使用嵌套集合属性动态生成 LINQ 选择

Question

问题和Dynamically generate LINQ select with nested properties很相似

public static Expression<Func<TSource, TTarget>> BuildSelector<TSource, TTarget>(string members) =>
    BuildSelector<TSource, TTarget>(members.Split(',').Select(m => m.Trim()));

public static Expression<Func<TSource, TTarget>> BuildSelector<TSource, TTarget>(IEnumerable<string> members)
{
    var parameter = Expression.Parameter(typeof(TSource), "e");
    var body = NewObject(typeof(TTarget), parameter, members.Select(m => m.Split('.')));
    return Expression.Lambda<Func<TSource, TTarget>>(body, parameter);
}

static Expression NewObject(Type targetType, Expression source, IEnumerable<string[]> memberPaths, int depth = 0)
{
    var bindings = new List<MemberBinding>();
    var target = Expression.Constant(null, targetType);
    foreach (var memberGroup in memberPaths.GroupBy(path => path[depth]))
    {
        var memberName = memberGroup.Key;
        var targetMember = Expression.PropertyOrField(target, memberName);
        var sourceMember = Expression.PropertyOrField(source, memberName);
        var childMembers = memberGroup.Where(path => depth + 1 < path.Length);
        var targetValue = !childMembers.Any() ? sourceMember :
            NewObject(targetMember.Type, sourceMember, childMembers, depth + 1);
        bindings.Add(Expression.Bind(targetMember.Member, targetValue));
    }
    return Expression.MemberInit(Expression.New(targetType), bindings);
}

Ivan Stoev 提供的不错的通用解决方案工作正常，但问题是它不支持 collection 属性。

例如，source.Property1.Property2 - 如果 Property1 是用户集合，则代码不起作用，因为 Property2 不是集合的属性，但属于 Property1 类型。

class Shipment {
   // other fields...

   public int Id;
   public Address Sender;
   public List<Address> Recipients;
}

class Address {
    public string AddressText;
    public string CityName;
    public string CityId;
}

上面的类和代码我可以查询

var test = BuildSelector<Shipment, Shipment>(
    "Id, Sender.CityId, Sender.CityName");

但如果我只想获取收件人的城市名称，我不能执行以下操作（因为收件人是集合）：

var test = BuildSelector<Shipment, Shipment>(
    "Recipients.CityName");

我是 C# 表达式的新手，不知道如何改进上述解决方案以使其适用于集合属性。

Answer 1

这是解决方案。现在它处理集合。

public static class ExpressionHelpers
{
    public static Expression<Func<TSource, TTarget>> BuildSelector<TSource, TTarget>(string members) =>
        BuildSelector<TSource, TTarget>(members.Split(',').Select(m => m.Trim()));

    public static Expression<Func<TSource, TTarget>> BuildSelector<TSource, TTarget>(IEnumerable<string> members)
    {
        var parameter = Expression.Parameter(typeof(TSource), "e");
        var body = NewObject(typeof(TTarget), parameter, members.Select(m => m.Split('.')));
        return Expression.Lambda<Func<TSource, TTarget>>(body, parameter);
    }

    static Expression NewObject(Type targetType, Expression source, IEnumerable<string[]> memberPaths, int depth = 0)
    {
        var bindings = new List<MemberBinding>();
        var target = Expression.Constant(null, targetType);
        foreach (var memberGroup in memberPaths.GroupBy(path => path[depth]))
        {
            var memberName = memberGroup.Key;
            var targetMember = Expression.PropertyOrField(target, memberName);
            var sourceMember = Expression.PropertyOrField(source, memberName);
            var childMembers = memberGroup.Where(path => depth + 1 < path.Length).ToList();

            Expression targetValue = null;
            if (!childMembers.Any())
                targetValue = sourceMember;
            else
            {
                if (IsEnumerableType(targetMember.Type, out var sourceElementType) &&
                    IsEnumerableType(targetMember.Type, out var targetElementType))
                {
                    var sourceElementParam = Expression.Parameter(sourceElementType, "e");
                    targetValue = NewObject(targetElementType, sourceElementParam, childMembers, depth + 1);
                    targetValue = Expression.Call(typeof(Enumerable), nameof(Enumerable.Select),
                        new[] { sourceElementType, targetElementType }, sourceMember,
                        Expression.Lambda(targetValue, sourceElementParam));

                    targetValue = CorrectEnumerableResult(targetValue, targetElementType, targetMember.Type);
                }
                else
                {
                    targetValue = NewObject(targetMember.Type, sourceMember, childMembers, depth + 1);
                }
            }

            bindings.Add(Expression.Bind(targetMember.Member, targetValue));
        }
        return Expression.MemberInit(Expression.New(targetType), bindings);
    }

    static bool IsEnumerableType(Type type, out Type elementType)
    {
        foreach (var intf in type.GetInterfaces())
        {
            if (intf.IsGenericType && intf.GetGenericTypeDefinition() == typeof(IEnumerable<>))
            {
                elementType = intf.GetGenericArguments()[0];
                return true;
            }
        }

        elementType = null;
        return false;
    }

    static bool IsSameCollectionType(Type type, Type genericType, Type elementType)
    {
        var result = genericType.MakeGenericType(elementType).IsAssignableFrom(type);
        return result;
    }

    static Expression CorrectEnumerableResult(Expression enumerable, Type elementType, Type memberType)
    {
        if (memberType == enumerable.Type)
            return enumerable;

        if (memberType.IsArray)
            return Expression.Call(typeof(Enumerable), nameof(Enumerable.ToArray), new[] { elementType }, enumerable);

        if (IsSameCollectionType(memberType, typeof(List<>), elementType)
            || IsSameCollectionType(memberType, typeof(ICollection<>), elementType)
            || IsSameCollectionType(memberType, typeof(IReadOnlyList<>), elementType)
            || IsSameCollectionType(memberType, typeof(IReadOnlyCollection<>), elementType))
            return Expression.Call(typeof(Enumerable), nameof(Enumerable.ToList), new[] { elementType }, enumerable);

        throw new NotImplementedException($"Not implemented transformation for type '{memberType.Name}'");
    }
}