GROUP BY 的 COUNT(*)答案

【问题标题】：COUNT(*) of GROUP BYGROUP BY 的 COUNT(*)
【发布时间】：2012-01-09 15:55:25
【问题描述】：

我有一个查询，它返回 644 行，按几列分组。我需要计算这些行数，644。

这是查询：

SELECT DISTINCT ho.ID,ho.honame,ho.ho_status,SUM(Properties.Value) as [sales value], COUNT(Properties.record_id) as [property count]
FROM HeadOffice ho INNER JOIN Properties ON Properties.head_office_code = ho.id
WHERE Somecondition
GROUP BY ho.ID,ho.honame,ho_status ORDER BY ho_status

尽管尝试了COUNT(*)，将其包装在另一个查询中并删除了GROUP BY，但我无法取回“644”。我最接近的是 644 行，全部包含“1”。这可能吗？

【问题讨论】：

select count(1) from (select distinct ID, honame, ho_status from HeadOffice) ignore 返回什么？
如果您的原始查询返回正确的行数，为什么要在 Select count(*) from () 中换行时更改它？

标签： sql sql-server sql-server-2008 count group-by

【解决方案1】：

这样做的简单方法是：

SELECT count(1) as NumRows from
(SELECT DISTINCT ho.ID,ho.honame,ho.ho_status,SUM(Properties.Value) as [sales value], COUNT(Properties.record_id) as [property count]
FROM HeadOffice ho INNER JOIN Properties ON clients.head_office_code = ho.id
WHERE Somecondition
GROUP BY ho.ID,ho.honame,ho_status ORDER BY ho_status) x

如果您想要计数加上您的列，请使用over：

SELECT 
    count(1) over () as NumRows,
    x.ID,
    x.ho_status,
    x.[sales value],
    x.[property count]
from
(SELECT DISTINCT ho.ID,ho.honame,ho.ho_status,SUM(Properties.Value) as [sales value], COUNT(Properties.record_id) as [property count]
FROM HeadOffice ho INNER JOIN Properties ON clients.head_office_code = ho.id
WHERE Somecondition
GROUP BY ho.ID,ho.honame,ho_status ORDER BY ho_status) x

【讨论】：

我在上面做 COUNT(*) ，但这很有效。谢谢。

【解决方案2】：

您将在列表中多获得 1 行，其中包含除属性计数和属性值之外的所有内容的空值。该记录将包含一个计数和所有属性值的总和。

SELECT DISTINCT ho.ID,ho.honame,ho.ho_status, 
SUM(Properties.Value) as [sales value], 
COUNT(Properties.record_id) as [property count]
FROM HeadOffice ho 
INNER JOIN Properties 
  ON clients.head_office_code = ho.id
WHERE Somecondition
GROUP BY Grouping sets((ho.ID,ho.honame,ho_status),()) 
ORDER BY ho_status

【讨论】：

【解决方案3】：

这应该可以工作

SELECT COUNT(ID) 
FROM (
SELECT DISTINCT ho.ID,ho.honame,ho.ho_status,
       SUM(Properties.Value) as [sales value], 
       COUNT(Properties.record_id) as [property count]
FROM HeadOffice ho 
     INNER JOIN Properties ON clients.head_office_code = ho.id
WHERE Somecondition
GROUP BY ho.ID, ho.honame, ho_status ) t

【讨论】：