SQL - 计算包含两个不同字段的行数

Question

是否有任何获取fieldB计数的方法，但只有fieldA和fieldB组合是唯一的？

例如

fieldA  fieldB 
X   A 
X   B 
Y   C 
X   B 
Y   A 

场B的计数将返回：

A = 2 
B = 1 
C = 1 

我一直在使用 'SELECT DISTINCT FIELDA，COUNT（fieldB）AS数量FROM表GROUP BY fieldB' 尝试。这似乎返回fieldB中所有值的计数，而不仅仅是具有唯一fieldA的值。

我希望我想要做的事情有意义，我很难说出来。我试图用SQL来做什么？

感谢

Answer 1

应该

SELECT fieldB, COUNT(DISTINCT fieldA, fieldB) AS count 
FROM table 
GROUP BY fieldB; 

Answer 2

SELECT fieldA, COUNT(fieldB) AS count from 
(SELECT DISTINCT fieldA, fieldB FROM table)a 
FROM a GROUP BY a.fieldB 

Answer 3

我觉得这让你有

Declare @mytbl table (fieldA char(1), fieldB char(1)) 
insert into @mytbl 
select 'X','A' 
union all select 'X','B' 
union all select 'Y','C' 
union all select 'X','B' 
union all select 'Y','A' 

select COUNT(*), LEFT(myfields,1) 
from 
(
select distinct fieldb + fielda as myfields 
from @mytbl 
group by fieldb + fielda 
) a 
group by LEFT(myfields,1) 

返回

2一 1B中 1 C

Answer 4

DECLARE @mytb table 
(
fieldA varchar(10), 
fieldB varchar(10) 
); 

INSERT into @mytb values('X','A'); 
INSERT into @mytb values('X','B'); 
INSERT into @mytb values('Y','C'); 
INSERT into @mytb values('X','B'); 
INSERT into @mytb values('Y','A'); 

SELECT fieldB,count(DISTINCT fieldA+fieldB) as count 
from @mytb 
group BY fieldB; 


    fieldB  count 
---------- ----------- 
    A   2 
    B   1 
    C   1