2023年6月23日发(作者：)

MYSQL经典SQL之时长统计概述假设我们有⼀张表，记录了⽤户的登⼊、登出信息，我们需要根据这些信息统计⽤户的在线时长，相应的设置⽤户等级，就像挂QQ升级表数据如下：mysql> select * from user_sessions;+----+----------+-------+----------+----------+| id | platform | usr | start | end |+----+----------+-------+----------+----------+| 1 | web | user1 | 08:30:00 | 10:30:00 || 2 | web | user2 | 08:30:00 | 08:45:00 || 3 | web | user1 | 09:00:00 | 09:30:00 || 4 | web | user2 | 09:15:00 | 10:30:00 || 5 | web | user1 | 09:15:00 | 09:30:00 || 6 | web | user2 | 10:30:00 | 14:30:00 || 7 | web | user1 | 10:45:00 | 11:30:00 || 8 | web | user2 | 11:00:00 | 12:30:00 || 9 | app | user1 | 08:30:00 | 08:45:00 || 10 | app | user2 | 09:00:00 | 09:30:00 || 11 | app | user1 | 11:45:00 | 12:00:00 || 12 | app | user2 | 12:30:00 | 14:00:00 || 13 | app | user1 | 12:45:00 | 13:30:00 || 14 | app | user2 | 13:00:00 | 14:00:00 || 15 | app | user1 | 14:00:00 | 16:30:00 || 16 | app | user2 | 15:30:00 | 17:00:00 |+----+----------+-------+----------+----------+我们的站点有两个⼊⼝：web端和app端很明显，这些数据时间上是有重叠的，例如user1在08:30～10:30这个时间段内⽤三个web端和⼀个app端登录，但他的时长应该只能算作2个⼩时要正确统计，我们需要三步：

1. 获取⽤户的重叠时间段中最早的起始时间

2. 获取⽤户的重叠时间段中最晚的结束时间

3. 前两步的起始时间与结束时间合并，得到综合的时间段，再进⾏统计第⼀步：查询⽤户名、起始时间

其中起始时间需要满⾜的条件是：不包含于其它时间段，即不存在⼀条记录来满⾜条件“s>start且s<=end”

得到的结果可能存在重复，所以⽤distinct去重

这样找到的s就是⽤户重叠时间段的最早时间。如下：mysql> select distinct usr,start from user_sessions as a where not exists(select * from user_sessions as b where = and > an+-------+----------+| usr | start |+-------+----------+| user1 | 08:30:00 || user2 | 08:30:00 || user1 | 10:45:00 || user2 | 09:00:00 || user1 | 11:45:00 || user1 | 12:45:00 || user1 | 14:00:00 || user2 | 15:30:00 |+-------+----------+第⼆步：查询⽤户名、结束时间

这步于第⼀步⼀样道理，结束时间满⾜的条件是：不包含于其它时间段，即不存在⼀条记录满⾜“e>=start且emysql> select distinct usr,end from user_sessions as a where not exists(select * from user_sessions as b where = and >= an+-------+----------+| usr | end |+-------+----------+| user1 | 10:30:00 || user2 | 08:45:00 || user2 | 14:30:00 || user1 | 11:30:00 || user1 | 12:00:00 || user1 | 13:30:00 || user1 | 16:30:00 || user2 | 17:00:00 |+-------+----------+第三步：合并前两步的结果

为了⽅便查询和叙述，我们将前两步建⽴对应的视图：mysql> create view v_s as select distinct usr,start from user_sessions as a where not exists(select * from user_sessions as b where = andmysql> create view v_e as select distinct usr,end from user_sessions as a where not exists(select * from user_sessions as b where = and查看v_s和v_e两个视图：mysql> select * from v_s;+-------+----------+| usr | start |+-------+----------+| user1 | 08:30:00 || user2 | 08:30:00 || user1 | 10:45:00 || user2 | 09:00:00 || user1 | 11:45:00 || user1 | 12:45:00 || user1 | 14:00:00 || user2 | 15:30:00 |+-------+----------+mysql> select * from v_e;+-------+----------+| usr | end |+-------+----------+| user1 | 10:30:00 || user2 | 08:45:00 || user2 | 14:30:00 || user1 | 11:30:00 || user1 | 12:00:00 || user1 | 13:30:00 || user1 | 16:30:00 || user2 | 17:00:00 |+-------+----------+可以看到，两个视图总⾏数是相等的，但并不是按⾏对应的，所以并不是单纯的将两个结果横向合并！⽅案⼀：对于v_s中的起始时间，我们应该在v_e中找“usr相同且end>start”的最⼩值

