通過ID

Question

返回一組值的計數的專欄中，我有一個這樣的表：

ID Type 
---------- 
1  sent 
1  sent 
1  open 
1  bounce 
1  click 
2  sent 
2  sent 
2  open 
2  open 
2  click 

我想要一個查詢返回這樣的結果：

ID sent open bounce click 
1  2  1  1   1 
2  2  2  0   1 

就不能研究如何去做。謝謝。

Answer 1

您可以通過使用PIVOT或GROUP BY得到這樣的結果，你甚至可以得到結果，如果你在Type列有變量值：

測試數據：

CREATE TABLE #t(ID INT, Type VARCHAR(100)) 
INSERT #t 
VALUES 
(1,  'sent'), 
(1,  'sent'), 
(1,  'open'), 
(1,  'bounce'), 
(1,  'click'), 
(2,  'sent'), 
(2,  'sent'), 
(2,  'open'), 
(2,  'open'), 
(2,  'click') 

PIVOT方法：

SELECT pvt.* 
FROM #t 
PIVOT 
(
    COUNT(Type) FOR Type IN ([sent], [open], [bounce], [click]) 
) pvt 

如果Type還有其他可能的值，並且您不知道它們提前使用動態PIVOT：

DECLARE @cols NVARCHAR(1000) = STUFF(
(
    SELECT DISTINCT ',[' + Type + ']' 
    FROM #t 
    FOR  XML PATH('') 
), 1, 1, '') 

DECLARE @query NVARCHAR(2000) = 
' 
SELECT pvt.* 
FROM #t 
PIVOT 
(
    COUNT(Type) FOR Type IN ('+@cols+') 
) pvt 
' 

EXEC(@query) 

如果你已經知道的固定值Type，你也可以使用：

SELECT ID, 
     COUNT(CASE WHEN Type = 'sent' THEN 1 END) [sent], 
     COUNT(CASE WHEN Type = 'open' THEN 1 END) [open], 
     COUNT(CASE WHEN Type = 'bounce' THEN 1 END) bounce, 
     COUNT(CASE WHEN Type = 'click' THEN 1 END) click 
FROM #t 
GROUP BY ID 

Answer 2

嘗試PIVOT

SELECT ID,[sent],[open],[bounce],[click] 
FROM your_table 
PIVOT (COUNT([Type]) 
FOR [Type] in ([sent],[open],[bounce],[click]))p 

SQL Fiddle Demo

Answer 3

Select Id, 
count(case When type='sent' then 1 else 0 end) as sent, 
count(case when type='open' then 1 else 0 end) as open 
From table 
Group by Id 

如果不會給你確切的答案，然後嘗試次數（不同的情況....）:)