mysql实现SQL统计的实例

/*Table structure for table `stuscore` */

DROP TABLE IF EXISTS `stuscore`;

CREATE TABLE `stuscore` (
 `id` int(11) NOT NULL AUTO_INCREMENT,
 `name` varchar(20) DEFAULT NULL,
 `subject` varchar(20) DEFAULT NULL,
 `score` varchar(20) DEFAULT NULL,
 `stuid` varchar(10) DEFAULT NULL,
 PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=13 DEFAULT CHARSET=utf8;

/*Data for the table `stuscore` */

insert into `stuscore`(`id`,`name`,`subject`,`score`,`stuid`) values 
(1,'张三','数学','','1'),
(2,'张三','语文','80','1'),
(3,'张三','英语','70','1'),
(4,'李四','数学','90','2'),
(5,'李四','语文','70','2'),
(6,'李四','英语','80','2'),
(7,'王五','数学','55','3'),
(8,'王五','语文','92','3'),
(9,'王五','英语','74','3'),
(10,'赵六','数学','62','4'),
(11,'赵六','语文','81','4'),
(12,'赵六','英语','93','4');

建表语句

1 SELECT a.name, SUM(score) sum_score FROM stuscore a
2 GROUP BY a.name ORDER BY sum_score DESC

1 SELECT a.stuid, a.name, SUM(score) sum_score FROM stuscore a
2 GROUP BY a.name ORDER BY sum_score DESC

1 SELECT a.stuid, a.name, a.subject, a.score FROM stuscore a
2 JOIN (
3 SELECT stuid, MAX(score) max_score FROM stuscore GROUP BY stuid4 )b ON a.stuid=b.stuid5 WHERE a.score=b.max_score

1 SELECT DISTINCT a.stuid, a.name, b.avg_score FROM stuscore a
2 JOIN (
3 SELECT stuid, AVG(score) avg_score FROM stuscore GROUP BY stuid
4 )b ON a.stuid=b.stuid

1 SELECT DISTINCT a.stuid, a.name, a.subject, a.score FROM stuscore a
2 JOIN (
3 SELECT subject, MAX(score) max_score FROM stuscore GROUP BY subject
4 )b ON a.subject=b.subject5 WHERE a.score=b.max_score

1 SELECT a.stuid, a.name, a.subject, a.score FROM stuscore a
2 WHERE (
3 SELECT COUNT(1) FROM stuscore b
4 WHERE a.subject=b.subject AND b.score>=a.score
5 ) <= 2
6 ORDER BY a.subject ASC, a.score DESC

1 SELECT stuid 学号, NAME 姓名,
2 SUM(CASE WHEN SUBJECT='语文' THEN score ELSE 0 END) 语文,
3 SUM(CASE WHEN SUBJECT='数学' THEN score ELSE 0 END) 数学,
4 SUM(CASE WHEN SUBJECT='英语' THEN score ELSE 0 END) 英语,
5 SUM(score) 总分, (SUM(score)/COUNT(1)) 平均分
6 FROM stuscore GROUP BY 学号

1 SELECT SUBJECT, AVG(score) avg_score FROM stuscore GROUP BY SUBJECT

1 SELECT a.*, @var:=@var+1 rank 
2 FROM(
3 SELECT stuid, NAME, score FROM stuscore 
4 WHERE SUBJECT='数学' ORDER BY score DESC
5 )a, (SELECT @var:=0)b

1 SELECT b.* FROM(
2 SELECT a.* FROM(
3 SELECT stuid, NAME, score FROM stuscore 
4 WHERE SUBJECT='数学' ORDER BY score DESC 
5 LIMIT 3
6 )a ORDER BY score ASC LIMIT 2
7 )b ORDER BY score DESC
8 
9 #注：当数学成绩只有2条以下数据时，此方法失效！

1 SELECT a.*, @var:=@var+1 rank 
2 FROM(
3 SELECT stuid, NAME, score FROM stuscore 
4 WHERE SUBJECT='数学' ORDER BY score DESC5 )a, (SELECT @var:=0)b
6 WHERE a.name='李四'

1 SELECT a.subject 课程,
2 (SELECT COUNT(1) FROM stuscore WHERE score<60 AND SUBJECT=a.subject)不及格,
3 (SELECT COUNT(1) FROM stuscore WHERE score BETWEEN 60 AND 80 AND SUBJECT=a.subject)良,
4 (SELECT COUNT(1) FROM stuscore WHERE score>80 AND SUBJECT=a.subject)优
5 FROM stuscore a GROUP BY SUBJECT