按最多重复记录排序

Order by most duplicate record first

create table tbl_dup
(
    name varchar(100)
);

insert into tbl_dup values('Arsel Rous'),('Oram Rock'),('Oram Rock'),('Brown Twor'),
                          ('John Mak'),('Mak Dee'),('Smith Will'),('Mak Dee'),
                          ('John Mak'),('Oram Rock'),('John Mak'),('Oram Rock');

查询:我正在寻找排序重复记录应该显示在结果集中的第一级。

select * 
from
(
select name,row_number() over(partition by name order by name) rn 
from tbl_dup
) a
order by name,rn;

获取:

name        rn
--------------
Arsel Rous  1
Brown Twor  1
John Mak    1
John Mak    2
John Mak    3
Mak Dee     1
Mak Dee     2
Oram Rock   1
Oram Rock   2
Oram Rock   3
Oram Rock   4
Smith Will  1

预期结果:

name        rn
---------------
Oram Rock   1
Oram Rock   2
Oram Rock   3
Oram Rock   4
John Mak    1
John Mak    2
John Mak    3
Mak Dee     1
Mak Dee     2
Arsel Rous  1
Brown Twor  1

您需要先按出现次数 desc 和行号 asc 排序。

select name,rn 
from
(
    select name,row_number() over(partition by name order by name) rn ,
    count(*) over(partition by name) ct
    from tbl_dup
) a
order by ct desc, rn asc

尝试在 ORDER BY 子句中使用 COUNT 作为解析函数:

SELECT name, ROW_NUMBER() OVER (PARTITION by name ORDER BY name) rn
FROM tbl_dup
ORDER BY COUNT(*) OVER (PARTITION BY name) DESC, rn;

Demo