SQL查詢 - 計算一組問題的yes答案數

Question

我有一個調查相關應用程序的表，其中有5個問題都需要是（1）或否（0）答案。該表設計如下：

CREATE TABLE score ( project_id int NOT NULL, resp_id int NULL, q1 int, q2 int, q3 int, q4 int, q5 int, );

的樣本數據：

project_id resp_id q1 q2 q3 q4 q5 
1   86  1 1 1 1 1 
1   114  1 1 1 1 1 
1   118  0 1 1 1 0 
2   154  1 1 0 1 1 
2   178  1 1 1 0 1 
3   182  1 0 1 1 1 
3   190  1 1 1 0 1 
3   208  1 1 1 1 1 
3   300  1 1 1 0 0 
3   329  1 1 1 1 1 

我需要做的是編寫一個查詢，將生成一個報告，我通過PROJECT_ID擊穿，受訪人數提供0個「是」答案（5箇中的0個），1個「是」答案（5箇中的1個），2個「是」答案（5箇中的2個）等。換句話說，有多少答覆者回答「是」所有問題，是5個問題中的4個，等等。

可行嗎？任何建議表示讚賞:)

Answer 1

創建一個存儲過程並將值設置爲變量，然後您可以使用一個select語句在sproc中傳遞它們，並有條件地使用where子句。

Answer 2

SELECT 
    project_ID, 
    q1 + q2 + q3 + q4 + q5 AS NumYesAnswers, 
    COUNT(*) AS NumResponses 
FROM 
    score 
GROUP BY 
    project_ID, q1 + q2 + q3 + q4 + q5 

編輯

我想補充這爲路寶的答案註釋，但格式是行不通的。你可以多一點優雅與PIVOT達到同樣的效果：

所有交易答案

SELECT 
    Project_ID, "0" AS NoToAll, "1", "2", "3", "4", "5" 
FROM 
    (
    SELECT Project_ID, Q1 + Q2 + Q3 + Q4 + Q5 AS NumYeses, COUNT(*) AS Answers 
    FROM Score 
    GROUP BY Project_ID, Q1 + Q2 + Q3 + Q4 + Q5 
    ) AS X 
PIVOT 
    (SUM(Answers) FOR NumYeses IN ("0", "1", "2", "3", "4", "5")) AS Y 

Answer 3

喬恩是很好的。另外，如果您不想重複sum語句，則使用CTE（公用表表達式）。

With 
ProYes as (
    select project_id, 
    q1 + q2 + q3 + q4 + q5 as NumYes 
    from score) 
select project_id, NumYes, COUNT(*) as NumAnswers 
from ProYes 
group by project_id, NumYes 

Answer 4

-- Be careful about NULL fields(the following is assuming that the q* are NOT NULL fields) 
SELECT project_id, 
    SUM(allzeroes) AS NoToAll, SUM(onlyone) AS YesToOne, SUM(two) AS YesToTwo, SUM(three) AS YesToThree, SUM(four) AS YesToFour, SUM(five) AS YesToAll 
FROM (
    SELECT project_id, 
     (CASE WHEN (q1 + q2 + q3 + q4 + q5) = 0 THEN COUNT(*) ELSE 0 END) AS allzeroes, 
     (CASE WHEN (q1 + q2 + q3 + q4 + q5) = 1 THEN COUNT(*) ELSE 0 END) AS onlyone, 
     (CASE WHEN (q1 + q2 + q3 + q4 + q5) = 2 THEN COUNT(*) ELSE 0 END) AS two, 
     (CASE WHEN (q1 + q2 + q3 + q4 + q5) = 3 THEN COUNT(*) ELSE 0 END) AS three, 
     (CASE WHEN (q1 + q2 + q3 + q4 + q5) = 4 THEN COUNT(*) ELSE 0 END) AS four, 
     (CASE WHEN (q1 + q2 + q3 + q4 + q5) = 5 THEN COUNT(*) ELSE 0 END) AS five 
    FROM score 
    GROUP BY project_id, (q1 + q2 + q3 + q4 + q5)) temp_table 
GROUP BY project_id 

PS：請修正採樣數據報頭;我無法編輯你的問題。謝謝！