List<int> ListInt = new List<int> { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };


                int count = ListInt.Count;
                int index = 1;
                foreach (var item in ListInt)
                {
                    if (index != count)
                    {
                        Console.WriteLine("do something at index number  " + index);
                    }
                    else
                    {
                        Console.WriteLine("Foreach loop, this is the last iteration of the loop " + index);
                    }
                    index++;

                }
 //OR
                int count = ListInt.Count;
                int index = 1;
                foreach (var item in ListInt)
                {
                    if (index < count)
                    {
                        Console.WriteLine("do something at index number  " + index);
                    }
                    else
                    {
                        Console.WriteLine("Foreach loop, this is the last iteration of the loop " + index);
                    }
                    index++;

                }

你可以这样做:

foreach (DataGridViewRow dgr in product_list.Rows)
{
    if (dgr.Index == dgr.DataGridView.RowCount - 1)
    {
        //do something
    }
}

在某些类型上使用Last()将遍历整个集合! 这意味着如果执行foreach并调用Last()，则循环两次!我相信你会尽量避免大量收藏。

那么解决方案是使用while循环:

using var enumerator = collection.GetEnumerator();

var last = !enumerator.MoveNext();
T current;

while (!last)
{
  current = enumerator.Current;        

  //process item

  last = !enumerator.MoveNext();        
  if(last)
  {
    //additional processing for last item
  }
}

因此，除非集合类型为IList<T>，否则Last()函数将遍历所有集合元素。

Test

如果你的集合提供了随机访问(例如实现了IList<T>)，你也可以像下面这样检查你的项目。

if(collection is IList<T> list)
  return collection[^1]; //replace with collection.Count -1 in pre-C#8 apps

     List<int> ListInt = new List<int> { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };


                int count = ListInt.Count;
                int index = 1;
                foreach (var item in ListInt)
                {
                    if (index != count)
                    {
                        Console.WriteLine("do something at index number  " + index);
                    }
                    else
                    {
                        Console.WriteLine("Foreach loop, this is the last iteration of the loop " + index);
                    }
                    index++;

                }
 //OR
                int count = ListInt.Count;
                int index = 1;
                foreach (var item in ListInt)
                {
                    if (index < count)
                    {
                        Console.WriteLine("do something at index number  " + index);
                    }
                    else
                    {
                        Console.WriteLine("Foreach loop, this is the last iteration of the loop " + index);
                    }
                    index++;

                }

还有更简单的方法吗?

Item last = null;
foreach (Item result in Model.Results)
{
    // do something with each item

    last = result;
}

//Here Item 'last' contains the last object that came in the last of foreach loop.
DoSomethingOnLastElement(last);

如果你只需要对最后一个元素做一些事情(而不是对最后一个元素做一些不同的事情)，那么在这里使用LINQ会有所帮助:

Item last = Model.Results.Last();
// do something with last

如果你需要对最后一个元素做一些不同的事情，那么你需要这样的东西:

Item last = Model.Results.Last();
foreach (Item result in Model.Results)
{
    // do something with each item
    if (result.Equals(last))
    {
        // do something different with the last item
    }
    else
    {
        // do something different with every item but the last
    }
}

不过，您可能需要编写一个自定义比较器，以确保您可以判断该项与Last()返回的项相同。

这种方法应该谨慎使用，因为Last可能必须遍历集合。虽然这对于小型集合可能不是问题，但如果它变得很大，可能会对性能产生影响。如果列表包含重复的项，也会失败。在这种情况下，这样的语句可能更合适:

int totalCount = result.Count();
for (int count = 0; count < totalCount; count++)
{
    Item result = Model.Results[count];

    // do something with each item
    if ((count + 1) == totalCount)
    {
        // do something different with the last item
    }
    else
    {
        // do something different with every item but the last
    }
}