需要我的 mysql 表結果在 2 條件下的結果的一些解決方案

Question

我有這樣的桌子

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    |

我需要在這里通過 2 個條件顯示一些結果，

$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);

我需要在這里做的是，我將id_session按 ID 從 URL SEGMENT （它可以），但通過從 foreach id_question循環的數組中輸入 id_question。

那就是查詢僅由id_question=1顯示，我如何id_question by [1,2,3,4]結果

我所期望的實際上是這樣的

|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      |

任何解決方案？ 或者我應該改變什么？

Answer 1

您可以使用sum()而不是count()然后按session和question ID 分組。

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