日期的间隔的SQL

Question

问题:
1 2009-4-1 -2000
1 2009-4-3 -2000
1 2009-4-7 16550
1 2009-4-11 4400
1 2009-4-13 -5000
1 2009-4-14 5050结果数据:
1 2009-4-1 -2000
1       2009-4-2        -2000
1 2009-4-3 -4000
1 2009-4-4 -4000
1 2009-4-5 -4000
1 2009-4-6 -4000
1 2009-4-7 16550  12550
1 2009-4-8 12550
1 2009-4-9 12550
1 2009-4-10 12550
1 2009-4-11 4400   16950
1 2009-4-12 16950
1 2009-4-13 -5000  11950
1 2009-4-14 5050   17000

Answer 1

设日期字段为rq，数字字段num
try it
select b.rq,sum(a.num)
from taba as a,
(select to_date('2009-4-1','yyyy-mm-dd')+rowNUM-1 as rq from dual connect by rownum<=14) as b
where a.rq<=b.rq
group by b.rq