表达式树lambda不得包含空传播运算符 [英] an expression tree lambda may not contain a null propagating operator

问题描述

问题:以下代码中的行price = co?.price ?? 0,给了我上面的错误.但是，如果我从co.?中删除?，则可以正常工作.我试图遵循此MSDN示例他们在select new { person.FirstName, PetName = subpet?.Name ?? String.Empty };行上使用?的位置因此，似乎我需要了解何时将?与??一起使用，何时不使用.

Question: The line price = co?.price ?? 0, in the following code gives me the above error. but if I remove ? from co.? it works fine. I was trying to follow this MSDN example where they are using ? on line select new { person.FirstName, PetName = subpet?.Name ?? String.Empty }; So, it seems I need to understand when to use ? with ?? and when not to.

错误:

表达式树lambda不得包含空传播运算符

an expression tree lambda may not contain a null propagating operator

public class CustomerOrdersModelView
{
    public string CustomerID { get; set; }
    public int FY { get; set; }
    public float? price { get; set; }
    ....
    ....
}
public async Task<IActionResult> ProductAnnualReport(string rpt)
{
    var qry = from c in _context.Customers
              join ord in _context.Orders
                on c.CustomerID equals ord.CustomerID into co
              from m in co.DefaultIfEmpty()
              select new CustomerOrdersModelView
              {
                  CustomerID = c.CustomerID,
                  FY = c.FY,
                  price = co?.price ?? 0,
                  ....
                  ....
              };
    ....
    ....
 }

推荐答案

您引用的示例使用LINQ to Objects，其中查询中的隐式lambda表达式转换为 delegates ...而您使用的是EF或类似版本，并带有IQueryable<T>查询，其中lambda表达式将转换为表达式树.表达式树不支持空条件运算符(或元组).

The example you were quoting from uses LINQ to Objects, where the implicit lambda expressions in the query are converted into delegates... whereas you're using EF or similar, with IQueryable<T> queryies, where the lambda expressions are converted into expression trees. Expression trees don't support the null conditional operator (or tuples).

只需按照旧方法即可:

price = co == null ? 0 : (co.price ?? 0)

(我相信在表达式树中可以使用null-coalescing运算符.)

(I believe the null-coalescing operator is fine in an expression tree.)

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