有3個表S，C，SC

S（SNO，SNAME）代表（學號，姓名）

C（CNO，CNAME，CTEACHER）代表（課號，課名，教師）

SC（SNO，CNO，SCGRADE）代表（學號，課號，成績）

問題：

1，找出沒選過“黎明”老師的所有學生姓名。

2，列出2門以上（含2門）不及格學生姓名及平均成績。

3，既學過1號課程又學過2號課所有學生的姓名。

CREATE TABLE SC
(
SNO           VARCHAR(200),
CNO           VARCHAR(200),
SCGRADE    VARCHAR(200)
);
CREATE TABLE S
(
SNO           VARCHAR(200),
SNAME       VARCHAR(200)
);
CREATE TABLE C
(
CNO           VARCHAR(200),
CNAME       VARCHAR(200),
CTEACHER   VARCHAR(200)
);

INSERT INTO C ( CNO, CNAME, CTEACHER ) VALUES ( '1', '語文', '張'); 
INSERT INTO C ( CNO, CNAME, CTEACHER ) VALUES ( '2', '政治', '王'); 
INSERT INTO C ( CNO, CNAME, CTEACHER ) VALUES ( '3', '英語', '李'); 
INSERT INTO C ( CNO, CNAME, CTEACHER ) VALUES ( '4', '數學', '趙'); 
INSERT INTO C ( CNO, CNAME, CTEACHER ) VALUES ( '5', '物理', '黎明'); 
commit;


INSERT INTO S ( SNO, SNAME ) VALUES ( '1', '學生1'); 
INSERT INTO S ( SNO, SNAME ) VALUES ( '2', '學生2'); 
INSERT INTO S ( SNO, SNAME ) VALUES ( '3', '學生3'); 
INSERT INTO S ( SNO, SNAME ) VALUES ( '4', '學生4'); 
commit;


INSERT INTO SC ( SNO, CNO, SCGRADE ) VALUES ( '1', '1', '40'); 
INSERT INTO SC ( SNO, CNO, SCGRADE ) VALUES ( '1', '2', '30'); 
INSERT INTO SC ( SNO, CNO, SCGRADE ) VALUES ( '1', '3', '20'); 
INSERT INTO SC ( SNO, CNO, SCGRADE ) VALUES ( '1', '4', '80'); 
INSERT INTO SC ( SNO, CNO, SCGRADE ) VALUES ( '1', '5', '60'); 
INSERT INTO SC ( SNO, CNO, SCGRADE ) VALUES ( '2', '1', '60'); 
INSERT INTO SC ( SNO, CNO, SCGRADE ) VALUES ( '2', '2', '60'); 
INSERT INTO SC ( SNO, CNO, SCGRADE ) VALUES ( '2', '3', '60'); 
INSERT INTO SC ( SNO, CNO, SCGRADE ) VALUES ( '2', '4', '60'); 
INSERT INTO SC ( SNO, CNO, SCGRADE ) VALUES ( '2', '5', '40'); 
INSERT INTO SC ( SNO, CNO, SCGRADE ) VALUES ( '3', '1', '60'); 
INSERT INTO SC ( SNO, CNO, SCGRADE ) VALUES ( '3', '3', '80'); 
commit;
 

select sname from s where sno not in
(
  select sno from sc where cno in
  (
    select distinct cno from c where cteacher='黎明'
  )
);

Select sname from s where sno in (select sno from sc where scgrade<60 group by sno having count(*)>=2) 

select s.sname, avg_grade from s
join
(select sno from sc where scgrade < 60 group by sno having count(*) >= 2) t1
on s.sno = t1.sno
join
(select sno, avg(scgrade) as avg_grade from sc group by sno ) t2
on s.sno = t2.sno;
 

select sname from s where
  sno in (select sno from sc where cno = 1)
  and
  sno in (select sno from sc where cno = 2);