postgresql rank() over, dense_rank(), row_number()用法区别
(编辑：jimmy 日期: 2024/11/18 浏览：3 次 )
如下学生表student，学生表中有姓名、分数、课程编号，需要按照课程对学生的成绩进行排序
select * from jinbo.student; id | name | score | course ----+-------+-------+-------- 5 | elic | 70 | 1 4 | dock | 100 | 1 3 | cark | 80 | 1 2 | bob | 90 | 1 1 | alice | 60 | 1 10 | jacky | 80 | 2 9 | iris | 80 | 2 8 | hill | 60 | 1 7 | grace | 50 | 2 6 | frank | 70 | 2 6 | test | | 2 (11 rows)
1、rank over () 可以把成绩相同的两名是并列，如下course = 2 的结果rank值为：1 2 2 4 5
select name, score, course, rank() over(partition by course order by score desc) as rank from jinbo.student; name | score | course | rank -------+-------+--------+------ dock | 100 | 1 | 1 bob | 90 | 1 | 2 cark | 80 | 1 | 3 elic | 70 | 1 | 4 hill | 60 | 1 | 5 alice | 60 | 1 | 5 test | | 2 | 1 iris | 80 | 2 | 2 jacky | 80 | 2 | 2 frank | 70 | 2 | 4 grace | 50 | 2 | 5 (11 rows)
2、dense_rank()和rank over()很相似，可以把学生成绩并列不间断顺序排名，如下course = 2 的结果rank值为：1 2 2 3 4
select name,score, course, dense_rank() over(partition by course order by score desc) as rank from jinbo.student; name | score | course | rank -------+-------+--------+------ dock | 100 | 1 | 1 bob | 90 | 1 | 2 cark | 80 | 1 | 3 elic | 70 | 1 | 4 hill | 60 | 1 | 5 alice | 60 | 1 | 5 test | | 2 | 1 iris | 80 | 2 | 2 jacky | 80 | 2 | 2 frank | 70 | 2 | 3 grace | 50 | 2 | 4 (11 rows)
3、row_number 可以把相同成绩的连续排名，如下 course = 2 的结果rank值为：1 2 3 4 5
select name,score, course, row_number() over(partition by course order by score desc) as rank from jinbo.student; name | score | course | rank -------+-------+--------+------ dock | 100 | 1 | 1 bob | 90 | 1 | 2 cark | 80 | 1 | 3 elic | 70 | 1 | 4 hill | 60 | 1 | 5 alice | 60 | 1 | 6 test | | 2 | 1 iris | 80 | 2 | 2 jacky | 80 | 2 | 3 frank | 70 | 2 | 4 grace | 50 | 2 | 5 (11 rows)
使用rank over()的时候，空值是最大的，如果排序字段为null, 可能造成null字段排在最前面，影响排序结果，可以如下：
rank over(partition by course order by score desc nulls last)
4、总结
partition by 用于结果集分组，如果没有指定，会把整个结果集作为一个分组
rank 、dense_rank 、row_numer 都是不同方式的结果集组内排序，一般都结合over 字句出现，over 字句里 会有 partition by、order by、last、first 的任意组合，如下：
rank() over(partition by a,b order by a, order by b desc); rank() over(partition by a order by b nulls first) rank() over(partition by a order by b nulls last)
补充：Oracle或者PostgreSQL的row_number over 排名语法
PostgreSQL 和Oracle 都提供了 row_number() over() 这样的语句来进行对应的字段排名，很是方便。MySQL却没有提供这样的语法。
这次我提供的表结构如下，
Table "ytt.t1"
Column | Type | Modifiers
--------+-----------------------+-----------
i_name | character varying(10) | not null
rank | integer | not null
我模拟了20条数据来做演示。
t_girl=# select * from t1 order by i_name; i_name | rank ---------+------ Charlie | 12 Charlie | 12 Charlie | 13 Charlie | 10 Charlie | 11 Lily | 6 Lily | 7 Lily | 7 Lily | 6 Lily | 5 Lily | 7 Lily | 4 Lucy | 1 Lucy | 2 Lucy | 2 Ytt | 14 Ytt | 15 Ytt | 14 Ytt | 14 Ytt | 15 (20 rows)
在PostgreSQL下，我们来对这样的排名函数进行三种不同的执行方式1：
第一种：
完整的带有排名字段以及排序。
t_girl=# select i_name,rank, row_number() over(partition by i_name order by rank desc) as rank_number from t1; i_name | rank | rank_number ---------+------+------------- Charlie | 13 | 1 Charlie | 12 | 2 Charlie | 12 | 3 Charlie | 11 | 4 Charlie | 10 | 5 Lily | 7 | 1 Lily | 7 | 2 Lily | 7 | 3 Lily | 6 | 4 Lily | 6 | 5 Lily | 5 | 6 Lily | 4 | 7 Lucy | 2 | 1 Lucy | 2 | 2 Lucy | 1 | 3 Ytt | 15 | 1 Ytt | 15 | 2 Ytt | 14 | 3 Ytt | 14 | 4 Ytt | 14 | 5 (20 rows)
第二种：
带有完整的排名字段但是没有排序。
t_girl=# select i_name,rank, row_number() over(partition by i_name ) as rank_number from t1; i_name | rank | rank_number ---------+------+------------- Charlie | 12 | 1 Charlie | 12 | 2 Charlie | 13 | 3 Charlie | 10 | 4 Charlie | 11 | 5 Lily | 6 | 1 Lily | 7 | 2 Lily | 7 | 3 Lily | 6 | 4 Lily | 5 | 5 Lily | 7 | 6 Lily | 4 | 7 Lucy | 1 | 1 Lucy | 2 | 2 Lucy | 2 | 3 Ytt | 14 | 1 Ytt | 15 | 2 Ytt | 14 | 3 Ytt | 14 | 4 Ytt | 15 | 5 (20 rows)
第三种：
没有任何排名字段，也没有任何排序字段。
t_girl=# select i_name,rank, row_number() over() as rank_number from t1; i_name | rank | rank_number ---------+------+------------- Lily | 7 | 1 Lucy | 2 | 2 Ytt | 14 | 3 Ytt | 14 | 4 Charlie | 12 | 5 Charlie | 13 | 6 Lily | 7 | 7 Lily | 4 | 8 Ytt | 14 | 9 Lily | 6 | 10 Lucy | 1 | 11 Lily | 7 | 12 Ytt | 15 | 13 Lily | 6 | 14 Charlie | 11 | 15 Charlie | 12 | 16 Lucy | 2 | 17 Charlie | 10 | 18 Lily | 5 | 19 Ytt | 15 | 20 (20 rows)
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