复杂，并且该表包含的记录是下面的查询的一半。

另一种方法是将关系保持在单独的行中。

1 2
2 1
2 5
5 2
1 3
3 1

SELECT u.name, f.friendid 
FROM friends f inner join user u ON ( u.userid = f.friendid ) 
WHERE f.userid = '$userid'

该查询很简单，尽管表占用的空间是原来的两倍。

我的担心是；假设有数百万用户；哪种方法会更快？

两种方式的优缺点是什么？

对于这些方式，我应该牢记或改变什么？两种方式我都可能面对什么问题？

mysql  relational-theory 
— 肯特·伊柳克
source
您今天问的这个问题很好。为您的问题+1。
— RolandoMySQLDBA 2012年

Answers:

4

引起我注意的第一件事是的索引设置friends。

您现在有这个：

friends
-------
userid
friendid
primary key (`userid`,`friendid`),
key `friendid` (`friendid`)

在相互检查相互友谊时，这可能会产生一些开销，因为遍历friendid索引时可能会从表中检索到用户ID 。也许您可以索引如下：

friends
-------
userid
friendid
primary key (`userid`,`friendid`),
unique key `friendid` (`friendid`,`userid`)

这可能消除了访问表和仅搜索索引的任何需求。

现在，就查询而言，它们都可以通过新的唯一索引得到改善。创建唯一索引也无需插入(A,B)并(B,A)到表中，因为(A,B)和(B,A)将是该指数反正。因此，第二个查询将不必遍历表格以查看某人是否是其他人的朋友，因为另一个人发起了友谊。这样，如果友谊仅被一个人破坏，那么就不会有单面的孤儿友谊（这些日子似乎很像生活，不是吗？）

您的第一个查询似乎将从唯一索引中受益更多。即使有数百万行，仅使用索引查找朋友也可以避免触摸表格。不过，由于您没有提出UNION查询，因此我建议您使用UNION查询：

SET @givenuserid = ?;
SELECT B.name "Friend's Name"
FROM 
(
    SELECT userid FROM friends WHERE friendid=@givenuserid
    UNION
    SELECT friendid FROM friends WHERE userid=@givenuserid
) A INNER JOIN user B USING (userid);

这将使您看到谁是每个用户标识的朋友

要查看所有友谊，请运行以下命令：

SELECT A.userid,A.name,B.friendid,C.name
FROM user A
INNER JOIN friends B ON A.userid=B.userid
INNER JOIN user C on B.friendid=C.userid;

首先，这是一些示例数据：

mysql> drop database if exists key_ilyuk;
Query OK, 2 rows affected (0.01 sec)

mysql> create database key_ilyuk;
Query OK, 1 row affected (0.00 sec)

mysql> use key_ilyuk
Database changed
mysql> create table user
    -> (
    ->     userid INT NOT NULL AUTO_INCREMENT,
    ->     name varchar(20),
    ->     primary key(userid)
    -> ) ENGINE=MyISAM;
Query OK, 0 rows affected (0.04 sec)

mysql> insert into user (name) values
    -> ('rolando'),('pamela'),('dominique'),('carlik'),('diamond');
Query OK, 5 rows affected (0.01 sec)
Records: 5  Duplicates: 0  Warnings: 0

mysql> create table friends
    -> (
    ->     userid INT NOT NULL,
    ->     friendid INT NOT NULL,
    ->     primary key (userid,friendid),
    ->     unique key (friendid,userid)
    -> ) ENGINE=MyISAM;
Query OK, 0 rows affected (0.03 sec)

mysql> insert into friends values (1,2),(2,5),(1,3);
Query OK, 3 rows affected (0.00 sec)
Records: 3  Duplicates: 0  Warnings: 0

mysql> select * from user;
+--------+-----------+
| userid | name      |
+--------+-----------+
|      1 | rolando   |
|      2 | pamela    |
|      3 | dominique |
|      4 | carlik    |
|      5 | diamond   |
+--------+-----------+
5 rows in set (0.00 sec)

mysql> select * from friends;
+--------+----------+
| userid | friendid |
+--------+----------+
|      1 |        2 |
|      1 |        3 |
|      2 |        5 |
+--------+----------+
3 rows in set (0.00 sec)

mysql>

让我们看一下所有的关系

mysql> SELECT A.userid,A.name,B.friendid,C.name
    -> FROM user A
    -> INNER JOIN friends B ON A.userid=B.userid
    -> INNER JOIN user C on B.friendid=C.userid
    -> ;
+--------+---------+----------+-----------+
| userid | name    | friendid | name      |
+--------+---------+----------+-----------+
|      1 | rolando |        2 | pamela    |
|      1 | rolando |        3 | dominique |
|      2 | pamela  |        5 | diamond   |
+--------+---------+----------+-----------+
3 rows in set (0.00 sec)

