計數值的數量並返回一個列作爲

我試圖寫一個查詢，有效地，結合了下面兩個查詢，但我認爲我有作家阻止。我試過寫一個子查詢，但無濟於事。我已經考慮過一個交叉表查詢，但我想避免任何特定的Access。我試過了一個聯盟，但是這似乎也不起作用。我高度懷疑這只是我的垃圾。我很想嘗試一下SQL PIVOT，但是我之前沒有嘗試過，並且根據我今天的行程來判斷，我懷疑我會成功。計數值的數量並返回一個列作爲

無論如何，這對我的工作的查詢是：

SELECT Operative.NAME, COUNT(*) AS [Yes to Left Tidy?] 
FROM Operative INNER JOIN ([Survey Results] INNER JOIN JOBS ON [Survey Results].Job_Number = JOBS.Job_Number) ON Operative.USERID = JOBS.SCHEDULEITEMS_ASSIGNEDWORKERID 
WHERE Q5 = 'yes' 
GROUP BY Operative.NAME 

SELECT Operative.NAME, COUNT(*) AS [No to Left Tidy?] 
FROM Operative INNER JOIN ([Survey Results] INNER JOIN JOBS ON [Survey Results].Job_Number = JOBS.Job_Number) ON Operative.USERID = JOBS.SCHEDULEITEMS_ASSIGNEDWORKERID 
WHERE Q5 = 'no' 
GROUP BY Operative.NAME

如果你要我詳細的表結構等，請讓我知道。

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2015-05-19 Matt

你嘗試的聯合查詢，併發生了什麼？ –

我想他希望Operative.NAME，count（no），count（是）在一行中，提供q5的表上的雙連接就足夠了，儘管 –

可以給我們你的表結構和預期的查詢結果請@馬特 –

你嘗試這個

select * FROM 
    (SELECT Operative.NAME, COUNT(*) AS [Yes to Left Tidy?],'YES' AS Y/N 
    FROM Operative INNER JOIN ([Survey Results] 
    INNER JOIN JOBS ON [Survey Results].Job_Number = JOBS.Job_Number) 
    ON Operative.USERID = JOBS.SCHEDULEITEMS_ASSIGNEDWORKERID 
    WHERE Q5 = 'yes' 
    GROUP BY Operative.NAME)A, 
    UNION 
    (SELECT Operative.NAME, COUNT(*) AS [Yes to Left Tidy?],'NO' AS Y/N 
    FROM Operative INNER JOIN ([Survey Results] 
    INNER JOIN JOBS ON [Survey Results].Job_Number = JOBS.Job_Number) 
    ON Operative.USERID = JOBS.SCHEDULEITEMS_ASSIGNEDWORKERID 
    WHERE Q5 = 'no' 
    GROUP BY Operative.NAME)B

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2015-05-19 12:25:55 Pu297

好像要組由Q5，如：

SELECT Operative.NAME, Q5 [Left Tidy?], COUNT(*) 
FROM Operative INNER JOIN ([Survey Results] INNER JOIN JOBS ON [Survey Results].Job_Number = JOBS.Job_Number) ON Operative.USERID = JOBS.SCHEDULEITEMS_ASSIGNEDWORKERID 
WHERE Q5 IN ('yes', 'no') 
GROUP BY Q5, Operative.NAME

..如果你想要一個對每行數，你可以把這些結果，並它們的樞軸，例如：

SELECT * FROM (
SELECT Operative.NAME, Q5 [Left Tidy?], COUNT(*) c 
FROM Operative INNER JOIN ([Survey Results] INNER JOIN JOBS ON [Survey Results].Job_Number = JOBS.Job_Number) ON Operative.USERID = JOBS.SCHEDULEITEMS_ASSIGNEDWORKERID 
WHERE Q5 IN ('yes', 'no') 
GROUP BY Q5, Operative.NAME) p 
PIVOT (SUM(c) FOR [Left Tidy?] IN ([no], [yes])) x;

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2015-05-19 12:28:28 gordy

在標準的SQL這將是SUM(CASE WHEN Q5 = 'yes' THEN 1 ELSE 0 END)，這簡單地總結一個或零基於o情況。 Access不支持CASE，但IIF應該提供相同的結果：

SELECT Operative.NAME, 
    SUM(IIF(Q5 = 'yes', 1, 0)) AS [Yes to Left Tidy?], 
    SUM(IIF(Q5 = 'no', 1, 0)) AS AS [No to Left Tidy?] 
FROM Operative INNER JOIN ([Survey Results] INNER JOIN JOBS ON [Survey Results].Job_Number = JOBS.Job_Number) ON Operative.USERID = JOBS.SCHEDULEITEMS_ASSIGNEDWORKERID 
WHERE Q5 IN ('yes', 'no') -- only needed if other values exist 
GROUP BY Operative.NAME

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2015-05-19 12:36:46 dnoeth

計數值的數量並返回一個列作爲

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