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。
答案 0 :(得分:6)
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
答案 1 :(得分: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