下面是更通用的解决方案。它本质上做相同的事情(以O(N)顺序)，但对任何IEnumerable类型，并且可以与属性选择器可以返回null的类型混合。

public static class LinqExtensions
{
    public static T MinBy<T>(this IEnumerable<T> source, Func<T, IComparable> selector)
    {
        if (source == null)
        {
            throw new ArgumentNullException(nameof(source));
        }
        if (selector == null)
        {
            throw new ArgumentNullException(nameof(selector));
        }

        return source.Aggregate((min, cur) =>
        {
            if (min == null)
            {
                return cur;
            }

            var minComparer = selector(min);

            if (minComparer == null)
            {
                return cur;
            }

            var curComparer = selector(cur);

            if (curComparer == null)
            {
                return min;
            }

            return minComparer.CompareTo(curComparer) > 0 ? cur : min;
        });
    }
}

测试:

var nullableInts = new int?[] {5, null, 1, 4, 0, 3, null, 1};
Assert.AreEqual(0, nullableInts.MinBy(i => i));//should pass

不幸的是，没有内置的方法来做到这一点，但它很容易为自己实现。以下是它的核心内容:

public static TSource MinBy<TSource, TKey>(this IEnumerable<TSource> source,
    Func<TSource, TKey> selector)
{
    return source.MinBy(selector, null);
}

public static TSource MinBy<TSource, TKey>(this IEnumerable<TSource> source,
    Func<TSource, TKey> selector, IComparer<TKey> comparer)
{
    if (source == null) throw new ArgumentNullException("source");
    if (selector == null) throw new ArgumentNullException("selector");
    comparer ??= Comparer<TKey>.Default;

    using (var sourceIterator = source.GetEnumerator())
    {
        if (!sourceIterator.MoveNext())
        {
            throw new InvalidOperationException("Sequence contains no elements");
        }
        var min = sourceIterator.Current;
        var minKey = selector(min);
        while (sourceIterator.MoveNext())
        {
            var candidate = sourceIterator.Current;
            var candidateProjected = selector(candidate);
            if (comparer.Compare(candidateProjected, minKey) < 0)
            {
                min = candidate;
                minKey = candidateProjected;
            }
        }
        return min;
    }
}

使用示例:

var firstBorn = People.MinBy(p => p.DateOfBirth ?? DateTime.MaxValue);

请注意，如果序列为空，将抛出异常，如果有多个元素，则返回第一个具有最小值的元素。

或者，你也可以在MinBy.cs中使用MoreLINQ中的实现。(当然，有一个相应的MaxBy。)

通过包管理器控制台安装:

PM>安装包morelinq

注意:我之所以包含这个答案是为了完整性，因为OP没有提到数据源是什么，我们不应该做任何假设。

这个查询给出了正确的答案，但可能会慢一些，因为它可能必须对People中的所有项进行排序，这取决于People是什么数据结构:

var oldest = People.OrderBy(p => p.DateOfBirth ?? DateTime.MaxValue).First();

UPDATE: Actually I shouldn't call this solution "naive", but the user does need to know what he is querying against. This solution's "slowness" depends on the underlying data. If this is a array or List<T>, then LINQ to Objects has no choice but to sort the entire collection first before selecting the first item. In this case it will be slower than the other solution suggested. However, if this is a LINQ to SQL table and DateOfBirth is an indexed column, then SQL Server will use the index instead of sorting all the rows. Other custom IEnumerable<T> implementations could also make use of indexes (see i4o: Indexed LINQ, or the object database db4o) and make this solution faster than Aggregate() or MaxBy()/MinBy() which need to iterate the whole collection once. In fact, LINQ to Objects could have (in theory) made special cases in OrderBy() for sorted collections like SortedList<T>, but it doesn't, as far as I know.

试试下面的方法:

var firstBornDate = People.GroupBy(p => p.DateOfBirth).Min(g => g.Key).FirstOrDefault();

这是一个获取最小值和最大值的简单方法:

    `dbcontext.tableName.Select(x=>x.Feild1).Min()`
    

public class Foo {
    public int bar;
    public int stuff;
};

void Main()
{
    List<Foo> fooList = new List<Foo>(){
    new Foo(){bar=1,stuff=2},
    new Foo(){bar=3,stuff=4},
    new Foo(){bar=2,stuff=3}};

    Foo result = fooList.Aggregate((u,v) => u.bar < v.bar ? u: v);
    result.Dump();
}