RANK 函数不适用于 postgresql 中的 CTE table？

Question

我正在尝试根据评论对我的 CTE table (a) 输出进行排名，并按两列（risk_full_nm 和 exam_year_nb）划分。

下面是我的代码：

\`WITH a AS
(
select rr.risk_full_nm ,e.exam_year_nb,COUNT(DISTINCT e.exam_dmn_xtrnl_id) as "reviews"
from examwrksp.dsh_exam_base e
left join star.firm f ON f.firm_id = e.crd_nb --firm
left join examwrksp.dsh_mlstn_event_dates d ON d.exam_id  = e.exam_id --exam milestones
left join examwrksp.dsh_scope_risk_cntnt rr on rr.exam_id = e.exam_id
where e.exam_year_nb \>= '2018' and e.exam_ctgry_cd = 'CYCLE' and e.exam_st_ds in ('Open','Closed') and e.dstrt_nm != 'MAP Group' and e.exam_type_ds not like 'Funding%'
and e.exam_dmn_prmry_fl = 'Y' and f.main_ofc = 'Y' and d.crd_nb is null
and rr.scope_actv_st is not null and rr.scope_actv_st = 'ACTIVE' and rr.unit_type_label_tx not in ('Discovery Review','Risk Identification Review')
GROUP BY rr.risk_full_nm ,e.exam_year_nb
)
select a.risk_full_nm,a.exam_year_nb,a.reviews,
RANK () OVER(PARTITION BY a.risk_full_nm,a.exam_year_nb ORDER BY a.reviews desc)
from a
order by reviews desc\`

我希望我的排名列对我的记录进行排名，但我在每一行中都得到“1”。我错过了什么？

Answer 1

我分析了你的查询。您的查询中有一些不正确的逻辑代码。我给你解释一下：

Window 每次为每个分组字段计算聚合函数。但是，分组字段来自 over (partition by 语句之后。在您的查询中，您对两个字段进行了分组，a.risk_full_nm 和 a.exam_year_nb。对这些字段进行分组后，结果数据的所有行都不同，因为您不必重复（相同）行。因此，RANK() 函数将针对所有行再次计算。为了更好地理解，我使用示例查询向您解释。

例如：

with test_data as materialized 
(
    select 1 as id, 'user1' as username, 'admin' as type_of, '2022-01-14'::date as login_date 
    union all 
    select 2 as id, 'user2' as username, 'user' as type_of, '2022-01-06'::date as login_date
    union all 
    select 3 as id, 'user1' as username, 'admin' as type_of, '2022-01-29'::date as login_date
    union all 
    select 4 as id, 'user3' as username, 'user' as type_of, '2022-02-11'::date as login_date
    union all 
    select 5 as id, 'user2' as username, 'user' as type_of, '2022-01-16'::date as login_date
    union all 
    select 6 as id, 'user2' as username, 'user' as type_of, '2022-01-17'::date as login_date
    union all 
    select 7 as id, 'user2' as username, 'user' as type_of, '2022-01-18'::date as login_date
)
select 
    id, 
    username, 
    type_of,  
    count(*) over (partition by id, username, type_of) as login_count
from 
    test_data;

-- Result of this query 
id  username    type_of     login_count
-----------------------------------------
1   user1       admin       1
2   user2       user        1
3   user1       admin       1
4   user3       user        1
5   user2       user        1
6   user2       user        1
7   user2       user        1

在此查询中，我想获取每个用户的登录次数。这是不正确的语法。所以，我使用唯一的主键字段 'id' 进行分组，我的行将是不同的。因此，对于这些行中的每一行，函数 count() 仅获得 1 个值。当我们从搜索字段列表中删除 'id' 字段时，我们返回了正确的数据并且 count() 计算正确。例如：

select 
    id, 
    username, 
    type_of,  
    count(*) over (partition by username, type_of) as login_count
from 
    test_data

-- Result this query: 

id  username    type_of     login_count
----------------------------------------
3   user1       admin       2
1   user1       admin       2
7   user2       user        4
2   user2       user        4
5   user2       user        4
6   user2       user        4
4   user3       user        1

RANK 函数不适用于 postgresql 中的 CTE table？

RANK function doesn't work on CTE table in posgresql?

postgresql

ranking

common-table-expression

rank

partition