如何获得这个特定"查询"的结果

Question

注意:我不小心在这里提出了另一个问题的句子(对我来说是非常道歉),我已经在3月14日星期三下午23:21更新了这篇文章,并提出了正确的问题.

我花了几个小时试图在没有任何人帮助的情况下弄清楚这个问题,但是我意识到我浪费了太多的生产时间,应该早点问别人.我有一个不错的裂缝,并且已经如此接近,但无法得到我需要的最终解决方案.我应该得到的是:

对于同一评论者对同一部电影进行两次评分并且第二次给予更高评级的所有情况,请返回评论者的姓名和电影的标题.

这是我设法到达的查询:

SELECT reviewer.name, movie.title, rating.stars  
FROM (reviewer JOIN rating ON reviewer.rid = rating.rid)  
JOIN movie ON movie.mid = rating.mid  
GROUP BY reviewer.name 
HAVING COUNT(*) >= 2  
ORDER BY reviewer.name DESC

(我觉得上面的查询中缺少WHERE子句,但不知道放在哪里)

(据我所知,SQLite目前不支持RIGHT和FULL OUTER JOIN)

以下是表格和数据(图片中)......

......还有DB代码......

/* Delete the tables if they already exist */
drop table if exists Movie;
drop table if exists Reviewer;
drop table if exists Rating;

/* Create the schema for our tables */
create table Movie(mID int, title text, year int, director text);
create table Reviewer(rID int, name text);
create table Rating(rID int, mID int, stars int, ratingDate date);

/* Populate the tables with our data */
insert into Movie values(101, 'Gone with the Wind', 1939, 'Victor Fleming');
insert into Movie values(102, 'Star Wars', 1977, 'George Lucas');
insert into Movie values(103, 'The Sound of Music', 1965, 'Robert Wise');
insert into Movie values(104, 'E.T.', 1982, 'Steven Spielberg');
insert into Movie values(105, 'Titanic', 1997, 'James Cameron');
insert into Movie values(106, 'Snow White', 1937, null);
insert into Movie values(107, 'Avatar', 2009, 'James Cameron');
insert into Movie values(108, 'Raiders of the Lost Ark', 1981, 'Steven Spielberg');

insert into Reviewer values(201, 'Sarah Martinez');
insert into Reviewer values(202, 'Daniel Lewis');
insert into Reviewer values(203, 'Brittany Harris');
insert into Reviewer values(204, 'Mike Anderson');
insert into Reviewer values(205, 'Chris Jackson');
insert into Reviewer values(206, 'Elizabeth Thomas');
insert into Reviewer values(207, 'James Cameron');
insert into Reviewer values(208, 'Ashley White');

insert into Rating values(201, 101, 2, '2011-01-22');
insert into Rating values(201, 101, 4, '2011-01-27');
insert into Rating values(202, 106, 4, null);
insert into Rating values(203, 103, 2, '2011-01-20');
insert into Rating values(203, 108, 4, '2011-01-12');
insert into Rating values(203, 108, 2, '2011-01-30');
insert into Rating values(204, 101, 3, '2011-01-09');
insert into Rating values(205, 103, 3, '2011-01-27');
insert into Rating values(205, 104, 2, '2011-01-22');
insert into Rating values(205, 108, 4, null);
insert into Rating values(206, 107, 3, '2011-01-15');
insert into Rating values(206, 106, 5, '2011-01-19');
insert into Rating values(207, 107, 5, '2011-01-20');
insert into Rating values(208, 104, 3, '2011-01-02');

我有另一个相对类似的问题,但是如果我在这个问题上得到一些帮助,我应该可以将这个模式和技术应用到下一个.

提前致谢!:)

Answer 1

我添加了一个带有派生表的内连接,它返回每部电影的最大星数.由于电影和评级之间的内部联接,仅检索具有评级的电影.将其加入主查询以获得每部电影的最大明星数.

注意:您声明您希望通过电影标题订购,但您的查询订单由评论者订购.

SELECT reviewer.name, movie.title, rating.stars, maxStarsPerMovie.MaxStars
FROM (reviewer JOIN rating ON reviewer.rid = rating.rid)  
JOIN movie ON movie.mid = rating.mid  
join
(
   select movie.mid, max(rating.stars) MaxStars
   from movie
   inner join rating
      on movie.mid = rating.mid
   group by movie.mid
) maxStarsPerMovie
on movie.mid = maxStarsPerMovie.mid
ORDER BY reviewer.name DESC

编辑:要求改变了.此查询将返回评论者列表,这些评论者在以后更改他们的观点以支持电影.它是通过第二次加入评级,在星星和日期加入两个过滤器来实现的.

SELECT reviewer.name, movie.title, rating.ratingDate, rating.stars,
       newRating.ratingDate newRatingDate, newRating.Stars newRatingStars
FROM (reviewer JOIN rating ON reviewer.rid = rating.rid)  
JOIN movie ON movie.mid = rating.mid 
inner join rating newRating
     on newRating.mid = movie.mid
        and newRating.rid = reviewer.rid
        and newRating.ratingdate > rating.ratingdate
        and newRating.stars > rating.stars
ORDER BY reviewer.name, movie.title