我有一个foreach循环,需要执行一些逻辑时,最后一项是从列表中选择,例如:

 foreach (Item result in Model.Results)
 {
      //if current result is the last item in Model.Results
      //then do something in the code
 }

我可以知道哪个循环是最后没有使用循环和计数器吗?


当前回答

接受的答案将不适用于集合中的副本。如果设置在foreach上,则可以添加自己的索引变量。

int last = Model.Results.Count - 1;
int index = 0;
foreach (Item result in Model.Results)
{
    //Do Things

    if (index == last)
        //Do Things with the last result

    index++;
}

其他回答

进一步改进Daniel Wolf的答案,你可以堆叠在另一个IEnumerable上,以避免多次迭代和lambda,例如:

var elements = new[] { "A", "B", "C" };
foreach (var e in elements.Detailed())
{
    if (!e.IsLast) {
        Console.WriteLine(e.Value);
    } else {
        Console.WriteLine("Last one: " + e.Value);
    }
}

扩展方法实现:

public static class EnumerableExtensions {
    public static IEnumerable<IterationElement<T>> Detailed<T>(this IEnumerable<T> source)
    {
        if (source == null)
            throw new ArgumentNullException(nameof(source));

        using (var enumerator = source.GetEnumerator())
        {
            bool isFirst = true;
            bool hasNext = enumerator.MoveNext();
            int index = 0;
            while (hasNext)
            {
                T current = enumerator.Current;
                hasNext = enumerator.MoveNext();
                yield return new IterationElement<T>(index, current, isFirst, !hasNext);
                isFirst = false;
                index++;
            }
        }
    }

    public struct IterationElement<T>
    {
        public int Index { get; }
        public bool IsFirst { get; }
        public bool IsLast { get; }
        public T Value { get; }

        public IterationElement(int index, T value, bool isFirst, bool isLast)
        {
            Index = index;
            IsFirst = isFirst;
            IsLast = isLast;
            Value = value;
        }
    }
}

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

只需存储之前的值,并在循环中使用它。然后在最后,“previous”值将是最后一项,让您以不同的方式处理它。不需要计数或特殊库。

bool empty = true;
Item previousItem;

foreach (Item result in Model.Results)
{
    // Alternatively, check if previousItem == null
    // if your Enumerable can't contain nulls
    if (!empty)
    {
        // We know this isn't the last item because
        // it came from the previous iteration
        handleRegularItem(previousItem);
    }

    previousItem = result;
    empty = false;
}

if (!empty)
{
    // We know this is the last item because the loop is finished
    handleLastItem(previousItem);
}

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);

如何将foreach转换为对最后一个元素做出反应:

List<int> myList = new List<int>() {1, 2, 3, 4, 5};
Console.WriteLine("foreach version");
{
    foreach (var current in myList)
    {
        Console.WriteLine(current);
    }
}
Console.WriteLine("equivalent that reacts to last element");
{
    var enumerator = myList.GetEnumerator();
    if (enumerator.MoveNext() == true) // Corner case: empty list.
    {
        while (true)
        {
            int current = enumerator.Current;

            // Handle current element here.
            Console.WriteLine(current);

            bool ifLastElement = (enumerator.MoveNext() == false);
            if (ifLastElement)
            {
                // Cleanup after last element
                Console.WriteLine("[last element]");
                break;
            }
        }
    }
    enumerator.Dispose();
}