在Oracle中使用sql查找日期范围中的差距

我需要使用sql查找基本日期范围和测试范围之间的差距 SD和ED是开始日期和结束日期。A和B的所有行都在同一个表中

A的约会

B的日期

缺失间隙范围

我在这里看了一些例子 但他们不像我

select 
   to_char(SD, 'yyyymmdd') as SD,
   to_char(ED, 'yyyymmdd') as ED
from
   (  -- prepare gaps in each A after removing all B
      select
         BED + 1 as SD,
         lead(BSD, 1, AED + 1) over (partition by ASD,AED order by BSD) - 1 as ED
      from
         (  -- prepare all intersections between A and B
            select AA.sd as ASD, AA.ed as AED, BB.sd as BSD, BB.ed as BED
            from AA join BB on least(AA.ed, BB.ed) >= greatest(AA.sd, BB.sd)
            union all
            select AA.sd, AA.ed, to_date('1000','yyyy'), AA.sd - 1
            from AA
         )
   )   
where SD <= ED  -- exclude empty gaps
order by 1

---

这太棒了。谢谢！！！！。我的数据是整数，不是数据库中的日期，而是一个表中的日期，你能在这里添加注释吗？在这里，它是非常酷的sql，但不确定它是如何运行的working@NatashaThapa-将数字转换为日期：转换为日期编号“yyyymmdd”，将日期转换为数字：转换为日期，“yyyymmdd”如果我必须更改如何处理A中的重叠，并且希望避免B中的重叠，请您添加一些注释，说明sql的每个部分正在做什么

 ID  SD            ED
 B   20130120      20130420
 B   20130601      20130830
 B   20130910      20131130

Output should be: 
the Dates that are in A but are not in B with no dates overlaps

SD           ED
20130101     20130119
20130421     20130531
20130831     20130909

select 
   to_char(SD, 'yyyymmdd') as SD,
   to_char(ED, 'yyyymmdd') as ED
from
   (  -- prepare gaps in each A after removing all B
      select
         BED + 1 as SD,
         lead(BSD, 1, AED + 1) over (partition by ASD,AED order by BSD) - 1 as ED
      from
         (  -- prepare all intersections between A and B
            select AA.sd as ASD, AA.ed as AED, BB.sd as BSD, BB.ed as BED
            from AA join BB on least(AA.ed, BB.ed) >= greatest(AA.sd, BB.sd)
            union all
            select AA.sd, AA.ed, to_date('1000','yyyy'), AA.sd - 1
            from AA
         )
   )   
where SD <= ED  -- exclude empty gaps
order by 1