用一条SQL语句查出每门课都大于80分的学生的姓名

 1 SET FOREIGN_KEY_CHECKS=0;
 2 
 3 -- ----------------------------
 4 -- Table structure for grade
 5 -- ----------------------------
 6 DROP TABLE IF EXISTS `grade`;
 7 CREATE TABLE `grade` (
 8   `name` varchar(255) NOT NULL,
 9   `class` varchar(255) NOT NULL,
10   `score` tinyint(4) NOT NULL
11 ) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4;
12 
13 -- ----------------------------
14 -- Records of grade
15 -- ----------------------------
16 INSERT INTO `grade` VALUES (‘张三‘, ‘语文‘, ‘81‘);
17 INSERT INTO `grade` VALUES (‘张三‘, ‘数学‘, ‘75‘);
18 INSERT INTO `grade` VALUES (‘李四‘, ‘语文‘, ‘76‘);
19 INSERT INTO `grade` VALUES (‘李四‘, ‘数学‘, ‘90‘);
20 INSERT INTO `grade` VALUES (‘王五‘, ‘语文‘, ‘81‘);
21 INSERT INTO `grade` VALUES (‘王五‘, ‘数学‘, ‘100‘);
22 INSERT INTO `grade` VALUES (‘王五‘, ‘英语‘, ‘90‘);
23 SET FOREIGN_KEY_CHECKS=1;

1 select distinct name from grade where name not in (select distinct name from grade where score<=80);

1 select name from grade group by name having min(score)>80;

1 select name from (
2    select count(*) t, sum(score) num, name from grade group by name       
3 ) as a where a.num>80*t;

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