C# 使用Dapper映射多个复杂类型属性

我有以下简洁的问题：

var orderModels = db.Query<OrderModel>(@"
SELECT
    o.[Id],
    o.[CustomerId],

    o.[DeliveryAddress_FirstName],
    o.[DeliveryAddress_LastName],
    o.[DeliveryAddress_Line1],
    o.[DeliveryAddress_Line2],
    o.[DeliveryAddress_City],
    o.[DeliveryAddress_State],
    o.[DeliveryAddress_PostCode],
    o.[DeliveryAddress_Country],

    o.[BillingAddress_FirstName],
    o.[BillingAddress_LastName],
    o.[BillingAddress_Line1],
    o.[BillingAddress_Line2],
    o.[BillingAddress_City],
    o.[BillingAddress_State],
    o.[BillingAddress_PostCode],
    o.[BillingAddress_Country]
FROM
    [Order] o
");

我已经试着查过了，有一个问题。然而，当我不连接来自多个表的结果时，我不知道如何在我的用例中使用它

对于像Dapper这样的ORM来说，这是一个非常常见的场景，所以我确信我只是错过了一些明显的东西。如果你对整洁更了解，请帮忙

---

实现这一点的最佳方法是什么？

Dapper不关心连接，您只需要为它提供正确的

splitOn

选项字段，如：

    var orderModels = db.Query<OrderModel, AddressModel, AddressModel, OrderModel>(@"
    SELECT
        o.[Id],
        o.[CustomerId],

        o.[DeliveryAddress_FirstName] AS [FirstName], // Field names should match properties of your model
        o.[DeliveryAddress_LastName] AS [LastName],
        o.[DeliveryAddress_Line1] AS [Line1],
        o.[DeliveryAddress_Line2] AS [Line2],
        o.[DeliveryAddress_City] AS [City],
        o.[DeliveryAddress_State] AS [State],
        o.[DeliveryAddress_PostCode] AS [PostCode],
        o.[DeliveryAddress_Country] AS [Country],

        o.[BillingAddress_FirstName] AS [FirstName],
        o.[BillingAddress_LastName] AS [LastName],
        o.[BillingAddress_Line1] AS [Line1],
        o.[BillingAddress_Line2] AS [Line2],
        o.[BillingAddress_City] AS [City],
        o.[BillingAddress_State] AS [State],
        o.[BillingAddress_PostCode] AS [PostCode],
        o.[BillingAddress_Country] AS [Country]
    FROM
        [Order] o
    ", 
(order, deliveryAddress,  billingAddress) => {
   order.DeliveryAddress = deliveryAddress; 
   order.BillingAddress = billingAddress; 
   return order; 
},
splitOn: "FirstName,FirstName");

var orderModels=db.Query（@）
挑选
o、 [Id]，
o、 [客户ID]，
o、 [DeliveryAddress_FirstName]为[FirstName]，//字段名应与模型的属性匹配
o、 [DeliveryAddress\u LastName]作为[LastName]，
o、 [DeliveryAddress_Line1]作为[Line1]，
o、 [DeliveryAddress_Line2]作为[Line2]，
o、 [DeliveryAddress_City]作为[City]，
o、 [DeliveryAddress_State]作为[State]，
o、 [送货地址\邮政编码]为[邮政编码]，
o、 [DeliveryAddress_Country]作为[Country]，
o、 [BillingAddress_FirstName]作为[FirstName]，
o、 [BillingAddress\u LastName]作为[LastName]，
o、 [BillingAddress_Line1]作为[Line1]，
o、 [BillingAddress_Line2]作为[Line2]，
o、 [BillingAddress_City]作为[City]，
o、 [BillingAddress_State]作为[State]，
o、 [BillingAddress_PostCode]作为[PostCode]，
o、 [BillingAddress_Country]作为[Country]
从…起
[命令]o
", 
（订单、发货地址、账单地址）=>{
order.DeliveryAddress=交货地址；
order.BillingAddress=BillingAddress；
退货单；
},
斯普利顿：“名字，名字”）；

这一点在本文中进行了解释

---

此外，select的字段名必须与模型属性名匹配，以便Dapper自动计算映射。

感谢您的帮助。我认为您的类型参数有误，它可能应该是

。但即使我这样做了，账单地址和送货地址的所有属性仍然为空。是的，修复了参数order。。。我想问题可能是这些列的前缀仍然是

DeliveryAddress.

。。。我尝试为列添加别名以删除前缀，结果成功了。但是我不得不说，重命名所有这些列（并且有多个同名列）感觉有点奇怪。我真的不喜欢为了整洁而做特殊的投影。如果可以配置Dapper，那就更好了，因为映射结果是Dapper的职责，而且不会让我更改查询。如果这些结果是在存储过程中生成的呢？你可以看看，但我自己还没有尝试过。谢谢，我不认为自定义映射在这种情况下会有所帮助，但值得一试。无论如何，使用列别名的解决方案是有效的，即使它并不完美。

    var orderModels = db.Query<OrderModel, AddressModel, AddressModel, OrderModel>(@"
    SELECT
        o.[Id],
        o.[CustomerId],

        o.[DeliveryAddress_FirstName] AS [FirstName], // Field names should match properties of your model
        o.[DeliveryAddress_LastName] AS [LastName],
        o.[DeliveryAddress_Line1] AS [Line1],
        o.[DeliveryAddress_Line2] AS [Line2],
        o.[DeliveryAddress_City] AS [City],
        o.[DeliveryAddress_State] AS [State],
        o.[DeliveryAddress_PostCode] AS [PostCode],
        o.[DeliveryAddress_Country] AS [Country],

        o.[BillingAddress_FirstName] AS [FirstName],
        o.[BillingAddress_LastName] AS [LastName],
        o.[BillingAddress_Line1] AS [Line1],
        o.[BillingAddress_Line2] AS [Line2],
        o.[BillingAddress_City] AS [City],
        o.[BillingAddress_State] AS [State],
        o.[BillingAddress_PostCode] AS [PostCode],
        o.[BillingAddress_Country] AS [Country]
    FROM
        [Order] o
    ", 
(order, deliveryAddress,  billingAddress) => {
   order.DeliveryAddress = deliveryAddress; 
   order.BillingAddress = billingAddress; 
   return order; 
},
splitOn: "FirstName,FirstName");