我怎样才能做到这一点Oracle查询

Question

声明：

SERIES: 
title   
total_seasons 

SEASONS: 
title   
season  
avgduration 
episodes 
PRIMARY KEY (title, season), 
FOREIGN KEY (title) REFERENCES SERIES 

LIC_SERIES: 
id_client 
datetime 
title 
season 
episode 
PRIMARY KEY (client,title,season,episode) 
FOREIGN KEY (title,season) REFERENCES seasons 

系列例子是这样的：

Title  Total_Seasons 
House  8 
Lost  6 

四季如

表 “谁购买了全套系列许可证的客户”看起来像这样：

Title Season AvgDuration Episodes 
House 1  44   22 
Mad Men 3  47   13 

Lic_Series例子是这样的：

Client Datetime Title Season Episode 
92/657/T 01/01/2017 House 8  18 

我试过吗？

WITH PART AS (SELECT DISTINCT EPISODES, TITLE 
FROM SEASONS INNER JOIN SERIES 
ON Seasons.Title = Series.Title 
GROUP BY Seasons.Title) 

SELECT CLIENT 
FROM LIC_SERIES INNER JOIN PART 
ON LIC_SERIES.TITLE = PART.TITLE 
ORDER BY title DESC; 

我不知道如何检查，如果一个客户有所有每个赛季

Answer 1

这将返回所有和没有发作的次数所拥有的情节，并且有问题的一系列的四季由每个季节的每个客户。要检查是否拥有所有剧集，请对episodes_not_owned为零的摘要执行额外的检查。

WITH PART AS 
(SELECT DISTINCT s2.TITLE, s1.SEASON, s2.EPISODE 
FROM SEASONS s1 
INNER JOIN SERIES s2 
ON s1.Title = s2.Title 
) 
,summary as (
SELECT l.CLIENT 
     ,p.SEASON 
     ,SUM(CASE WHEN l.episode is not null THEN 1 ELSE NULL END) as episodes_owned 
     ,SUM(CASE WHEN l.episode is null THEN 1 ELSE NULL END) as episodes_not_owned 
FROM LIC_SERIES l 
RIGHT OUTER JOIN PART p 
ON l.TITLE = p.TITLE 
AND l.EPISODE = p.EPISODE 
AND l.SEASON = p.SEASON 
GROUP BY l.CLIENT, p.SEASON 
ORDER BY l.CLIENT, p.SEASON DESC 
) 
SELECT s.client 
     ,s.season 
     ,s.episodes_owned 
     ,s.episodes_not_owned 
     ,case when s.episodes_not_owned = 0 THEN 'Yes' ELSE 'No' END as all_owned 
from summary s;