Question

我有两张牌= subjects和stats。

  -- subjects --             
------------------          
|  id  | name    |          
------------------         
|  1  |  subjecta |        
|  2  |  subjectb |       
|  3  |  subjectc |         
|  4  |  subjectd |         
|  5  |  subjecte |        
|  6  |  subjectf |        
|  7  |  subjectg |      
|  8  |  subjecth |         
|  9  |  subjecte |       
| ... |  subjectf |       
------------------       

-- stats --
-----------------------------------
| user_id | subject_id | correct  |
-----------------------------------
|  1      |    1      |   false  |
|  1      |    1      |   false  | 
|  1      |    2      |   false  |
|  4      |    3      |   false  |
|  4      |    4      |   false  |
|  4      |    5      |   false  |
|  2      |    1      |    true  |
|  2      |    1      |    true  |
|  2      |    2      |    false |
|  2      |    2      |    true  |
|  2      |    3      |    false |
---------------------------------

我需要什么，例如给予user_id（如2），以获得所有主题（来自主题表）和他所做的（来自统计数据）正确真/假的计数这样：

--------------------------------------------------
|  id  | name     |  correct true | correct false|
----------------------------------|----------------  
|  1  |  subjecta |       2       |       0      |
|  2  |  subjectb |       1       |       1      |  
|  3  |  subjectc |       0       |       1      |  
|  4  |  subjectd |       0       |       0      | 
|  5  |  subjecte |       0       |       0      | 
|  6  |  subjectf |       0       |       0      | 
|  7  |  subjectg |       0       |       0      | 
|  8  |  subjecth |       0       |       0      | 
|  9  |  subjecte |       0       |       0      | 
| ... |  subjectf |       0       |       0      | 
----------------------------------|--------------|

我不知道该怎么做。

Answer 1

您可以通过两个表之间的连接以及一些条件聚合来实现此目的，以计算真假答案的数量。

SELECT
    t1.id,
    t1.name,
    SUM(CASE WHEN t2.correct = 'true'  THEN 1 ELSE 0 END) AS correct_true,
    SUM(CASE WHEN t2.correct = 'false' THEN 1 ELSE 0 END) AS correct_false
FROM subjects t1
LEFT JOIN stats t2
    ON t1.id = t2.subject_id AND
       t2.user_id = 2
GROUP BY t1.id, t1.name

Answer 2

select subjects.id, subjects.name, coalesce(t.correct_true, 0), coalesce(t.correct_false, 0)
from subjects
left join (
SELECT
    subject_id,
    SUM(CASE WHEN correct = 'true'  THEN 1 ELSE 0 END) AS correct_true,
    SUM(CASE WHEN correct = 'false' THEN 1 ELSE 0 END) AS correct_false
    FROM stats
    where user_id = 2
    group by subject_id
) t
on subjects.id = t.subject_id

计算其他表中的值

2 个答案: