如何编写SQL查询以通过频道和节目播放以下数据？ [关闭]

Posted 2021-05-03

我坚持使用查询逻辑，请找到表结构的图像。我有两个表，其中一个表是User，说明用户在哪个时间段看到的电视频道，还有另一个表我们有所有频道列表，在给定日期我们有所有播放的节目。

我需要用通道表来爆炸用户表，在用户之间，通道在给定时间内播放的节目是什么。

答案

请你检查以下SELECT语句的结果

select
*,
case 
    when ( u.starttime <= c.start and u.endtime >= c.[end])
    then datediff(mi, c.start,c.[end]) 
    when ( c.start <= u.starttime and c.[end] >= u.endtime)
    then datediff(mi, u.starttime,u.endtime) 
    when (u.endtime between c.start and c.[end]) 
    then datediff(mi, c.start, u.endtime) 
    when (c.[end] between u.starttime and u.endtime) 
    then datediff(mi, u.starttime,c.[end]) 
    end as period
from  [user] u
inner join channel c
    on  u.channel = c.channel and (
        u.starttime between c.start and c.[end] or
        u.endtime between c.start and c.[end] or 
        ( u.starttime <= c.start and u.endtime >= c.[end]) or
        ( c.start <= u.starttime and c.[end] >= u.endtime)
        )

这是可用于解决方案的元数据脚本

  create table [user] (
  name varchar(10),
  aga int,
  channel varchar(10),
  starttime time,
  endtime time
  )
    create table channel (
  channel varchar(10),
  title varchar(10),
  [start] time,
  [end] time
  )

insert into [user] select 'a',12,'abc','0:10','0:15'
insert into [user] select 'a',12,'abc','0:16','1:00'  
insert into [user] select 'a',12,'xyz','1:10','1:30'
insert into [user] select 'a',12,'xyz','1:40','1:50'
insert into [user] select 'a',12,'xyz','2:20','2:40'
insert into [user] select 'a',12,'xyz','2:40','3:00'
insert into [user] select 'a',12,'abc','5:30','6:40'
insert into [channel] select 'abc','a','0:00','0:12'
insert into [channel] select 'abc','b','0:12','0:15'
insert into [channel] select 'abc','c','0:15','0:30'
insert into [channel] select 'abc','1','0:30','0:55'
insert into [channel] select 'abc','2','0:45','1:15'
insert into [channel] select 'abc','3','5:00','5:35'
insert into [channel] select 'abc','4','5:35','6:35'
insert into [channel] select 'abc','5','6:35','7:35'
insert into [channel] select 'abc','6','7:35','7:45'
insert into [channel] select 'abc','7','7:45','8:45'

insert into [channel] select 'xyz','1','1:00','2:00'
insert into [channel] select 'xyz','2','2:00','2:10'
insert into [channel] select 'xyz','3','2:10','2:40'
insert into [channel] select 'xyz','4','2:40','3:00'
insert into [channel] select 'xyz','5','3:00','3:45'
insert into [channel] select 'xyz','6','3:45','3:50'

insert into [channel] select 'esk','N','2:00','3:00'
insert into [user] select 'M',10,'esk','1:00','4:00'

另一答案

当每个在另一个结束之前开始时范围重叠WHERE a.start < b.cease AND b.start < a.start

然后重叠的开始是两次开始的最高值，并且重叠的停止是两次停止中的最低值。

SELECT
    CASE WHEN c.start > u.start THEN c.start ELSE u.start END   overlap_start,
    CASE WHEN c.cease < u.cease THEN c.cease ELSE u.cease END   overlap_cease,
    *
FROM
    user_viewing        u
INNER JOIN
    channel_schedule    c
        ON  c.channel_id = u.channel_id
        AND c.start      < u.cease
        AND c.cease      > u.start
;

http://sqlfiddle.com/#!6/fd02e/3

在测试时要记住的重要案例包括在SQLFiddle中： - a在b开始之前开始并停止 - a在b开始之前开始，并在b结束 - a在b开始和停止 - a在b开始并在b停止后停止 - a在b停止后开始并停止 - a在b开始之前开始并在b停止之后停止

           |-------A------->

|--1--> |--2--> |--3--> |--4--> |--5-->
        |----------6---------->