需要我的 mysql 表结果在 2 条件下的结果的一些解决方案

Question

I have table like this

table_name : AnswerDetail

|id_session|id_question|value|
------------------------------
|1         |1          |4    |
|1         |2          |4    |
|1         |3          |4    |
|1         |4          |2    |
|1         |1          |3    |
|1         |2          |2    |
|1         |3          |1    |
|1         |4          |4    |
|2         |1          |3    |
|2         |2          |2    |
|2         |3          |2    |
|3         |1          |4    |

I need show some result by 2 condition here,

$query = $this->db->query("SELECT a.id_session, a.id_question,
(SELECT COUNT(*) FROM AnswerDetail WHERE id_session=1 AND value=4 AND id_question=1) AS great,
(SELECT COUNT(*) FROM AnswerDetail WHERE id_session=1 AND value=3 AND id_question=1) AS good"),
(SELECT COUNT(*) FROM AnswerDetail WHERE id_session=1 AND value=2 AND id_question=1) AS not bad
FROM (SELECT DISTINCT id_session,id_question FROM AnswerDetail) a WHERE id_session=1);

what i need to do here is, im put id_session by ID from URL SEGMENT (its ok),but id_question by an array by looping from foreach php.

and thats query just show by id_question=1 , how can i get result by id_question by [1,2,3,4]

what im expected is actual like this

|id_session|id_question|great|good|not bad|
-------------------------------------------
|1         |1          |1    |1   |0      |
|1         |2          |1    |0   |1      |
|1         |3          |1    |0   |0      |
|1         |4          |1    |0   |1      |

any solution? or what should i change?

Answer 1

you can use sum() instead of count() then group by session and question ids.

select id_session, id_question, sum(case when value = 4 then 1 else 0 end) as great,
    sum(case when value = 3 then 1 else 0 end) as good,
    sum(case when value = 2 then 1 else 0 end) as notbad
from AnswerDetail
where id_session = 1
group by id_session, id_question