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', '');

INSERT INTO `grade` VALUES ('张三', '数学', '');

INSERT INTO `grade` VALUES ('李四', '语文', '');

INSERT INTO `grade` VALUES ('李四', '数学', '');

INSERT INTO `grade` VALUES ('王五', '语文', '');

INSERT INTO `grade` VALUES ('王五', '数学', '');

INSERT INTO `grade` VALUES ('王五', '英语', '');

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 ;