今天分享一个非常常考，同时也不太容易的sql题目类型，求连续时间问题。
这类题的核心是：分组排序，用时间减去排序，如果连续的话他们的差会是相同值
记住下面的表，按照提供的三步思路理清楚里面的数据逻辑，连续时间问题再不会难到你
思路：
求解连续天数是面试时非常常考的问题，这类问题有非常巧妙的统一解法，分为3步：
1、为日期排序
- row number() over (partition by use_id order by date) as rank
2、求日期和排序的差值（diff）
- 因为日期存在周期，会存在跨月、跨年的情况，故使用datediff最保险
date_diff(date, rank) as diff
3、求diff出现最多的次数
- max(count(diff))
面试真题：
表1：audit_log
表2：staff_info
表3：category_dict
题目：
1）查询近一周分天各部门员工审核量、审核业务类型量、审核来源量，并按日期升序、审核量降序排列
2）查询近7天平均审核时长大于200s的部门中平均审核时长大于400的员工信息（审核时长=审核完成时间-开始审核时间）
3）查询各业务类型中，审核量Top3的员工姓名、员工部门，以及相应的审核通过占比
4）查询连续审核大于等于3天的员工信息：部门、姓名、id
-- 1）查询近一周分天各部门员工审核量、审核业务类型量、审核来源量，并按日期升序、审核量降序排列
select date(FROM_UNIXTIME(start_time)) 日期,
audito_staf_id,
count(*) 员工审核量,
count(distinct object_category) 审核业务类型量,
count(distinct origin) 审核来源量from audit_log al
WHERE date(FROM_UNIXTIME(start_time)) > DATE_SUB(CURDATE(), INTERVAL 7 DAY)
GROUP BY 日期, audito_staf_id
ORDER BY 日期, 审核来源量 desc;-- 2）查询近7天平均审核时长大于200s的部门中平均审核时长大于400的员工信息（审核时长=审核完成时间-开始审核时间）
select b.*
from
(select department_id
from audit_log al
join staff_info si on al.audito_staf_id = si.staff_id
WHERE date(FROM_UNIXTIME(start_time)) > DATE_SUB(CURDATE(), INTERVAL 7 DAY)
GROUP BY department_id HAVING avg(timestampdiff(SECOND,FROM_UNIXTIME(start_time),FROM_UNIXTIME(create_time))) > 200) a
join
(select department_id, si.staff_id, staff_name
from audit_log al
join staff_info si on al.audito_staf_id = si.staff_id
WHERE date(FROM_UNIXTIME(start_time)) > DATE_SUB(CURDATE(), INTERVAL 7 DAY)
GROUP BY department_id, si.staff_id HAVING avg(timestampdiff(SECOND,FROM_UNIXTIME(start_time),FROM_UNIXTIME(create_time))) > 400 ) b on a.department_id = b.department_id-- 3）查询各业务类型中，审核量Top3的员工姓名、员工部门，以及相应的审核通过占比
select a1.*
from
(select a.*,
row_number() over (partition by object_category order by 员工审核量 desc) as rank_num
from
(select
object_category,
audito_staf_id,
staff_name,
department_id,
count(*) 员工审核量,
-- sum(if(al.status = 1,1,0) ) 审核通过,
sum(if(al.status = 1,1,0) ) / count(*) 审核通过占比from audit_log al
join staff_info si on al.audito_staf_id = si.staff_id
GROUP BY object_category, audito_staf_id) a) a1
where rank_num >= 3;-- 4）查询连续审核大于等于3天的员工信息：部门、姓名、id
SELECT distinct department_name, staff_name, a3.audito_staf_id
from
(select audito_staf_id,
date_sub(date, INTERVAL rank_num DAY) as gap,
count(1) as 连续审核天数
(select *,
row_number() over (partition by user_id order by 日期) as rank_num
from
(select distinct audito_staf_id, date(FROM_UNIXTIME(start_time)) 日期
from audit_log ) a1
GROUP BY audito_staf_id) a2
GROUP BY audito_staf_id, gap ) a3
join staff_info si on a3.audito_staf_id = si.staff_id
WHERE 连续审核天数 >= 3;
