自己的解决方案：

Question

我有一个客户端模型，它有许多属性，如名字，姓氏，地址......

通过ClientEditViewModel，我可以更改TextBoxes中的属性值。

public class ClientEditViewModel : EditableViewModelBase
{
    public int ClientID
    {
        get { return this.currentClient.ClientID; }
        set { this.SetProperty(newValue => this.currentClient.ClientID = newValue, value); }
    }

    public string Title
    {
        get { return this.currentClient.Title; }
        set { this.SetProperty(newValue => this.currentClient.Title = newValue, value); }
    }

    public string FirstName
    {
        get { return this.currentClient.FirstName; }
        set { this.SetProperty(newValue => this.currentClient.FirstName = newValue, value); }
    }

    public string LastName
    {
        get { return this.currentClient.LastName; }
        set { this.SetProperty(newValue => this.currentClient.LastName = newValue, value); }
    }

    ...
}

当用户按下搜索按钮时，我想遍历所有属性。如果属性不为空或为null，我想将它们添加到具有“LIKE”限制的查询中。我没有手动检查每个属性，而是希望通过Reflection迭代所有属性。

public ICriterion GetSearchClientCriteria()
{
    var conjunction = Restrictions.Conjunction();

    if (this.ClientID != 0)
    {
        conjunction.Add(Restrictions.Where<Client>(x => x.ClientID == this.ClientID));
        return conjunction;
    }

    foreach (PropertyInfo propertyInfo in this.PropertyInfosWithPublicGetterAndSetter)
    {
        if (propertyInfo.PropertyType == typeof(string))
        {
            string currentPropertyValue = propertyInfo.GetValue(this) as string;

            if (!string.IsNullOrEmpty(currentPropertyValue))
            {
                /* This statement can be executed */
                // conjunction.Add(Restrictions.Where<Client>(c => c.FirstName.IsLike(this.FirstName, MatchMode.Anywhere)));
                conjunction.Add(Restrictions.Where<Client>(c => c.GetType().GetProperty(propertyInfo.Name).GetValue(c, null).ToString()
                .IsLike(this.GetType().GetProperty(propertyInfo.Name).GetValue(this).ToString())));

                return conjunction;
            }
        }
    }

    return conjunction;
}

不幸的是，当我在以下代码中使用此连接时，出现错误。如何在不手动检查每个属性的情况下迭代所有属性？

public class NHibernateRepository : IRepository
{
    public ICollection<T> GetByCriteria<T>(ICriterion criterion) where T : class
    {
        using (var session = this.sessionManager.OpenSession())
        {
            return session.QueryOver<T>().Where(criterion).List();
        }
    }
}

System.InvalidOperationException：Auf die变量“c”vom Typ   “Rechnungsprogramm.Model.Client”wird vom Bereich“”verwiesen，sie ist jedoch   nicht definiert。


System.InvalidOperationException：变量'c'     引用的类型'Rechnungsprogramm.Model.Client'     来自范围''，但未定义

自己的解决方案：

不是最漂亮的解决方案，但它确实有效。

private ICriterion GetClientSearchCriterion()
{
    Conjunction conjunction = Restrictions.Conjunction();

    if (this.CurrentClientDetailViewModel.ClientId != 0)
    {
        conjunction.Add(Restrictions.Where<Client>(c => c.ClientId == this.CurrentClientDetailViewModel.ClientId));
        return conjunction;
    }

    if (!string.IsNullOrEmpty(this.CurrentClientDetailViewModel.Title))
    {
        conjunction.Add(Restrictions.Where<Client>(c => c.Title.IsLike(this.CurrentClientDetailViewModel.Title, MatchMode.Anywhere)));
    }
    if (!string.IsNullOrEmpty(this.CurrentClientDetailViewModel.FirstName))
    {
        conjunction.Add(Restrictions.Where<Client>(c => c.FirstName.IsLike(this.CurrentClientDetailViewModel.FirstName, MatchMode.Anywhere)));
    }
    if (!string.IsNullOrEmpty(this.CurrentClientDetailViewModel.LastName))
    {
        conjunction.Add(Restrictions.Where<Client>(c => c.LastName.IsLike(this.CurrentClientDetailViewModel.LastName, MatchMode.Anywhere)));
    }
    if (!string.IsNullOrEmpty(this.CurrentClientDetailViewModel.Street))
    {
        conjunction.Add(Restrictions.Where<Client>(c => c.Street.IsLike(this.CurrentClientDetailViewModel.Street, MatchMode.Anywhere)));
    }
    if (!string.IsNullOrEmpty(this.CurrentClientDetailViewModel.HouseNumber))
    {
        conjunction.Add(Restrictions.Where<Client>(c => c.HouseNumber.IsLike(this.CurrentClientDetailViewModel.HouseNumber, MatchMode.Anywhere)));
    }
    if (!string.IsNullOrEmpty(this.CurrentClientDetailViewModel.PostalCode))
    {
        conjunction.Add(Restrictions.Where<Client>(c => c.PostalCode.IsLike(this.CurrentClientDetailViewModel.PostalCode, MatchMode.Anywhere)));
    }
    if (!string.IsNullOrEmpty(this.CurrentClientDetailViewModel.City))
    {
        conjunction.Add(Restrictions.Where<Client>(c => c.City.IsLike(this.CurrentClientDetailViewModel.City, MatchMode.Anywhere)));
    }
    if (!string.IsNullOrEmpty(this.CurrentClientDetailViewModel.DateOfBirth))
    {
        conjunction.Add(Restrictions.Where<Client>(c => c.DateOfBirth.IsLike(this.CurrentClientDetailViewModel.DateOfBirth, MatchMode.Anywhere)));
    }
    if (!string.IsNullOrEmpty(this.CurrentClientDetailViewModel.PhoneNumber1))
    {
        conjunction.Add(Restrictions.Where<Client>(c => c.PhoneNumber1.IsLike(this.CurrentClientDetailViewModel.PhoneNumber1, MatchMode.Anywhere)));
    }
    if (!string.IsNullOrEmpty(this.CurrentClientDetailViewModel.PhoneNumber2))
    {
        conjunction.Add(Restrictions.Where<Client>(c => c.PhoneNumber2.IsLike(this.CurrentClientDetailViewModel.PhoneNumber2, MatchMode.Anywhere)));
    }
    if (!string.IsNullOrEmpty(this.CurrentClientDetailViewModel.MobileNumber))
    {
        conjunction.Add(Restrictions.Where<Client>(c => c.MobileNumber.IsLike(this.CurrentClientDetailViewModel.MobileNumber, MatchMode.Anywhere)));
    }
    if (!string.IsNullOrEmpty(this.CurrentClientDetailViewModel.Telefax))
    {
        conjunction.Add(Restrictions.Where<Client>(c => c.Telefax.IsLike(this.CurrentClientDetailViewModel.Telefax, MatchMode.Anywhere)));
    }
    if (!string.IsNullOrEmpty(this.CurrentClientDetailViewModel.Email))
    {
        conjunction.Add(Restrictions.Where<Client>(c => c.Email.IsLike(this.CurrentClientDetailViewModel.Email, MatchMode.Anywhere)));
    }

    return conjunction;
}

Answer 1

你不能把任何你喜欢的东西放到lambda中，必须由ORM翻译成SQL 您的lambda必须能够转换为SQL，它与GetType，GetProperty，GetValue等等无关。

您可以尝试动态构建Expression，而不是使用lambda，有点像question个答案。

通过NHInnate会话的PropertyInfo迭代属性来创建ICriterion

自己的解决方案：

1 个答案: