Mysql 查询实现成绩排名

Mysql 查询实现成绩排名,相同分数名次相同,类似于rank()函数

近日系统要实现总分成绩排名,而且相同分数的学生排名要一样,在网上搜了一圈,没有找到合适的方法,只能靠自己实现了,这里提供两种方法

//还有其他排名方式可以借鉴https://blog.csdn.net/a9925/article/details/76804951

1、sql查询实现

测试如下:

mysql> select * from score ;
+----------+--------------+---------------------+--------------+-------+
| study_no | student_name | subject_id          | subject_name | score |
+----------+--------------+---------------------+--------------+-------+
| student1 | student1     | CodeCourseSubject_0 | 语文         |   120 |
| student2 | student2 | CodeCourseSubject_0 | 语文 | 110 | | student3 | student3 | CodeCourseSubject_0 | 语文 | 110 | | student4 | student4 | CodeCourseSubject_0 | 语文 | 80 | | student5 | student5 | CodeCourseSubject_0 | 语文 | 81 | | student1 | student1 | CodeCourseSubject_2 | 英语 | 150 | | student2 | student2 | CodeCourseSubject_2 | 英语 | 130 | | student3 | student3 | CodeCourseSubject_2 | 英语 | 130 | | student4 | student4 | CodeCourseSubject_2 | 英语 | 44 | | student5 | student5 | CodeCourseSubject_2 | 英语 | 45 | +----------+--------------+---------------------+--------------+-------+ 10 rows in set

首先这里对科目显示进行了行列转换,并且对以学号study_no进行分组计算总分,并且按排名排序, 
sql如下:

SELECT @rownum:=@rownum+1 AS rownum,
   if(@total=total,@rank,@rank:=@rownum)as rank,
   @total:=total,
   A.* 
   FROM (SELECT study_no AS studyNo,
              student_name AS studentName,
              SUM(score) AS total,
              SUM(IF(subject_id='CodeCourseSubject_0',score,0)) AS 语文,
              SUM(IF(subject_id='CodeCourseSubject_2',score,0)) AS 英语 
              FROM score GROUP BY study_no ORDER BY total DESC
       )A,(SELECT @rank:=0,@rownum:=0,@total:=null)B

  

结果:

+--------+------+---------------+----------+-------------+-------+-------+-------+
| rownum | rank | @total:=total | studyNo  | studentName | total | 语文  | 英语  |
+--------+------+---------------+----------+-------------+-------+-------+-------+
|      1 |    1 | 270.0 | student1 | student1 | 270.0 | 120.0 | 150.0 | | 2 | 2 | 240.0 | student2 | student2 | 240.0 | 110.0 | 130.0 | | 3 | 2 | 240.0 | student3 | student3 | 240.0 | 110.0 | 130.0 | | 4 | 4 | 126.0 | student5 | student5 | 126.0 | 81.0 | 45.0 | | 5 | 5 | 124.0 | student4 | student4 | 124.0 | 80.0 | 44.0 | +--------+------+---------------+----------+-------------+-------+-------+-------+ 5 rows in set

可见排名第二名有两个相同的分数,后面的排名自动往后移。

下面提供另外一种在程序中实现的方法:

2、程序实现

首先获得全班总分并且以降序排序

public List<Object[]> findAllTotalScore() {
        String sql = "SELECT study_no,sum(score) AS total from score s ";
        javax.persistence.Query query = em.createNativeQuery(sql);
        return query.getResultList();
    }

对获得的总分信息进行排序和封装成map(study_no,rank)

private Map<String, Long> rank(List<Object[]> objects) {
        long previousRank = 1;
        double total = 0;
        //记录排名的map,map<study_no,排名>
        Map<String, Long> rankMap = new HashMap<>();
        for (int i = 0; i < objects.size(); i++) {
            Object[] object = objects.get(i);
            //计算名次,相同分数排名一样
            if (total == (double) object[1]) {
                rankMap.put(object[0].toString(), previousRank);
            } else {
                rankMap.put(object[0].toString(), i + 1L);
                total = (double) object[1];
                previousRank = i + 1;
            }
        }
        return rankMap;
    }

使用map直接根据学号就可以获得成绩排名。

转自:https://blog.csdn.net/a56508820/article/details/49663069

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转载自www.cnblogs.com/SimonHu1993/p/9022960.html