c# - QueryOver 中的 SelectList

Question

我在通过 QueryOver 选择时遇到一些问题，这是我的模型类:

public class BaseModel
{
    public virtual int Id { get; set; }
    public virtual DateTime ModifyDate { get; set; }
    public virtual DateTime CreateDate { get; set; }
    public virtual User CreateUser { get; set; }
    public virtual User ModifyUser { get; set; }
    public virtual bool IsActive { get; set; }
}

public class BaseClassMap<T> : ClassMap<T> where T : BaseModel
{
    public BaseClassMap()
    {
        Id(x => x.Id).GeneratedBy.Identity();
        References(x => x.CreateUser, "CreateUser_Id");
        References(x => x.ModifyUser, "ModifyUser_Id");
        Map(x => x.CreateDate).Not.Nullable().Default("GETDATE()");
        Map(x => x.ModifyDate).Not.Nullable().Default("GETDATE()"); ;
        Map(x => x.IsActive).Not.Nullable().Default("1");
    }
}

public class Expenditure : BaseModel
{
    public virtual decimal Amount { get; set; }
    public virtual Category Category { get; set; }
    public virtual string Comment { get; set; }
    public virtual DateTime Month { get; set; }
}

public class ExpenditureMap : BaseClassMap<Expenditure>
{
    public ExpenditureMap()
    {
        Schema("[budget]");
        Table("[Expenditure]");
        Id(x => x.Id);
        Map(x => x.Amount).Not.Nullable(); 
        Map(x => x.Comment);
        Map(x => x.Month).Not.Nullable(); 
        References(x => x.Category, "Category_Id").Not.Nullable();
    }
}

public class User : BaseModel
{
    public virtual string Login { get; set; }
    public virtual string Email { get; set; }
    public virtual string Password { get; set; }
    public virtual string PasswordSalt { get; set; }
    public virtual string HashAlgorithm { get; set; }
}

public class UserMap : BaseClassMap<User>
{
    public UserMap()
    {
        Schema("[permission]");
        Table("[User]");
        Id(x => x.Id);
        Map(x => x.Login).Not.Nullable(); 
        Map(x => x.Email).Not.Nullable();
        Map(x => x.Password).Not.Nullable();
        Map(x => x.PasswordSalt);
        Map(x => x.HashAlgorithm).Not.Nullable();
    }
}

public class ExpenseDetailsItemVm
{
    public int Id { get; set; }
    public string Comment { get; set; }
    public decimal Amount { get; set; }
    public DateTime CreateDate { get; set; }
    public string CreateUser { get; set; }
}

这是我的查询:

var user = new User();
var result = new ExpenseDetailsItemVm();
var expenses = SessionManager.Session.QueryOver<Expenditure>()
                             .JoinAlias(x => x.CreateUser, () => user)
                             .Where(x => x.IsActive &&  
                                         x.Category.Id == categoryId && 
                                         x.Month == date.ToFirstDayOfMonth())
                             .SelectList(list => list
                                              .Select(x => x.Id).WithAlias(() => result.Id)
                                              .Select(x => x.Amount).WithAlias(() => result.Amount)
                                              .Select(x => x.Comment).WithAlias(() => result.Comment)
                                              .Select(x => x.CreateDate).WithAlias(() => result.CreateDate)
                                              .Select(()=> user.Login).WithAlias(() => result.CreateUser)
            ).TransformUsing(Transformers.AliasToBean<ExpenseDetailsItemVm>())
            .List<ExpenseDetailsItemVm>();

我收到这个错误:

Object reference not set to an instance of an object.

我确定问题出在这行代码上:

 .Select(()=> user.Login).WithAlias(() => result.CreateUser)

如果我注释掉这一行，查询工作正常并返回结果没有错误。

为了消除映射错误，我尝试了这个:

var u = SessionManager.Session.QueryOver<User>().Where(x => x.Id == 1).
            SelectList(list => list.
                Select(x => x.Login).WithAlias(() => result.CreateUser))
            .TransformUsing(Transformers.AliasToBean<ExpenseDetailsItemVm>())
            .List<ExpenseDetailsItemVm>();

这也很好用。

在我的Expenditure 表中，列CreateUser_Id 各处都有一个值

Answer 1

尝试将用作别名的对象作为null:

User user = null; 
ExpenseDetailsItemVm result = null;

因为它们仅用于创建具有强类型语法的 SQL。