mysql>

让我们看一下所有5个用户ID，看看关系是否正确显示

mysql> SET @givenuserid = 1;
Query OK, 0 rows affected (0.00 sec)

mysql> SELECT B.name "Friend's Name"
    -> FROM
    -> (
    ->     SELECT userid FROM friends WHERE friendid=@givenuserid
    ->     UNION
    ->     SELECT friendid FROM friends WHERE userid=@givenuserid
    -> ) A INNER JOIN user B USING (userid);
+---------------+
| Friend's Name |
+---------------+
| pamela        |
| dominique     |
+---------------+
2 rows in set (0.00 sec)

mysql> SET @givenuserid = 2;
Query OK, 0 rows affected (0.00 sec)

mysql> SELECT B.name "Friend's Name"
    -> FROM
    -> (
    ->     SELECT userid FROM friends WHERE friendid=@givenuserid
    ->     UNION
    ->     SELECT friendid FROM friends WHERE userid=@givenuserid
    -> ) A INNER JOIN user B USING (userid);
+---------------+
| Friend's Name |
+---------------+
| rolando       |
| diamond       |
+---------------+
2 rows in set (0.00 sec)

mysql> SET @givenuserid = 3;
Query OK, 0 rows affected (0.00 sec)

mysql> SELECT B.name "Friend's Name"
    -> FROM
    -> (
    ->     SELECT userid FROM friends WHERE friendid=@givenuserid
    ->     UNION
    ->     SELECT friendid FROM friends WHERE userid=@givenuserid
    -> ) A INNER JOIN user B USING (userid);
+---------------+
| Friend's Name |
+---------------+
| rolando       |
+---------------+
1 row in set (0.01 sec)

mysql> SET @givenuserid = 4;
Query OK, 0 rows affected (0.00 sec)

mysql> SELECT B.name "Friend's Name"
    -> FROM
    -> (
    ->     SELECT userid FROM friends WHERE friendid=@givenuserid
    ->     UNION
    ->     SELECT friendid FROM friends WHERE userid=@givenuserid
    -> ) A INNER JOIN user B USING (userid);
Empty set (0.00 sec)

mysql> SET @givenuserid = 5;
Query OK, 0 rows affected (0.00 sec)

mysql> SELECT B.name "Friend's Name"
    -> FROM
    -> (
    ->     SELECT userid FROM friends WHERE friendid=@givenuserid
    ->     UNION
    ->     SELECT friendid FROM friends WHERE userid=@givenuserid
    -> ) A INNER JOIN user B USING (userid);
+---------------+
| Friend's Name |
+---------------+
| pamela        |
+---------------+
1 row in set (0.00 sec)

mysql>

他们对我看来都是正确的。

现在，让我们使用第二个查询来查看它是否匹配...

mysql> SET @givenuserid = 1;
Query OK, 0 rows affected (0.00 sec)

mysql> SELECT u.name, f.friendid
    -> FROM friends f inner join user u ON ( u.userid = f.friendid )
    -> WHERE f.userid = @givenuserid;
+-----------+----------+
| name      | friendid |
+-----------+----------+
| pamela    |        2 |
| dominique |        3 |
+-----------+----------+
2 rows in set (0.00 sec)

mysql> SET @givenuserid = 2;
Query OK, 0 rows affected (0.01 sec)

mysql> SELECT u.name, f.friendid
    -> FROM friends f inner join user u ON ( u.userid = f.friendid )
    -> WHERE f.userid = @givenuserid;
+---------+----------+
| name    | friendid |
+---------+----------+
| diamond |        5 |
+---------+----------+
1 row in set (0.00 sec)

mysql> SET @givenuserid = 3;
Query OK, 0 rows affected (0.01 sec)

mysql> SELECT u.name, f.friendid
    -> FROM friends f inner join user u ON ( u.userid = f.friendid )
    -> WHERE f.userid = @givenuserid;
Empty set (0.00 sec)

mysql> SET @givenuserid = 4;
Query OK, 0 rows affected (0.00 sec)

mysql> SELECT u.name, f.friendid
    -> FROM friends f inner join user u ON ( u.userid = f.friendid )
    -> WHERE f.userid = @givenuserid;
Empty set (0.00 sec)

mysql> SET @givenuserid = 5;
Query OK, 0 rows affected (0.00 sec)

