2019独角兽企业重金招聘Python工程师标准>>>

-- 首先,建立表:
-- ----------------------------
-- Table structure for `s`
-- ----------------------------
DROP TABLE IF EXISTS `s`;
CREATE TABLE `s` (`sno` int(11) NOT NULL,`sname` varchar(32) DEFAULT NULL,PRIMARY KEY (`sno`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;-- ----------------------------
-- Records of s
-- ----------------------------
INSERT INTO `s` VALUES ('1', '奥巴马');
INSERT INTO `s` VALUES ('2', '小布什');
INSERT INTO `s` VALUES ('3', '本拉登');
INSERT INTO `s` VALUES ('4', '普京');
INSERT INTO `s` VALUES ('5', '奥斯特洛夫斯基');-- ----------------------------
-- Table structure for `c`
-- ----------------------------
DROP TABLE IF EXISTS `c`;
CREATE TABLE `c` (`cno` int(11) NOT NULL,`cname` varchar(32) DEFAULT NULL,`cteacher` varchar(32) DEFAULT NULL,PRIMARY KEY (`cno`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;-- ----------------------------
-- Records of c
-- ----------------------------
INSERT INTO `c` VALUES ('1', '语文', '李明');
INSERT INTO `c` VALUES ('2', '数学', '爱因斯坦');-- ----------------------------
-- Table structure for `sc`
-- ----------------------------
DROP TABLE IF EXISTS `sc`;
CREATE TABLE `sc` (`sno` int(11) NOT NULL,`cno` int(11) NOT NULL,`sc` int(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;-- ----------------------------
-- Records of sc
-- ----------------------------
INSERT INTO `sc` VALUES ('1', '2', '50');
INSERT INTO `sc` VALUES ('1', '1', '50');
INSERT INTO `sc` VALUES ('2', '1', '50');
INSERT INTO `sc` VALUES ('2', '2', '50');
INSERT INTO `sc` VALUES ('3', '1', '60');
INSERT INTO `sc` VALUES ('3', '2', '60');
INSERT INTO `sc` VALUES ('4', '1', '70');
INSERT INTO `sc` VALUES ('4', '2', '70');
INSERT INTO `sc` VALUES ('5', '2', '90');

select s.sname as "不参加李明课程" from s where s.sno not in (select sno from c,sc where c.cteacher="李明" and sc.cno=c.cno)

查询结果图:

--================初步答案=============
select s.sname from s
where s.sno in(select sc.sno from sc where sc.cno in (select cno from c where cname in ("语文","数学")) group by sc.sno having count(sc.cno)>1)
--================优化==============select s.sname from sc,s where sc.cno in (select cno from c where cname in ("语文","数学")) and sc.sno=s.sno group by sc.sno having count(sc.cno)>1

首先要修改数据,如图,左为修改后的数据库数据,将sno为4的一个cno改为一个并不存在的3,就是说我们在查询的时候不应该查出第四个来.查询结果显示是3个,是对的.

select s1.sno,s1.sname
from sc as sc1,sc as sc2,s as s1
where sc1.sno=sc2.sno and sc1.cno=1 and sc2.cno=2 and sc1.sc>sc2.sc  and s1.sno=sc1.sno

select s1.sno,s1.sname,sc1.sc as "1" ,sc2.sc as "2"
from sc as sc1,sc as sc2,s as s1
where sc1.sno=sc2.sno and sc1.cno=1 and sc2.cno=2 and sc1.sc>sc2.sc  and s1.sno=sc1.sno