題目：有一張使用者簽到表【t_user_attendence】，標記每天使用者是否簽到（說明：該錶包含所有使用者所有工作日的出勤記錄） ，包含三個字段：日期【fdate】，使用者id【fuser_id】，使用者當天是否簽到【fis_sign_in：0否1是】
問題1：請計算截至當前每個使用者已經連續簽到的天數
select
fuser_id
,datediff(
'20200415'
,fdate_max)
as fconsecutive_days
from
(select fuser_id
,max(fdate) fdate_max
from t_user_attendence
where fis_sign_in =
0group
by fuser_id
) t1;
select fuser_id
,max
(length(cut_fsign_record)
)as fmax_days
(select fuser_id
,fsign_record
,cut_fsign_record
from
(select fuser_id
,wm_concat(fis_sign_in) fsign_record
from t_user_attendence
group
by fuser_id
) t1
lateral view explode(split(fsign_record,
'0')
) t as cut_fsign_record
) t2
where cut_fsign_record<>
''group
by fuser_id;
據此解決思路，重新碼了取最大連續天數的**。
先按人分組按天進行自然連續排序；再只取簽到部分，按人分組進行自然連續排序，相差得到差值diff1；再按照差值diff1分組計數，得到每人連續簽到的天數，求最大值即可。
select
fuser_id,
max(ct)
as max_ct
from
(select
fuser_id
,diff1
,count
(diff1)
as ct
from
(select
*,row_number(
)over
(partition
by fuser_id order
by fdate)
as or2
,or1 - row_number(
)over
(partition
by fuser_id order
by fdate)
as diff1
from
(select
fdate
,fuser_id
,fis_sign_in
,row_number(
)over
(partition
by fuser_id order
by fdate)
as or1
from t_user_attendence
order
by fuser_id,fdate
) twhere fis_sign_in =
1) t2
group
by fuser_id,diff1
) t3
group
by fuser_id;
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