具有安全强制转换和空值验证的 Linq

Question

给定代码：

from i in this.GridViewFoo.SelectedItems
select new EmployeeEntity
{
    EmployeeID = (i as EmployeeDto).EmployeeID,
    Email = this.GetAllEmail((i as EmployeeDto).Email, (i as EmployeeDto).SecondaryEmails),
    EmployeeNumber = (i as EmployeeDto).EmployeeNumber,
    FirstName = (i as EmployeeDto).FirstName,
    LastName = (i as EmployeeDto).LastName
}

在安全转换(i as EmployeeDto) 之后，我可能会收到 NullReferenceException。我怎样才能确保代码的安全性，又不会因为大量的 null 检查而使他超负荷？

解决方案概述：

我做了一些测试来判断解决方案是否有效。两者都运行良好并带来相同的结果，您可以查看HERE。之后我用OfTypeSolution 和letSolution 做了一些性能测试。

由于 OfType 解决方案的平均时间更好，这就是答案！

Answer 1

你可以在Select之前使用OfType：

from i in this.GridViewFoo.SelectedItems.OfType<EmployeeDto>()
select new EmployeeEntity
{
    EmployeeID = i.EmployeeID,
    Email = this.GetAllEmail(i.Email, i.SecondaryEmails),
    EmployeeNumber = i.EmployeeNumber,
    FirstName = i.FirstName,
    LastName = i.LastName
}

它只会为您提供来自 SelectedItems 的 EmployeeDto 类型项目，因此无需强制转换和空值检查。

Answer 2

from si in this.GridViewFoo.SelectedItems
let i = si as EmployeeDto
where i != null
select new EmployeeEntity
{
    EmployeeID = i.EmployeeID,
    Email = this.GetAllEmail(i.Email, i.SecondaryEmails),
    EmployeeNumber = i.EmployeeNumber,
    FirstName = i.FirstName,
    LastName = i.LastName
}