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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 ;