每组访问查询计数器答案

【问题标题】：Access query counter per group每组访问查询计数器
【发布时间】：2021-12-31 06:39:30
【问题描述】：

是否可以在 Access Query 中添加按组排序（排序/组）字段以按照下表按值排序：

注意： ID 是唯一索引（自动编号）

【问题讨论】：

【解决方案1】：

在真正的 RDBMS 上，通常会为此使用窗口函数 ROW_NUMBER。

select *
, row_number() over (partition by Group1 order by Value, ID) as Rownum
from yourtable

但另一种方法是使用相关子查询。

select *, 
(select count(*) from yourtable t2 
 where t2.Group1 = t.Group1
   and (t2.Value < t.Value
        or (t2.Value = t.Value and t2.ID <= t.ID)) ) as Rownum
from yourtable t

额外：

显示ROW_NUMBER、RANK 和DENSE_RANK 之间区别的简单测试

create table yourtable (
 ID int identity(1,1) primary key, 
 Group1 int, 
 Value int
);

insert into yourtable (Group1, Value) values
  (10,100),(10,150),(10,150),(10,150),(10,360)
, (200,360),(200,420),(200,420),(200,500),(200,500)

--
-- ROW_NUMBER (to get a sequence per group) 
--
select *
, ROW_NUMBER() OVER (PARTITION BY Group1 ORDER BY Value, ID) as Rownum
from yourtable
order by Group1, Rownum

身份证 |组1 |价值 |朗姆 -: | -----: | ----: | -----： 1 | 10 | 100 | 1 2 | 10 | 150 | 2 3 | 10 | 150 | 3 4 | 10 | 150 | 4 5 | 10 | 360 | 5 6 | 200 | 360 | 1 7 | 200 | 420 | 2 8 | 200 | 420 | 3 9 | 200 | 500 | 4 10 | 200 | 500 | 5

--
-- Emulating ROW_NUMBER via a correlated sub-query
--
select *, 
(select count(*) from yourtable t2 
 where t2.Group1 = t.Group1
   and (t2.Value < t.Value 
        or (t2.Value = t.Value 
            and t2.ID <= t.ID))
) as Rownum
from yourtable t
order by Group1, Rownum
身份证 |组1 |价值 |朗姆 -: | -----: | ----: | -----： 1 | 10 | 100 | 1 2 | 10 | 150 | 2 3 | 10 | 150 | 3 4 | 10 | 150 | 4 5 | 10 | 360 | 5 6 | 200 | 360 | 1 7 | 200 | 420 | 2 8 | 200 | 420 | 3 9 | 200 | 500 | 4 10 | 200 | 500 | 5

--
-- RANK (same values get same rank, but with gaps)
--
select *
, RANK() over (partition by Group1 order by Value) as Ranknum
from yourtable
order by Group1, Ranknum
身份证 |组1 |价值 |秩数 -: | -----: | ----: | ------: 1 | 10 | 100 | 1 2 | 10 | 150 | 2 3 | 10 | 150 | 2 4 | 10 | 150 | 2 5 | 10 | 360 | 5 6 | 200 | 360 | 1 7 | 200 | 420 | 2 8 | 200 | 420 | 2 9 | 200 | 500 | 4 10 | 200 | 500 | 4

--
-- Emulating RANK via a correlated sub-query
--
select *, 
(select count(t2.value)+1 from yourtable t2 
 where t2.Group1 = t.Group1
   and t2.Value < t.Value) as Ranknum
from yourtable t
order by Group1, Ranknum
身份证 |组1 |价值 |秩数 -: | -----: | ----: | ------: 1 | 10 | 100 | 1 2 | 10 | 150 | 2 3 | 10 | 150 | 2 4 | 10 | 150 | 2 5 | 10 | 360 | 5 6 | 200 | 360 | 1 7 | 200 | 420 | 2 8 | 200 | 420 | 2 9 | 200 | 500 | 4 10 | 200 | 500 | 4

--
-- DENSE_RANK (same values get same rank, without gaps)
--
select *
, DENSE_RANK() over (partition by Group1 order by Value) as Ranknum
from yourtable
order by Group1, Ranknum
身份证 |组1 |价值 |秩数 -: | -----: | ----: | ------: 1 | 10 | 100 | 1 2 | 10 | 150 | 2 3 | 10 | 150 | 2 4 | 10 | 150 | 2 5 | 10 | 360 | 3 6 | 200 | 360 | 1 7 | 200 | 420 | 2 8 | 200 | 420 | 2 9 | 200 | 500 | 3 10 | 200 | 500 | 3

--
-- Emulating DENSE_RANK via a correlated sub-query
--
select *, 
(select count(distinct t2.Value) from yourtable t2 
 where t2.Group1 = t.Group1
   and t2.Value <= t.Value
) as Ranknum
from yourtable t
order by Group1, Ranknum
身份证 |组1 |价值 |秩数 -: | -----: | ----: | ------: 1 | 10 | 100 | 1 2 | 10 | 150 | 2 3 | 10 | 150 | 2 4 | 10 | 150 | 2 5 | 10 | 360 | 3 6 | 200 | 360 | 1 7 | 200 | 420 | 2 8 | 200 | 420 | 2 9 | 200 | 500 | 3 10 | 200 | 500 | 3

db小提琴here

【讨论】：

替代解决方案有效，但前提是值是唯一的。否则，重复值从上一组的下一个编号开始计数。
现在应该修复了。必须用 () 包围 OR
现在可以了，除了按Rownum排序，不能按它排序。
谢谢。好吧，要对其进行排序，可以始终将其包装在子查询中，然后在子查询之外对其进行排序。

【解决方案2】：

考虑：

SELECT Data.ID, Data.Group1, Data.Value, 
DCount("*","Data","Group1='" & [Group1] & "' AND Value<" & [Value])+1 AS GrpSeq
FROM Data
ORDER BY Data.Value;

或者

SELECT Data.ID, Data.Group1, Data.Value, (
       SELECT Count(*) AS Cnt FROM Data AS Dupe 
       WHERE Dupe.Value<Data.Value AND Dupe.Group1=Data.Group1)+1 AS GrpSeq
FROM Data
ORDER BY Data.Value;

【讨论】：

第二种方法运行良好，但第一种方法在 Grpseq 上出错。
为我工作。两者都有相同的输出。可能第二个在大型数据集上执行得更快。
现在好了，两者都工作正常我找到了错误的原因。