Question

我是SQL的初学者，我正在斯坦福拉古尼塔上讨论运动问题并遇到一个奇怪的行为，当我使用时，我会得到不同的结果！=而不是＆lt;对于不相等的值。

以下是问题：

“对于同一评论者对同一部电影进行过两次评分的所有情况第二次给它一个更高的评价，返回评论者的名字和电影的标题。“

这是架构：

/* 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');

我的工作解决方案：

SELECT Reviewer.name, Movie.title
FROM Rating r1, Rating r2, Movie, Reviewer
WHERE Reviewer.rID = r1.rID and Reviewer.rID = r2.rID and 
      Movie.mID = r1.mID and Movie.mID = r2.mID and 
      r1.rID = r2.rID and r1.mID = r2.mID and 
      r1.ratingDate < r2.ratingDate and
      r2.stars > r1.stars

现在，如果您查看架构，您会看到在同一部电影上有重复评论的评论者（只有评论者rID：201和203）有不同的评分日期，所以如果我改变 r1.ratingDate < r2.ratingDate到r1.ratingDate != r2.ratingDate，结果将包括Brittany Harris（rID：203），这也是错误的。

有人能告诉我为什么会这样吗？

由于

Answer 1

r1.ratingDate < r2.ratingDate and r2.stars > r1.stars

意味着：返回那些第二次评分并给予更多明星的人。

如果你没有强制执行日期顺序（！=）那么该对的另一个记录（由联接产生的交叉产品）将导致返回Brittany Harris。尝试将两者都设置为！=，您将在不同时间看到所有不同评级的记录对。

使用！=而不是＆lt;

1 个答案: