我有这个代码及其临时表,所以你可以操作它。 9 a# m$ W& m% [4 Jcreate 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NBA0 B+ w$ a& K4 w& A
测试50名最佳球员。) }$ G0 m8 C9 S8 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。( O6 w! `; H& h$ u
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? " h* |4 V! `9 g# {我的尝试:( \1 r. E# i U8 I5 w+ b1 f7 I7 ~# a
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 / C3 K q) c e; A. _ 解决方案: ' \: `! k/ a% r( P% | 这应该让你开始: * t6 ]" S2 n9 {4 E% c c5 H-- 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