select did_cnt, question_cnt from (select
    count(distinct device_id) as did_cnt,
    count(question_id) as question_cnt,
    (case when date like "2021-08%" then "2021-08" else "其他"end) as mouth 
    from question_practice_detail
group by mouth having mouth like "2021-08") as question;

不用where的写法