2017-01-02 2017-01-02 0
2017-01-03 2017-01-03 0
2017-01-04 null null
2017-01-05 null null
2017-01-06 null null
2017-01-07 2017-01-07 0
期望结果, T2 IS NULL 的时候,用最近的T2补
T1 T2 T32017-01-01 2017-01-01 02017-01-02 2017-01-02 0
2017-01-03 2017-01-03 0
2017-01-04 2017-01-03 12017-01-05 2017-01-03 2
2017-01-06 2017-01-03 3
2017-01-07 2017-01-07 0
假设你的表名为t,执行下面的SQL,但是这个是三层嵌套,耗费比较大
select T1,T2,value(T3,(T1-T2)) from (select tab1.t1 as t1,case when tab1.t2 is not null then tab1.t2 else (select max(tab2.t2) from t tab2 where tab2.t2 is not null and tab2.T1