2016-07-26 65 views
2

輸入表:市在表中查找值的總的和不同的值之間的差值

+-----------+ 
    | Name | 
    +-----------+ 
    | AAA  | 
    +-----------+ 
    | AAA  | 
    +-----------+ 
    | BBB  | 
    +-----------+ 
    | CCC  | 
    +-----------+ 

具有一列Name。我想要輸出將返回城市名稱總數和城市名稱數量之間的差異。

在該表的情況下,結果將是1

+0

您需要詳細說明您打算如何計算差異。根據需要放置更多數據。 –

回答

8

使用COUNTDISTINCTCOUNT

SELECT COUNT(name)-COUNT(DISTINCT name) AS nameDifference 
FROM city 

輸出:

nameDifference 
1 
1

試試這個

SELECT COUNT(Name)-COUNT(DISTINCT Name) AS NewName 
FROM city 
0

你可以使用DISTINCT COUNT

SELECT COUNT(name) - COUNT(DISTINCT name)FROM city