用一条sql语句查询出所有课程都大于80分的学生名单：

http://www.cnblogs.com/praglody/p/6854181.html

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 ;