用一条SQL语句查出每门课都大于80分的学生的姓名

Posted 2020-09-18 朱德平的技术博客

用一条SQL语句查出每门课都大于80分的学生的姓名

用一条SQL语句查出每门课都大于80分的学生的姓名，数据表结构如下：

建表SQL如下：

SET FOREIGN_KEY_CHECKS=0;

-- ----------------------------
-- Table structure for grade
-- ----------------------------
DROP TABLE IF EXISTS `grade`;
CREATE TABLE `grade` (
  `name` varchar(255) NOT NULL,
  `class` varchar(255) NOT NULL,
  `score` tinyint(4) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4;

-- ----------------------------
-- Records of grade
-- ----------------------------
INSERT INTO `grade` VALUES (\'张三\', \'语文\', \'81\');
INSERT INTO `grade` VALUES (\'张三\', \'数学\', \'75\');
INSERT INTO `grade` VALUES (\'李四\', \'语文\', \'76\');
INSERT INTO `grade` VALUES (\'李四\', \'数学\', \'90\');
INSERT INTO `grade` VALUES (\'王五\', \'语文\', \'81\');
INSERT INTO `grade` VALUES (\'王五\', \'数学\', \'100\');
INSERT INTO `grade` VALUES (\'王五\', \'英语\', \'90\');
SET FOREIGN_KEY_CHECKS=1;

 

查询每门课都大于80分的同学的姓名：

SELECT DISTINCT name FROM grade WHERE name NOT IN(SELECT DISTINCT name FROM grade WHERE score <=80);

  更简单的：

SELECT name FROM grade GROUP BY name HAVING MIN(score) > 80;

 

查询平均分大于80的学生的姓名：

SELECT name FROM (SELECT COUNT(*) AS t,SUM(score) AS num,name FROM `grade` GROUP BY name) AS a WHERE a.num > 80*t;

 更简单的：

select name, avg(score) as sc from grade g1 group by name having avg(score)>80 ;