Oracle SQL比较for循环中的两列
Oracle SQL comparing two columns in a for loop

您已为两个数据库系统标记了该问题-我的回答与Oracle有关。

此处无需使用循环，因为您可以使用SQL分析函数LAG来处理此类问题。

以下内容将提供您想要的结果：

SELECT DISTINCT T.NAME
  FROM (SELECT A.NAME,
               B.START_DATE,
               B.END_DATE,
               LAG(B.START_DATE) OVER (PARTITION BY A.NAME ORDER BY B.NAME, B.START_DATE) AS PREV_START,
               LAG(B.END_DATE) OVER (PARTITION BY A.NAME ORDER BY B.NAME, B.START_DATE) AS PREV_END
          FROM A
          INNER JOIN B
            ON B.NAME = A.NAME
          ORDER BY A.NAME) T
  WHERE (T.START_DATE BETWEEN T.PREV_START AND T.PREV_END) OR
        (T.END_DATE BETWEEN T.PREV_START AND T.PREV_END) OR
        (T.PREV_START BETWEEN T.START_DATE AND T.END_DATE) OR
        (T.PREV_END BETWEEN T.START_DATE AND T.END_DATE);

SQLFiddle在这里

这是我的建议。 要获得重叠，可以使用：

select b.*
from tableb b
where exists (select 1
              from tableb b2
              where b.aid = b2.aid and
                    b.startdate < b2.enddate and b.enddate > b.startdate
             );

逻辑很简单。 如果第一个时间间隔在第二个时间间隔之前开始，而第一个时间间隔在第二个时间间隔之后开始，则两个时间间隔重叠。

要获取表的值：

with overlaps as (
      select b.*
      from tableb b
      where exists (select 1
                    from tableb b2
                    where b.aid = b2.aid and
                          b.startdate < b2.enddate and b.enddate > b.startdate
                   )
    )
select a.*
from tablea a
where a.aid in (select o.aid from overlaps o);

注意：根据您定义“重叠”的方式，您可能需要<=和>=进行比较。