CREATE TABLE s(SNO VARCHAR(10),SNAME VARCHAR2(20));
CREATE TABLE c(CNO VARCHAR(10),CNAME VARCHAR2(20),CTEACHER VARCHAR2(10));
CREATE TABLE sc(SNO VARCHAR(10),CNO VARCHAR(10),SCGRADE NUMBER(4));
INSERT INTO s VALUES ('s001','jack');
INSERT INTO s VALUES ('s002','rose');
INSERT INTO s VALUES ('s003','mike');

INSERT INTO c VALUES ('1','history','t01');
INSERT INTO c VALUES ('2','English','t02');

INSERT INTO sc VALUES ('s001','1',90);
INSERT INTO sc VALUES ('s001','2',80);
INSERT INTO sc VALUES ('s002','1',85);
INSERT INTO sc VALUES ('s002','2',90);
INSERT INTO sc VALUES ('s003','1',80);
INSERT INTO sc VALUES ('s003','2',60);

SELECT sno FROM SC t WHERE t.cno='1' AND EXISTS (SELECT 1 FROM SC WHERE sno=t.sno AND cno='2' AND scgrade<t.scgrade);

输出：
SNO
s001
s003

------解决方案--------------------
select sno from (select t.SNO,t.SCGRADE-nvl(d.SCGRADE ) SCGRADE_c from (SELECT sc.sno, SCGRADE  FROM sc where CNO=1) t,(SELECT sc.sno, SCGRADE  FROM sc where CNO=2) d
where t.sno = d.sno(+)) t where SCGRADE_c>0