Example of mysql implementing SQL statistics

Create table statement

/*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','89','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

Question:

1. Calculate everyone’s total score and rank (required display fields: name, total score)

Answer

##

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

2. Calculate everyone’s total score and rank Ranking (fields required to be displayed: student number, name, total score)

Answer

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

3. Calculate each person’s highest score in a single subject (required to be displayed) Fields: student number, name, course, highest grade)

Answer

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

4. Calculate the average grade of each person (required to display fields : Student number, name, average grade)

Answer

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

5. List the students with the best grades in each course (required to display Fields: student number, name, subject, score)

Answer

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

6. List the two best performers in each course Students (required display fields: student number, name, subject, grade)

Answer

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

7. The statistics are as follows:

##Total scoreAverage score

Student Number	Name	Chinese	Mathematics	English

Answer

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

8. List the average grades of each course (required display fields: course, average grade)

Answer

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

9. List the rankings of mathematics scores (required display fields: student number, name, grades, ranking)

Answer

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

10. List the students who ranked 2-3 in mathematics (required display fields: student number, name, subject, grade)

Answer

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！

11. Find the ranking of John Doe’s math scores

Answer

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'

12. The statistics are as follows:

##Course

Failed (0-59)	Good (60-80)	Excellent (81-100)

##Answer

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

The above is the detailed content of Example of mysql implementing SQL statistics. For more information, please follow other related articles on the PHP Chinese website!