用一條SQL語句查出每門課都大於80分的學生的姓名

用一條SQL語句查出每門課都大於80分的學生的姓名，數據表結構如下：

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 ('張三', '語文', '81');
INSERT INTO `grade` VALUES ('張三', '數學', '75');
INSERT INTO `grade` VALUES ('李四', '語文', '76');
INSERT INTO `grade` VALUES ('李四', '數學', '90');
INSERT INTO `grade` VALUES ('王五', '語文', '81');
INSERT INTO `grade` VALUES ('王五', '數學', '100');
INSERT INTO `grade` VALUES ('王五', '英語', '90');
SET FOREIGN_KEY_CHECKS=1;

 

查詢每門課都大於80分的同學的姓名：

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

  更簡單的：

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

 

查詢平均分大於80的學生的姓名：

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

 更簡單的：

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