1、虚表(测试表和数据) create test_table as
select a.cust_no, a.r_date, a.yqts from (
select '123' as cust_no, '20231101' as r_date, 0 as yqts
union all
select '123' as cust_no, '20231102' as r_date, 1 as yqts
union all
select '123' as cust_no, '20231103' as r_date, 2 as yqts
union all
select '123' as cust_no, '20231104' as r_date, 3 as yqts
union all
select '123' as cust_no, '20231105' as r_date, 0 as yqts
union all
select '123' as cust_no, '20231106' as r_date, 0 as yqts
union all
select '123' as cust_no, '20231107' as r_date, 1 as yqts
) a
2、加入有表test_table,数据如上,0标识未逾期,1标识逾期,求连续最大无逾期和逾期的天数,以上数据可直接看出最大连续逾期天数为3、未逾期未2。
3、首先将测试数据排序,让数据连续
select a.cust_no, a.r_date, a.yqts from test_table a where 1=1 order by a.cust_no, a.r_date asc
4、将数据分组并给是否逾期打上标识1逾期,2未逾期,多行转一行
se