Question

这可能是一个简单的但我无法理解它。

我有一个MemberBusinessCats表，其中包含一个BusinessCatID和一个MemberID ..表格可以这样呈现：

+-----------------------+-----------------+------------+
|  MemberBusinessCatID  |  BusinessCatID  |  MemberID  |
+-----------------------+-----------------+------------+
|  27                   |  45             |  102       |
+-----------------------+-----------------+------------+
|  28                   |  55             |  102       |
+-----------------------+-----------------+------------+
|  29                   |  61             |  102       |
+-----------------------+-----------------+------------+
|  30                   |  45             |  33        |
+-----------------------+-----------------+------------+
|  31                   |  23             |  33        |
+-----------------------+-----------------+------------+
|  32                   |  45             |  73        |
+-----------------------+-----------------+------------+
|  32                   |  61             |  73        |
+-----------------------+-----------------+------------+
|  32                   |  45             |  73        |
+-----------------------+-----------------+------------+

如何制作脚本以显示以下数据

+-----------------+---------------------+
|  BusinessCatID  |  NumMembers In Cat  |
+-----------------+---------------------+
|  45             |  3                  |
+-----------------+---------------------+
|  55             |  1                  |
+-----------------+---------------------+
|  61             |  2                  |
+-----------------+---------------------+
|  23             |  1                  |
+-----------------+---------------------+

非常感谢提前。

neojakey

Answer 1

您需要使用带GROUP BY的聚合函数：

select BusinessCatID, count(*) NumMembersInCat
from MemberBusinessCats
group by BusinessCatID

请参阅SQL Fiddle with Demo

也可以使用count() over()：

编写

select distinct BusinessCatID, 
  count(*) over(partition by BusinessCatID) NumMembersInCat
from MemberBusinessCats

请参阅SQL Fiddle with Demo

如果您想计算每个类别中的成员数量，那么您可以使用：

select BusinessCatID, 
  count(distinct MemberID) NumMembersInCat
from MemberBusinessCats
group by BusinessCatID

请参阅SQL Fiddle with Demo

Answer 2

试试这个

   select BusinessCatID ,count(BusinessCatID) as NumMembers_In_Cat 
   from MemberBusinessCats 
   group by BusinessCatID

DEMO SQLFIDDLE

Answer 3

基于您拥有BusinessCatID＆amp ;;会员ID（45,73）列出两次，但只在您需要COUNT（DISTINCT x）时计算

SELECT BusinessCatID, COUNT(DISTINCT MemberID) as NumMembersInCat
FROM MemberBusinessCatID
GROUP BY BusinessCatID

这将计算每个BusinessCatID的不同成员（基于MemberID）。如果您不担心重复，那么使用COUNT（MemberID）甚至COUNT（1）将正常工作。

Answer 4

试试这个（或类似的东西）：

       select BusinessCatID ,count(*) as NumMembersInCat 
       from MemberBusinessCats 
       group by BusinessCatID

Answer 5

试试这个

select BusinessCatID, COUNT(BusinessCatID)
from MemberBusinessCatID
group by BusinessCatID

Answer 6

Select BusinessCatID, count(MemberID) as [NumMembers In Cat] 
from MemberBusinessCats 
group by BusinessCatID, MemberID

Answer 7

我理解这个问题来自不同的TRIPLE（MemberBusinessCATID，BusinessCATID，MemberID），总结了BusinessCATID。在这种情况下，查看不同的三元组或双元组是相同的，但在更大的数据集中可能会有所不同：

创建了一个视图：

create view dist_catView as
SELECT distinct MemberBusinessCATID, BusinessCATID, Member ID from cat_table

然后

SELECT BusinessCATID, count(MemberID) from dist_catView
group by BusinessCATID

计算记录问题

7 个答案: