如何使用 LINQ 显示列表中的项目，即使其中某些项目具有空值？ [重复]

Question

我正在尝试编写一个 LINQ 查询来显示与用户相关的地址。

但我也想显示引用无效用户的地址。在这种情况下，与用户相关的字段应设置为空。 词汇表：“无效用户”：在用户列表中找不到。

这是我的代码：

    class Program
    {
        public static void Main(string[] args)
        {
            var users = GetUsers();
            var addresses = GetAddresses();

            var result = from address in addresses
                         from user in users
                         where user.Id == address.UserId
                         select new
                         {
                             address.Street,
                             address.Zipcode,
                             Id = address.UserId,
                             Name = user.Name,
                             Email = user.Email,
                         };
            foreach(var item in result)
            {
                Console.WriteLine($"Street: {item.Street}");
                Console.WriteLine($"Zipcode: {item.Zipcode}");
                Console.WriteLine($"Id: {item.Id}");
                Console.WriteLine($"Name: {item.Name}");
                Console.WriteLine($"Email: {item.Email}");
            }
        }

        static List<User> GetUsers()
        {
            const string usersData = "[{\"id\":\"e135e5f9-64ed-4176-a543-ac7f09c15600\",\"name\":\"John Doe\",\"email\":\"[email protected]\"},{\"id\":\"d906db73-f711-4498-88b9-dc2d9d54ece5\",\"name\":\"James Scott\",\"email\":\"[email protected]\"}]";

            return JsonSerializer.Deserialize<List<User>>(usersData, new JsonSerializerOptions { PropertyNameCaseInsensitive = true })!;
        }

        static List<Address> GetAddresses()
        {
            const string addressData = "[{\"street\":\"206 Hazel St\",\"zipcode\":\"16853\",\"city\":\"Milesburg\",\"state\":\"PA\", \"country\": \"USA\",\"userId\":\"e135e5f9-64ed-4176-a543-ac7f09c15600\"},{\"street\":\"208 Mill St\",\"zipcode\":\"16853\",\"city\":\"Milesburg\",\"state\":\"PA\", \"country\": \"USA\",\"userId\":\"d906db73-f711-4498-88b9-dc2d9d54ece5\"},{\"street\":\"100 Turnpike St\",\"zipcode\":\"16853\",\"city\":\"Milesburg\",\"state\":\"PA\", \"country\": \"USA\",\"userId\":\"b84b9728-b657-4db9-8af7-45789983d931\"}]";

            return JsonSerializer.Deserialize<List<Address>>(addressData, new JsonSerializerOptions { PropertyNameCaseInsensitive = true })!;
        }
    }

    internal class Address
    {
        public string Street { get; set; }
        public string Zipcode { get; set; }
        public string City { get; set; }
        public string State { get; set; }
        public string Country { get; set; }
        public Guid UserId { get; set; }
    }

    internal class User
    {
        public Guid Id { get; set; }
        public string Name { get; set; }
        public string Email { get; set; }
    }

    
}

使用上面的代码，与无效用户相关的地址将被忽略。

我期待收到这样的列表：

  "Street": "206 Hazel St"
  "Zipcode": "16853",
  "Id": "e135e5f9-64ed-4176-a543-ac7f09c15600",
  "Name": "John Doe",
  "Email": "[email protected]"
},
{
  "Street": "208 Mill St"
  "Zipcode": "16853",
  "Id": "d906db73-f711-4498-88b9-dc2d9d54ece5",
  "Name": "James Scott",
  "Email": "[email protected]"
},
{
  "Street": "100 Turnpike St"
  "Zipcode": "16853",
  "Id": null,
  "Name": null,
  "Email": null
}]

Answer 1

首先，让我们更改您的查询以使用

join

:

var result =
    from address in addresses
    join user in users on address.UserId equals user.Id
    select new
    {
        address.Street,
        address.Zipcode,
        Id = address.UserId,
        Name = user.Name,
        Email = user.Email,
    };

我将

address.UserId

替换为

user.Id

。

现在我们可以将其更改为组

join

:

var result =
    from address in addresses
    join user in users on address.UserId equals user.Id into grouped_users
    let maybe_single_user = grouped_users.SingleOrDefault()
    select new
    {
        address.Street,
        address.Zipcode,
        Id = maybe_single_user?.Id,
        Name = maybe_single_user?.Name,
        Email = maybe_single_user?.Email,
    };

grouped_users.SingleOrDefault()

为我们提供了用户或

null

。然后使用 Elvis 运算符将

null

传播到输出中。

让我们精心设计

output

而不是抛弃一系列

Console.WriteLine

语句。

var outputs =
    result
        .Select(item =>
            String
                .Join(
                    Environment.NewLine,
                    new[]
                    {
                        $"Street: {item.Street}",
                        $"Zipcode: {item.Zipcode}",
                        $"Id: {item.Id}",
                        $"Name: {item.Name}",
                        $"Email: {item.Email}",
                    }));

Console.WriteLine(String.Join(Environment.NewLine + Environment.NewLine, outputs));

这给了我们：

Street: 206 Hazel St
Zipcode: 16853
Id: e135e5f9-64ed-4176-a543-ac7f09c15600
Name: John Doe
Email: [email protected]

Street: 208 Mill St
Zipcode: 16853
Id: d906db73-f711-4498-88b9-dc2d9d54ece5
Name: James Scott
Email: [email protected]

Street: 100 Turnpike St
Zipcode: 16853
Id:
Name:
Email: