select id,max(lxcs) as zdlxcs -- 最大连续次数

from (

select id,fenzu,count(1) as lxcs --连续次数

from (

select *

,qc-rownum as fenzu -- 分组

 from (

select id,qc

,row_number() over(partition by id order by id,qc) as rownum

from ( -- 表

select 1 as id,1 as qc union all

select 1 as id,6 as qc union all

select 1 as id,7 as qc union all

select 1 as id,8 as qc union all

select 1 as id,9 as qc union all

select 1 as id,10 as qc union all

select 1 as id,11 as qc union all

select 1 as id,12 as qc union all

select 2 as id,1 as qc union all

select 2 as id,2 as qc union all

select 2 as id,3 as qc union all

select 2 as id,4 as qc union all

select 2 as id,9 as qc union all

select 2 as id,10 as qc union all

select 2 as id,11 as qc union all

select 2 as id,12 as qc --union all

) a

) b

) c

group by id,fenzu

) d

group by id

你这个连续怎么定义？每天？每周？每月？

问题不够清晰，qc字段 6,7 | 7,8|8,9|9,10|……各算一次，还是6-12只算一次。。是否你截图中只有一次连续还款？

select tt.id,max(tt.flg) from (

select t.id,count(t.flag) flg from (

select id,qc,to_number(qc) -lag(to_number(qc),1,to_number(qc)-1) over (partition by id order by  to_number(qc)) flag

from testab ) t GROUP BY t.id 

)  tt

GROUP BY tt.id

因为我创建qc字段用了文本，但是不能计算 所以这里转换了下