SQL查詢至少連續n天登錄的用戶
以連續3天為例,使用工具:MySQL。
1.創建SQL表:
create table if not exists orde(id varchar(10),date datetime,orders varchar(10)); insert into orde values('1' , '2019/1/1',10 ); insert into orde values('1' , '2019/1/2',109 ); insert into orde values('1' , '2019/1/3',150 ); insert into orde values('1' , '2019/1/4',99); insert into orde values('1' , '2019/1/5',145); insert into orde values('1' , '2019/1/6',1455); insert into orde values('1' , '2019/1/7',199); insert into orde values('1' , '2019/1/8',188 ); insert into orde values('4' , '2019/1/1',10 ); insert into orde values('2' , '2019/1/2',109 ); insert into orde values('3' , '2019/1/3',150 ); insert into orde values('4' , '2019/1/4',99); insert into orde values('5' , '2019/1/5',145); insert into orde values('6' , '2019/1/6',1455); insert into orde values('7' , '2019/1/7',199); insert into orde values('8' , '2019/1/8',188 ); insert into orde values('9' , '2019/1/1',10 ); insert into orde values('9' , '2019/1/2',109 ); insert into orde values('9' , '2019/1/3',150 ); insert into orde values('9' , '2019/1/4',99); insert into orde values('9' , '2019/1/6',145); insert into orde values('9' , '2019/1/9',1455); insert into orde values('9' , '2019/1/10',199); insert into orde values('9' , '2019/1/13',188 );
查看數據表:
2.使用row_number() over() 排序函數計算每個id的排名,SQL如下:
select *,row_number() over(partition by id order by date ) 'rank' from orde where orders is not NULL;
查看數據表:
3.將date日期字段減去rank排名字段,SQL如下:
select *,DATE_SUB(a.date,interval a.rank day) 'date_sub' from( select *,row_number() over(partition by id order by date ) 'rank' from orde where orders is not NULL ) a;
查看數據:
4.根據id和date分組並計算分組後的數量(count)、計算最早登錄和最晚登錄的時間,SQL如下:
select b.id,min(date) 'start_time',max(date) 'end_time',count(*) 'date_count' from( select *,DATE_SUB(a.date,interval a.rank day) 'date_sub' from( select *,row_number() over(partition by id order by date ) 'rank' from orde where orders is not NULL ) a ) b group by b.date_sub,id having count(*) >= 3 ;
查看數據:
參考資料:
SQL查詢至少連續七天下單的用戶
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