例如，对于v_s中user1起始时间为08:30:00，我们应该找到v_e中end>08:30:00且usr=user1的记录，然后取min得到结果为10:30:00作为与之对应的结束时间

SQL及结果如下：mysql> select distinct v_,v_,(select min(end) from v_e where v_>v_ and v_=v_) as end from v_s,v_e where v_+-------+----------+----------+| usr | start | end |+-------+----------+----------+| user1 | 08:30:00 | 10:30:00 || user2 | 08:30:00 | 08:45:00 || user1 | 10:45:00 | 11:30:00 || user2 | 09:00:00 | 14:30:00 || user1 | 11:45:00 | 12:00:00 || user1 | 12:45:00 | 13:30:00 || user1 | 14:00:00 | 16:30:00 || user2 | 15:30:00 | 17:00:00 |+-------+----------+----------+在此基础上，按usr进⾏分组，统计⽤户的总时长，得到最终结果：mysql> select usr,sec_to_time(sum(timestampdiff(second,start,end))) as time from (select distinct v_,v_,(select min(end) from v_e wher+-------+----------+| usr | time |+-------+----------+| user1 | 06:15:00 || user2 | 07:15:00 |+-------+----------+其中，timestampdiff函数是计算两个时间的差值，第⼀个参数⽤来指定其单位，这⾥指定为秒；

sec_to_time函数是将求和后的秒数再转为时间格式来显⽰⽅案⼆既然v_s与v_e的数据不是按⾏对应的，那么我们可以对其排序，使其按⾏对应后，就可以直接按⾏合并

为了按⾏合并，我们同时加上⾏号，让两个⼦查询⾏号匹配即可：mysql> set @sno=0;mysql> set @eno=0;+-------+----------+----------+| usr | start | end |+-------+----------+----------+| user1 | 08:30:00 | 10:30:00 || user1 | 10:45:00 | 11:30:00 || user1 | 11:45:00 | 12:00:00 || user1 | 12:45:00 | 13:30:00 || user1 | 14:00:00 | 16:30:00 || user2 | 08:30:00 | 08:45:00 || user2 | 09:00:00 | 14:30:00 || user2 | 15:30:00 | 17:00:00 |+-------+----------+----------+mysql> select ,, from (select @sno:=@sno+1 as sno,usr,start from v_s order by usr,start) as s, (select @eno:=@eno+1 as eno,usr,end from之后的分组统计与⽅案⼀相同扩展实际场景中，我们经常将不同的⽅式的在线时长加上奖励倍数，来⿎励⽤户更多的使⽤某种⼊⼝。

例如，我们为了让⽤户更多的使⽤app登录，规定app的在线时长奖励2倍，即app在线⼀⼩时相当于web在线两⼩时。

这时，我们就需要将不同的登录⽅式区分开来统计。要区分开也很简单，只需要添加SQL条件让platform相同即可，例如第⼀步修改为：mysql> select distinct usr,platform,start from user_sessions as a where not exists(select * from user_sessions as b where = and r+-------+----------+----------+| usr | platform | start |+-------+----------+----------+| user1 | web | 08:30:00 || user2 | web | 08:30:00 || user2 | web | 09:15:00 || user1 | web | 10:45:00 || user1 | app | 08:30:00 || user2 | app | 09:00:00 || user1 | app | 11:45:00 || user2 | app | 12:30:00 || user1 | app | 12:45:00 || user1 | app | 14:00:00 || user2 | app | 15:30:00 |+-------+----------+----------+之后的步骤类似，在最终统计时分别统计web和app的时长，并将app的时长乘以2再与web时长相加即可，不再赘述。