mysql> SELECT u.name, f.friendid
    -> FROM friends f inner join user u ON ( u.userid = f.friendid )
    -> WHERE f.userid = @givenuserid;
Empty set (0.00 sec)

mysql>

为什么不匹配？那是因为我没有(B,A)为每个加载(A,B)。让我加载(B,A)关系，然后再次尝试您的第二个查询。

mysql> insert into friends values (2,1),(5,2),(3,1);
Query OK, 3 rows affected (0.02 sec)
Records: 3  Duplicates: 0  Warnings: 0

mysql> SET @givenuserid = 1;
Query OK, 0 rows affected (0.01 sec)

mysql> SELECT u.name, f.friendid
    -> FROM friends f inner join user u ON ( u.userid = f.friendid )
    -> WHERE f.userid = @givenuserid;
+-----------+----------+
| name      | friendid |
+-----------+----------+
| pamela    |        2 |
| dominique |        3 |
+-----------+----------+
2 rows in set (0.00 sec)

mysql> SET @givenuserid = 2;
Query OK, 0 rows affected (0.00 sec)

mysql> SELECT u.name, f.friendid
    -> FROM friends f inner join user u ON ( u.userid = f.friendid )
    -> WHERE f.userid = @givenuserid;
+---------+----------+
| name    | friendid |
+---------+----------+
| rolando |        1 |
| diamond |        5 |
+---------+----------+
2 rows in set (0.00 sec)

mysql> SET @givenuserid = 3;
Query OK, 0 rows affected (0.00 sec)

mysql> SELECT u.name, f.friendid
    -> FROM friends f inner join user u ON ( u.userid = f.friendid )
    -> WHERE f.userid = @givenuserid;
+---------+----------+
| name    | friendid |
+---------+----------+
| rolando |        1 |
+---------+----------+
1 row in set (0.00 sec)

mysql> SET @givenuserid = 4;
Query OK, 0 rows affected (0.00 sec)

mysql> SELECT u.name, f.friendid
    -> FROM friends f inner join user u ON ( u.userid = f.friendid )
    -> WHERE f.userid = @givenuserid;
Empty set (0.00 sec)

mysql> SET @givenuserid = 5;
Query OK, 0 rows affected (0.00 sec)

mysql> SELECT u.name, f.friendid
    -> FROM friends f inner join user u ON ( u.userid = f.friendid )
    -> WHERE f.userid = @givenuserid;
+--------+----------+
| name   | friendid |
+--------+----------+
| pamela |        2 |
+--------+----------+
1 row in set (0.00 sec)

mysql>

他们仍然不匹配。那是因为您的第二个查询仅检查一侧。

让我们针对仅包含（A，B）而不包含（B，A）的每个值检查您的第一个查询：

mysql> SET @givenuserid = 1;
SELECT u.name, f.friendid userid, IF(f.userid = @givenuserid, f.friendid, f.userid) friendid
FROM friends f
    inner join user u  ON ( u.userid = IF(f.userid = @givenuserid, f.friendid, f.userid))
Query OK, 0 rows affected (0.00 sec)

mysql> SELECT u.name, f.friendid userid, IF(f.userid = @givenuserid, f.friendid, f.userid) friendid
    -> FROM friends f
    ->     inner join user u  ON ( u.userid = IF(f.userid = @givenuserid, f.friendid, f.userid))
    -> WHERE ( f.userid = @givenuserid or f.friendid = @givenuserid  );
+-----------+--------+----------+
| name      | userid | friendid |
+-----------+--------+----------+
| pamela    |      2 |        2 |
| dominique |      3 |        3 |
+-----------+--------+----------+
2 rows in set (0.00 sec)

mysql> SET @givenuserid = 2;
FROM friends f
    inner join user u  ON ( u.userid = IF(f.userid = @givenuserid, f.friendid, f.userid))
WHERE ( f.userid = @givenuserid or f.friendid = @givenuserid  );
Query OK, 0 rows affected (0.01 sec)

mysql> SELECT u.name, f.friendid userid, IF(f.userid = @givenuserid, f.friendid, f.userid) friendid
    -> FROM friends f
    ->     inner join user u  ON ( u.userid = IF(f.userid = @givenuserid, f.friendid, f.userid))
    -> WHERE ( f.userid = @givenuserid or f.friendid = @givenuserid  );
+---------+--------+----------+
| name    | userid | friendid |
+---------+--------+----------+
| rolando |      2 |        1 |
| diamond |      5 |        5 |
+---------+--------+----------+
2 rows in set (0.00 sec)

mysql> SET @givenuserid = 3;
SELECT u.name, f.friendid userid, IF(f.userid = @givenuserid, f.friendid, f.userid) friendid
FROM friends f
    inner join user u  ON ( u.userid = IF(f.userid = @givenuserid, f.friendid, f.userid))
WHERE ( f.userid = @givenuserid or f.friendid = @givenuserid  );
Query OK, 0 rows affected (0.00 sec)

mysql> SELECT u.name, f.friendid userid, IF(f.userid = @givenuserid, f.friendid, f.userid) friendid
    -> FROM friends f
    ->     inner join user u  ON ( u.userid = IF(f.userid = @givenuserid, f.friendid, f.userid))
    -> WHERE ( f.userid = @givenuserid or f.friendid = @givenuserid  );
+---------+--------+----------+
| name    | userid | friendid |
+---------+--------+----------+
| rolando |      3 |        1 |
+---------+--------+----------+
1 row in set (0.00 sec)

mysql> SET @givenuserid = 4;
FROM friends f
    inner join user u  ON ( u.userid = IF(f.userid = @givenuserid, f.friendid, f.userid))
WHERE ( f.userid = @givenuserid or f.friendid = @givenuserid  );
Query OK, 0 rows affected (0.00 sec)

mysql> SELECT u.name, f.friendid userid, IF(f.userid = @givenuserid, f.friendid, f.userid) friendid
    -> FROM friends f
    ->     inner join user u  ON ( u.userid = IF(f.userid = @givenuserid, f.friendid, f.userid))
    -> WHERE ( f.userid = @givenuserid or f.friendid = @givenuserid  );
Empty set (0.01 sec)

mysql> SET @givenuserid = 5;
FROM friends f
Query OK, 0 rows affected (0.00 sec)

    inner join user u  ON ( u.userid = IF(f.userid = @givenuserid, f.friendid, f.userid))
mysql> SELECT u.name, f.friendid userid, IF(f.userid = @givenuserid, f.friendid, f.userid) friendid
    -> FROM friends f
    ->     inner join user u  ON ( u.userid = IF(f.userid = @givenuserid, f.friendid, f.userid))
    -> WHERE ( f.userid = @givenuserid or f.friendid = @givenuserid  );
+--------+--------+----------+
| name   | userid | friendid |
+--------+--------+----------+
| pamela |      5 |        2 |
+--------+--------+----------+
1 row in set (0.00 sec)

mysql>

您的第一个效果很好。我敢肯定，正如我之前所说，它受益于唯一索引，但是恕我直言，我认为UNION更简单。通过唯一索引，就执行和输出而言，它似乎是其他六个索引中的六个。

您将不得不根据我的建议UNION对您的第一个查询进行基准测试，然后看看。

您今天问的这个问题很好。为您的问题+1。

— 罗兰多·MySQLDBA
source
我进行了一些测试，以了解当前设置的速度。我没有改变桌子的方案。第一个查询1,000,000行（用户表）2,045,007行（朋友表-每个关系一行。为10.000个用户随机创建友谊）第一个查询需要0.01094秒才能返回600行。用UNION更改的同一查询需要0.0086返回600行。第二个查询1,000,000行（用户表）4,048,781行（friends_twoway表-每个关系两行）我的第一篇文章中的第二个查询花费0.0090秒。返回600行。您如何看待这些结果？
— 肯特·伊柳克
经过一堆测试，我将更改表设置，并根据您的建议添加不同的索引。
— 肯特·伊柳克
在您的第一个测试中，.0086（使用UNION）要好于.01094（不使用UNION）。实际上，这快了27.21％。数据量增加一倍的首次查询的性能降低了0.0004秒。即使有给定的数字，我仍然倾向于仅使用数据并创建唯一索引的UNION，因为索引将在查询中完全使用，而无需考虑数据。
— RolandoMySQLDBA 2012年
我已经取代friendid键来唯一键（friendid，userid）和现在的结果是约0.00794这是尽可能快能得到什么？查看结果，您认为第一种方法更好（每个关系一行）？因为它的空间是第二个空间的两倍，并且结果与当前设置大致相同。
— 肯特·伊柳克
在您的特定情况下，由于依赖于索引，因此较少的数据是好的。索引过大，但出于有益的目的。这是一个叫做覆盖索引的概念，其目的是将其创建的索引WHERE，GROUP BY以及ORDER BY条款导致从只索引读取的数据。这里有一些很好的链接，它们证明使用唯一键和主键作为覆盖索引是合理的：1）peter-zaitsev.livejournal.com/6949.html，2） mysqlperformanceblog.com/2006/11/23/…，3） ronaldbradford .com / blog / tag / covering-index
— RolandoMySQLDBA 2012年
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