Question

我有这张桌子：

idSection   | idQuestion    | title     | enunciation | idScale
1           | 1             | title 1   | question 1  |    3      
1           | 1             | title 1   | question 1  |    3      
1           | 1             | title 1   | question 1  |    3      
1           | 1             | title 1   | question 1  |    2
1           | 1             | title 1   | question 1  |    5
1           | 2             | title 2   | question 2  |    1      
1           | 2             | title 2   | question 2  |    3      
1           | 3             | title 3   | question 3  |    1

并有这张表：

idScale   |       name 
   1      |      Very Bad
   2      |         Bad
   3      |         Good
   4      |      Very Good
   5      |      Excellent

我想要一张这样的桌子：

idSection   | idQuestion    | title     | enunciation | Total  | Name
1           | 1             | title 1   | question 1  |    0   |  Very Bad
1           | 1             | title 1   | question 1  |    0   |  Bad
1           | 1             | title 1   | question 1  |    3   |  Good
1           | 1             | title 1   | question 1  |    0   |  Very Good
1           | 1             | title 1   | question 1  |    1   |  Excellent
1           | 2             | title 2   | question 2  |    0   |  Very Bad
1           | 2             | title 2   | question 2  |    1   |   Bad
1           | 2             | title 2   | question 2  |    3   |  Good 
1           | 2             | title 2   | question 2  |    0   |  Very Good
1           | 2             | title 2   | question 2  |    0   |  Excellent

查询：

SELECT 
    t1.idSection, t1.idQuestion, t1.title, t1.enunciation, 
    COUNT(t1.idScale) as Total, t2.name 
FROM 
    table1 AS t1
INNER JOIN 
    table2 as t2 ON t2.idScale = t1.idScale
GROUP BY 
    t1.idSection, t1.idQuestion, t1.title, t1.enunciation, t2.name

结果是查询：

idSection   | idQuestion    | title     | enunciation | Total  | Name
1           | 1             | title 1   | question 1  |    3   |  Good
1           | 1             | title 1   | question 1  |    1   |  Excellent
1           | 2             | title 2   | question 2  |    1   |   Bad
1           | 2             | title 2   | question 2  |    3   |  Good

这个问题是没有出现0的查询值。

Answer 1

我认为你正在寻找这个：

SELECT
  t1.idSection,
  t1.idQuestion,
  t1.title,
  t1.enunciation, 
  SUM(case when t1.idScale=t2.idScale then 1 else 0 end) as Total,
  t2.name
FROM
  table1 AS t1, table2 as t2
GROUP BY t1.idSection, t1.idQuestion, t1.title, t1.enunciation, t2.name, t2.idScale
ORDER BY t1.idSection, t1.idQuestion, t2.idScale

这不是一个INNER JOIN，而是一个笛卡尔连接（table1的每一行都乘以table2的每一行）。我正在使用SUM来计算INNER JOIN成功的行数。

Answer 2

试试这个：

(SELECT t1.idSection, t1.idQuestion, t1.title, t1.enunciation, 
   COUNT(t1.idScale) as Total, t2.name 
 FROM table1 t1
      JOIN table2 t2 
       ON t2.idScale=t1.idScale
GROUP BY t1.idSection, t1.idQuestion, t1.title, t1.enunciation, t2.name)
UNION 
 (SELECT DISTINCT 
  t1.idSection, idQuestion, title, enunciation, 0 as Total , t2.name
  FROM table1 t1,table2 t2
   WHERE NOT EXISTS
  (SELECT  *
   FROM table1 
        JOIN table2 
         ON table2.idScale=table1.idScale
   WHERE t1.idSection = table1.idSection
     AND t1.idQuestion = table1.idQuestion
     AND t2.idScale= table2.idScale)
)

http://sqlfiddle.com/#!3/7332c/15

Answer 3

试试这个：

SELECT t1.idSection, t1.idQuestion, t1.title, t1.enunciation, 
COUNT(t1.idScale) as Total, t2.name FROM table1 AS t1
RIGHT JOIN table2 as t2 ON t2.idScale=t1.idScale
GROUP BY t1.idSection, t1.idQuestion, t1.title, t1.enunciation, t2.name

SQL查询内部联接，值为0

3 个答案: