我有这个代码及其临时表，所以你可以操作它。 $ t) f4 v2 b1 @- f8 s0 ocreate table #student( id int identity firstname varchar lastname varchar(50))create table #quiz( id int identity quiz_name varchar(50))create table #quiz_details( id int identity quiz_id int, student_id int)insert into #student(firstname,lastname)values ('LeBron','James'),('Stephen','Curry')insert into #quiz(quiz_name)values('NBA 50 Greatest Player Quiz'),('NBA Top 10 3 point shooters')insert into #quiz_details(quiz_id,student_id)values (1，2)drop table #studentdrop table #quizdrop table #quiz_details所以你可以看到勒布朗·詹姆斯（Lebron James）参加了NBA斯蒂芬·库里（Stephen Curry）则获得了NBA5 s2 f7 v) p) l
测试50名最佳球员。. {7 X4 Z/ D% k5 K
我只想得到他们还没有得到的东西，比如勒布朗还没有参加过50个最伟大的球员，所以我想要的就是这样。3 W: R1 j* K6 _4 ~0 }
id quiz_name firstname lastname----------------------------------------------------1 NBA 50 Greatest Player Quiz NULL NULL我需要两个参数，即lebron的ID和测验的ID，以便我知道lebron或stephen它还没有被接受，但如果值student_id还是空的，我该怎么办？3 @/ G# q, S R) g) a
我的尝试：& ~6 | l$ v, p& q4 ~/ F$ {" S
select QD.id, Q.quiz_name, S.firstname, S.lastnamefrom #quiz_details QDinner join #quiz Q on Q.id = QD.quiz_idinner join #student S on S.id = QD.student_id 1 C3 w$ X h ^! R解决方案: 6 Y' C2 a$ [2 G4 s- [6 L9 o 这应该让你开始： 4 h s M+ [, j: y% n& b F7 F' i+ B: ^-- filter out the student and quiz you wantDECLARE @qid INT = 1DECLARE @sid INT = 1SELECT * FROM #student AS sINNER JOIN #quiz AS q -- you want the quiz ON 1=1LEFT OUTER JOIN #quiz_details AS qd -- left join here to get result where rows not found ON qd.id = q.id AND qd.student_id=s.idWHERE s.id = @sid AND q.id = @qid AND qd.id IS NULL -- only return quizes not taken
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2023-09-14
