返回具有重叠日期的行

Question

我试图找到那些与其他行重叠日期的行。

示例数据：

Period: 13, Start Date: 9/1/2015, End Date: 9/30/2015

Period: 14, Start Date: 9/15/2015, End Date: 10/30/2015

Period: 15, Start Date: 11/1/2015, End Date: 11/30/2015

这是我当前的查询：

select pd1.periodNumber AS 'Period Seq A', pd1.startDate, pd1.endDate,   
pd2.periodNumber AS 'Period Seq B', pd2.startDate, pd2.endDate,    
pd2.adjustment 
from dbo.PeriodDefinition pd1 
inner join dbo.PeriodDefinition pd2 on pd2.startDate > pd1.startDate and   
pd2.startDate < pd2.endDate 
where (pd1.adjustment =1 AND pd2.adjustment = 1)

它应该只返回1行，并显示第13期（期间A）和期间14（期间B）的数据。现在它返回3行，并在结果中显示13和15。我错过了什么？谢谢！

Answer 1

and  pd2.startDate < pd2.endDate

应该是

and pd2.startDate < pd1.endDate  

我会写

on pd2.startDate between pd1.startDate and pd1.endDate

你也可能想要一个

and pd1.periodNumber != pd2.periodNumber

（假设periodNumber是唯一的行标识符。）