SQL查询内连接0值

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      

还有这张桌子：

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

这样的结果是查询：

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 

这样的问题是查询值为0的不会出现。

Answer 1

我想你正在寻找这个：

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 INNER JOIN，而是笛卡尔连接（table1 的每一行都与 table2 的每一行相乘）。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