Question

所以这是我所面对的问题 - 我有一张桌子＆＃34; A＆＃34;如下

  Name     AmountPaid    Type
  NAME1        5          1
  NAME1        10         1
  NAME1        3          1
  NAME2        5          2
  NAME2        9          2
  NAME2        8          2

这里我们有两个名字 - 名字1和名字2，类型1和2

我正在尝试获取一个查询，该查询应仅返回数字，该数字应显示表中存在多少次唯一名称集合。在上面我希望结果为 - 2 我试过了查询

Select NAME, count(Name) from A group by Name

然而，这将返回2个以下的记录。

  NAME1    3
  NAME2    3

非常感谢任何帮助。

Answer 1

请尝试以下查询：

Select count(distinct(NAME)) from A

Answer 2

如果没有错，这就是你需要的。考虑到count如果不重复，则name

SELECT Count(DISTINCT NAME)
FROM   (SELECT *,
               Row_number()OVER(partition BY NAME ORDER BY type) rn
        FROM   Yourtable) a
WHERE  rn > 1

Answer 3

您可以使用DISTINCT执行此操作：

SELECT COUNT (DISTINCT NAME)
FROM A

Answer 4

使用它：

--TEST TABLE
Declare @sometable as table (Name Varchar(10))
insert into @sometable
values  ('Name1'),
        ('Name1'),
        ('Name1'),
        ('Name2'),
        ('Name2'),
        ('Name2'),
        ('Name3'),
        ('Name4')
--VARIANT 1
SELECT COUNT(*)
FROM   (
            SELECT NAME,
                   COUNT(*) AS Cnt
            FROM   @sometable
            GROUP BY NAME
        ) AS T
WHERE  Cnt > 1
--VARIANT 2
SELECT COUNT(*)
FROM   ( SELECT NAME
         FROM   @sometable
         GROUP BY NAME
         HAVING COUNT(*) > 1 ) AS T
--VARIANT 3
SELECT COUNT(*)
FROM   (SELECT DISTINCT NAME, COUNT(*) OVER (PARTITION BY NAME) AS Cnt
        FROM   @sometable
        ) AS T
WHERE  Cnt > 1

仅计算重复记录的行数

4 个答案: