我怎樣才能做到這一點Oracle查詢

Question

聲明：

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 

系列例子是這樣的：

Title  Total_Seasons 
House  8 
Lost  6 

四季如

表 「誰購買了全套系列許可證的客戶」看起來像這樣：

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

Lic_Series例子是這樣的：

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

我試過嗎？

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; 

我不知道如何檢查，如果一個客戶有所有每個賽季

Answer 1

這將返回所有和沒有發作的次數所擁有的情節，並且有問題的一系列的四季由每個季節的每個客戶。要檢查是否擁有所有劇集，請對episodes_not_owned爲零的摘要執行額外的檢查。

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;