'How to use Automapper to project object to Dto in LINQ
I am working on .NET Core 6 along with EF Core. I want to convert Customer
and Order
object that I have inside LINQ select
to CustomerDto
and OrderDto
using ProjectTo
.
Using libraries
using AutoMapper;
using AutoMapper.QueryableExtensions;
I am aware how to map IQueryable
using the code shown here:
var x = (from customer in db.Customers
where customer.CustomerId == CustomerId
select customer).ProjectTo<CustomerDto>(_mapper.ConfigurationProvider);
but I'm not sure how to project inside new or if it should be done differently? Below is LINQ code that I want customer to CustomerDto
& Order
to OrderDto
var customerOrdersQuery =
(from customer in db.Customers.Where(x => x.CustomerId == CustomerId)
join orders in db.Orders on customer.CustomerId equals orders.CustomerId into cst_Ord
from customerOrders in cst_Ord.DefaultIfEmpty()
select new {
customer, //ProjectTo<CustomerDto>
customerOrders //ProjectTo<OrderDto>
}).AsEnumerable();
Solution 1:[1]
Afaik you need an intermediary class CustomerWithOrdersSource which you map to CustomerWithOrdersDto, using Automapper's ProjectTo.
Mapping
public class CustomerWithOrdersSource
{
public Customer Customer {get;set;}
public ICollection<Order> Orders {get;set;}
}
public class CustomerWithOrdersDto
{
public CustomerDto Customer {get;set;}
public ICollection<OrderDto> Orders {get;set;}
}
CreateMap<CustomerWithOrdersSource, CustomerWithOrdersDto>();
Query
var customerOrdersQuery =
(from customer in db.Customers.Where(x => x.CustomerId == CustomerId)
join orders in db.Orders on customer.CustomerId equals orders.CustomerId into cst_Ord
from customerOrders in cst_Ord.DefaultIfEmpty()
select new CustomerWithOrdersSource() {
Customer = customer,
Orders = customerOrders
}).ProjectTo<CustomerWithOrdersDto>(_mapper.ConfigurationProvider);
Edit: According to @Lucian Bargaoanu you can project directly from anonymous object as well. But this way you could easily customize the mapping by editing the mapping profile, if necessary later on.
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
Solution | Source |
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Solution 1 |