我认为，在以下情况下，两个范围重叠就足够了：

(StartDate1 <= EndDate2) and (StartDate2 <= EndDate1)

这里有一个可以在本地使用的通用方法。

    // Takes a list and returns all records that have overlapping time ranges.
    public static IEnumerable<T> GetOverlappedTimes<T>(IEnumerable<T> list, Func<T, bool> filter, Func<T,DateTime> start, Func<T, DateTime> end)
    {
        // Selects all records that match filter() on left side and returns all records on right side that overlap.
        var overlap = from t1 in list
                      where filter(t1)
                      from t2 in list
                      where !object.Equals(t1, t2) // Don't match the same record on right side.
                      let in1 = start(t1)
                      let out1 = end(t1)
                      let in2 = start(t2)
                      let out2 = end(t2)
                      where in1 <= out2 && out1 >= in2
                      let totover = GetMins(in1, out1, in2, out2)
                      select t2;

        return overlap;
    }

    public static void TestOverlap()
    {
        var tl1 = new TempTimeEntry() { ID = 1, Name = "Bill", In = "1/1/08 1:00pm".ToDate(), Out = "1/1/08 4:00pm".ToDate() };
        var tl2 = new TempTimeEntry() { ID = 2, Name = "John", In = "1/1/08 5:00pm".ToDate(), Out = "1/1/08 6:00pm".ToDate() };
        var tl3 = new TempTimeEntry() { ID = 3, Name = "Lisa", In = "1/1/08 7:00pm".ToDate(), Out = "1/1/08 9:00pm".ToDate() };
        var tl4 = new TempTimeEntry() { ID = 4, Name = "Joe", In = "1/1/08 3:00pm".ToDate(), Out = "1/1/08 8:00pm".ToDate() };
        var tl5 = new TempTimeEntry() { ID = 1, Name = "Bill", In = "1/1/08 8:01pm".ToDate(), Out = "1/1/08 8:00pm".ToDate() };
        var list = new List<TempTimeEntry>() { tl1, tl2, tl3, tl4, tl5 };
        var overlap = GetOverlappedTimes(list, (TempTimeEntry t1)=>t1.ID==1, (TempTimeEntry tIn) => tIn.In, (TempTimeEntry tOut) => tOut.Out);

        Console.WriteLine("\nRecords overlap:");
        foreach (var tl in overlap)
            Console.WriteLine("Name:{0} T1In:{1} T1Out:{2}", tl.Name, tl.In, tl.Out);
        Console.WriteLine("Done");

        /*  Output:
            Records overlap:
            Name:Joe T1In:1/1/2008 3:00:00 PM T1Out:1/1/2008 8:00:00 PM
            Name:Lisa T1In:1/1/2008 7:00:00 PM T1Out:1/1/2008 9:00:00 PM
            Done
         */
    }

此处发布的解决方案不适用于所有重叠范围。。。

----------------------|-------A-------|----------------------
    |----B1----|
           |----B2----|
               |----B3----|
               |----------B4----------|
               |----------------B5----------------|
                      |----B6----|
----------------------|-------A-------|----------------------
                      |------B7-------|
                      |----------B8-----------|
                         |----B9----|
                         |----B10-----|
                         |--------B11--------|
                                      |----B12----|
                                         |----B13----|
----------------------|-------A-------|----------------------

我的工作解决方案是：

AND (
  ('start_date' BETWEEN STARTDATE AND ENDDATE) -- caters for inner and end date outer
  OR
  ('end_date' BETWEEN STARTDATE AND ENDDATE) -- caters for inner and start date outer
  OR
  (STARTDATE BETWEEN 'start_date' AND 'end_date') -- only one needed for outer range where dates are inside.
) 

为了涵盖PHP/LaravelCarbon包中的所有重叠情况，您可以在Charles的答案中展开逻辑，如下所示。

if ( ($startTime1->between($startTime2, $endTime2, true) || $endTime1->between($startTime2, $endTime2, true)) || (($startTime1 <= $endTime2) && ($endTime2 <= $endTime1)) ){
//Complete Overlap, Partial Left Overlap, Partial Right Overlap. 
}

这将检查StartTime1是否在（StartTime2-EndTime2）范围之间，或者EndTime1是否位于（StartTime2-EndTime2的范围之间。

其余部分完全重叠，如其他答案所述。

由于对于不同的语言和环境有多种答案，这里有一个标准ANSISQL的答案。

在标准SQL中，它非常简单

(StartDate1, EndDate1) overlaps (StartDate2, EndDate2)

假设所有四列都是DATE或TIMESTAMP列。如果两个范围至少有一天相同（假设DATE值），则返回true

（但并非所有DBMS产品都支持此功能）

在PostgreSQL中，使用日期范围也很容易测试是否包含

daterange(StartDate1, EndDate1) @> daterange(StartDate2, EndDate2)

如果第二个范围完全包含在第一个范围中（与“重叠”不同），则上述值返回真

这里是我的简化。假设时间间隔

       | |
     | |

不要重叠，xa<xb和ya<yb可以简化处理。上述条件可以简单检查。然后你会得到

      ya       yb
      |--------|
xa |--|xb
   |----|
         |----|
ya > xa => return ya - xa < xb - xa
else    => return xa < yb

正如您可能注意到的，第二个间隔仅在xa>=ya的情况下相关。