Question

所以我有一个表格，我填充了我所做的测验中的数据。为什么？因此，我可以发现具有特定高/低成功率的问题，以确定是否需要重写问题。我认为它会派上用场=）任何方式，我都会遇到一些问题＆＃34;分组＆＃34;一行的答案，并计算＆＃34;正确＆＃34;的出现次数。和＆＃34;错误＆＃34;。

表格如下：

+----+---------+----------+-------------+---------+------------+------------+
| id | user_id | stack_id | question_id | answer  | created_at | updated_at |
+----+---------+----------+-------------+---------+------------+------------+
|  1 |      34 |        1 |          45 | wrong   | NOW()      | NOW()      |
|  2 |      55 |        1 |          45 | correct | NOW()      | NOW()      |
|  3 |       2 |        3 |          13 | wrong   | NOW()      | NOW()      |
|  4 |      99 |        1 |          46 | correct | NOW()      | NOW()      |
|  5 |      12 |        3 |          13 | wrong   | NOW()      | NOW()      |
|  6 |      12 |        2 |          10 | correct | NOW()      | NOW()      |
+----+---------+----------+-------------+---------+------------+------------+

列名正确，数据是模型。但我想要的是对所有question_id进行分组，并计算wrong存在多少correct和question_id个。{/ p>

（我没有sql的家伙，所以请，如果还有其他方式可以免费显示这些数据！）

但预期的结果应该是：

+----------+-------------+-------+---------+
| stack_id | question_id | wrong | correct |
+----------+-------------+-------+---------+
|        1 |          45 |     1 |       1 |
|        3 |          13 |     1 |       0 |
|        1 |          46 |     0 |       1 |
|        2 |          10 |     0 |       1 |
+----------+-------------+-------+---------+

这是我到目前为止所做的，并不是很多：SELECT answer, COUNT(*) FROM answers GROUP BY answer，它似乎也显示错误的信息。

Answer 1

试试这个：

    select stack_id,question_id,sum(case when answer='wrong' then 1 else 0 end) as wrong,
           sum(case when answer='correct' then 1 else 0 end) as correct
    from your_table
    group by stack_id,question_id;

让我知道它是否有效。

Answer 2

您可以使用以下SQL来获得所需的结果：

select stack_id,question_id,
sum(case when answer='wrong' then 1 else 0 end) as wrong,
sum(case when answer='correct' then 1 else 0 end) as correct
from your_table_name
group by stack_id,question_id;

Answer 3

如果这是MySQL，我会使用这个逻辑：

select stack_id, question_id,
       sum(answer = 'correct') as correct, sum(answer = 'wrong') as wrong
from t
group by stack_id, question_id;

Answer 4

SELECT
    Answer
    , QuestionID
    , Count(Answer)
GROUP BY
   Answer, 
   QuesitonID

如何计算特定ID的两个不同值？

4 个答案: