我是linq的新手,无法编写两个简单的查询。出于某种原因,我无法绕过它。
它的结构简单:Order有OrderItems。每个orderItem都有一个productID。
我想:
获取订购productId 3的所有订单
获取在同一订单上订购productId 4和5的所有订单。
我已经尝试了很多方法。这两个查询位于小测试应用程序的底部。
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace test
{
class Program
{
static void Main(string[] args)
{
OrderService svc = new OrderService();
//find all orders that purchased ProductID 3
IEnumerable<Order> data = svc.GetOrdersWithProduct(3);
//find all orders that purchase product 4 AND 5
IEnumerable<Order> data2 = svc.GetOrdersWithProduct(new int[] { 4, 5} );
}
}
public class Order
{
public int OrderId { get; set; }
public IEnumerable<OrderItem> Items { get; set; }
}
public class OrderItem
{
public int OrderItemId { get; set; }
public int OrderId { get; set; }
public int ProductId { get; set; }
}
public class OrderService
{
private static List<Order> GetTestData()
{
List<Order> orders = new List<Order>();
//5 Orders, 3 items each (every orderitem has a unique product in this test set)
int orderitemid = 1;
int productid = 1;
for (int orderid = 1; orderid < 6; orderid++)
{
orders.Add(new Order
{
OrderId = orderid,
Items = new List<OrderItem>
{
new OrderItem() { OrderId = orderid, OrderItemId = orderitemid++, ProductId = productid ++ },
new OrderItem() { OrderId = orderid, OrderItemId = orderitemid++, ProductId = productid ++ },
new OrderItem() { OrderId = orderid, OrderItemId = orderitemid++, ProductId = productid ++ }
}
});
}
return orders;
}
public IEnumerable<Order> GetOrdersWithProduct(int productId)
{
List<Order> orders = OrderService.GetTestData();
// ?? not really what i want, since this returns only if all items are the same
var result = orders.Where(o => o.Items.All(i => i.ProductId == productId));
return result.ToList();
}
public IEnumerable<Order> GetOrdersWithProduct(IEnumerable<int> productIds)
{
List<Order> orders = OrderService.GetTestData();
//??
var result = orders.Where(o => o.Items.All(productIds.Contains(i => i.ProductId)));
return result.ToList();
}
}
}
答案 0 :(得分:6)
我会这样做:
获取订购productId 3的所有订单
var result = orders.Where(o => o.Items.Any(item => item.ProductId == 3));
获取订购productId 4和5的所有订单
var result = orders.Where(o => o.Items.Any(item => item.ProductId == 4))
.Where(o => o.Items.Any(item => item.ProductId == 5));
或者:
public static IEnumerable<Order> GetOrdersWithProduct(int id)
{
return orders.Where(o => o.Items.Any(item => item.ProductId == productId));
}
然后:
var result1 = GetOrdersWithProduct(3);
var result2 = GetOrdersWithProduct(4).Intersect(GetOrdersWithProduct(5));
另一种选择:
public static IEnumerable<Order> GetOrdersWithProducts(params int[] ids)
{
return GetOrdersWithProducts((IEnumerable<int>) ids);
}
public static IEnumerable<Order> GetOrdersWithProducts(IEnumerable<int> ids)
{
return orders.Where(o => !ids.Except(o.Items.Select(p => p.ProductId))
.Any());
}
var result1 = GetOrdersWithProducts(3);
var result2 = GetOrdersWithProduct(4, 5);
答案 1 :(得分:1)
没有Linq“语言集成查询”语法:
public IEnumerable<Order> GetOrdersWithProduct(int productId)
{
List<Order> orders = OrderService.GetTestData();
var result = orders.Where(o => o.Items.Any(i => i.ProductId == productId));
return result.ToList();
}
public IEnumerable<Order> GetOrdersWithProduct(IEnumerable<int> productIds)
{
List<Order> orders = OrderService.GetTestData();
var result = orders.Where(o => productIds.All(id => o.Items.Any(i => i.ProductId == id)));
return result.ToList();
}
似乎The Lame Duck正在进行“语言集成查询”版本,所以我不会这样做。
答案 2 :(得分:0)
检查这些:
1
from order in orders
where order.Items.Exists(item => item.OrderItemId == 3)
select order
非常相似:
从订单中订购
其中order.Items.Exists(item =&gt; item.OrderItemId == 3)&amp;&amp;
order.Items.Exists(item =&gt; item.OrderItemId == 4)
选择订单
答案 3 :(得分:0)
public IEnumerable<Order> GetOrdersWithProduct( int productId )
{
List<Order> orders = OrderService.GetTestData( );
// ?? not really what i want, since this returns only if all items are the same
//var result = orders.Where( o => o.Items.All( i => i.ProductId == productId ) );
var result = orders.Where(o => o.Items.Any(ii => ii.ProductId == productId));
return result.ToList( );
}