请任何人协助我们解决最终的挑战者查询。
表名：attendance_tbl
nota_id represents present or absent nota_id ='1' => PRESENT
priod_id
1=>Hour-1
2=>Hour-2
3=>Hour-3
4=>Hour-4
5=>Hour-5
如果priod_id包含值1或2或3或4，则根据条件我们将百分比设置为0.25
如果priod_id包含1或2或3或4或5，则根据条件我们将百分比设置为0.2。
如果priod_id大于4，如果在任何一天其值为5，那么我们将所有1,2,3,4的百分比都设置为0.2
如果priod_id为4，则必须将所有1,2,3,4的百分比设置为0.25
示例：
Date : 2018-08-31
priod_id =1 , priod_id =2 priod_id =5
Then set Percentage = 0.2 for 1,2,5
priod_id =1 , priod_id =2 priod_id =4
Then set Percentage = 0.25 for 1,2,4
Please refer to the query we have tried the least.
SELECT *,
IF(`nota_id` = '1', Percentage,'0.0') AS ABSENT
from (SELECT *, CASE WHEN (max(`priod_id`) > 4) THEN "0.2"
ELSE "0.25"
END as Percentage
FROM `attendance_tbl`
WHERE DATE(attend_date) BETWEEN '2018-08-01' AND '2018-08-31'
AND sid = '74'
GROUP BY sid, attend_date) t
答案 0 :(得分：0)
我们找到了答案。
SELECT *，
案件
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）> 4）然后
“ 0.2”
其他
“ 0.25”
以百分比结尾，
COUNT（priod_id）为HOURS，
COUNT（箱）
当nota_id ='1'时
1个
END）AS PR
来自priod_id A
DATE（'2018-08-01'）和DATE（'2018-08-31'）之间的DATE（attend_date）
AND sid ='60'
GROUP BY DATE（attend_date）
