Question

我会切入正确的追逐：我有一个选择我正在写一个相当冗长的where子句，我想要做的是计算百分比。

所以我需要的是所有结果的计数，然后是我的每个不同的计数。

SELECT distinct count(*) 
FROM mytable 
WHERE mywhereclause 
ORDER BY columnIuseInWhereClause

可以很好地获取每个单独的值，但我想避免做像

这样的事情

Select (Select count(*) from mytable WHERE mywhereclause),
       distinct count(*) 
FROM mytable 
WHERE mywhereclause

因为我两次使用相同的where子句，这似乎是不必要的。

这是针对OracleDB的，但我只使用标准的SQL语法，如果我可以提供帮助，我没有特定的数据库。

感谢任何想法。

编辑：样本数据

__ID__,__someValue__
  1   |      A      
  2   |      A
  3   |      B
  4   |      C

我希望A，B，C的出现是数字以及总数。

__CountAll__,__ACounts__,__BCounts__,__CCounts__
     4      |     2     |     1     |     1

所以我可以去

   100%     |    50%    |    25%    |    25%

最后一部分我可以自己解决。请原谅我缺乏经验甚至是逻辑思维，这是一大早。 ;）

EDIT2：我确实已经写了一个有效的查询，但是很笨拙，只要是所有神圣的哎呀，这个是用于尝试分组的。

Answer 1

尝试：

select count(*) as CountAll, 
       count(distinct SomeColumn) as CoundDistinct -- The DISTINCT goes inside the brackets
from myTable
where SomeOtherColumn = 'Something'

Answer 2

使用case表达式进行条件计数：

select count(*) as CountAll,
       count(case when someValue = 'A' then 1 end) as ACounts,
       count(case when someValue = 'B' then 1 end) as BCounts,
       count(case when someValue = 'C' then 1 end) as CCounts
FROM mytable 
WHERE mywhereclause

将其包装在派生表中以轻松完成％部分：

select 100,
       ACounts * 100 / CountAll,
       BCounts * 100 / CountAll,
       CCounts * 100 / CountAll
from
(
    select count(*) as CountAll,
           count(case when someValue = 'A' then 1 end) as ACounts,
           count(case when someValue = 'B' then 1 end) as BCounts,
           count(case when someValue = 'C' then 1 end) as CCounts
    FROM mytable 
    WHERE mywhereclause
) dt

Answer 3

这是使用窗口功能的另一种选择：

with data_table(ID, some_value)
AS
(SELECT 1,'A' UNION ALL
 SELECT 2,'A' UNION ALL
 SELECT 3,'B' UNION ALL
 SELECT 4,'C' 
)

SELECT DISTINCT [some_value],
       COUNT([some_value]) OVER () AS Count_All, 
       COUNT([some_value]) OVER (PARTITION BY [some_value]) AS 'Counts' FROM [data_table]
ORDER BY [some_value]

优点是您不必硬编码[some_value]

计数后选择不同的计数？

3 个答案: