Question

我有一个“Estate”实体，这个实体有一个集合“EstateFeatures”（类型：EstateFeature），而EstateFeature有一个属性“MyFeatureValue”。

注意：这些是问题的有限属性。所有实体都有Id和所有necesarry等

遗产

IList<EstateFeature> EstateFeatures;

EstateFeature

FeatureValue MyFeatureValue;

FeatureValue

public virtual long Id;

我正在尝试获得具有给定FeatureValue.Id

的Real Estates

DetachedCriteria query = DetachedCriteria.For<Estate>();
Conjunction and = new Conjuction();
foreach (var id in idCollection)
   and.Add(Expression.Eq("MyFeatureValue.Id",id);

query
     .CreateCriteria("EstateFeatures")
     .Add(and);
IList<Estate> estates = query.GetExecutableCriteria(session).List<Estate>();

此查询没有返回任何内容，我做错了什么？

由于

Answer 1

如果我理解正确，我认为这样的事可能有效

CreateCriteria(typeof(Estate))
     .CreateAlias("EstateFeatures", "estatefeature")
     .Add(Restrictions.In("estatefeature.MyFeatureValue.Id", ids))
     .List<Estate>();

Answer 2

NHibernate为您生成了什么查询？您可以使用show_sql配置属性来检查这一点。

在我看到您的查询时，您正在尝试获取具有一组给定功能的所有Estates。我想，这会产生一个看起来像

的查询

SELECT ....
FROM Estates
INNER JOIN Features
WHERE Feature.Id = 1 AND Feature.Id = 2 ...

如果要检索包含所有指定功能的所有属性，我认为您必须使用Disjunction，以便NHibernate检索具有至少一个这些功能的所有Estates。然后，在您的客户端代码中，您将检查“客户代码”中的每个Estate，以便最终最终获得具有所有功能的Estates。
我不知道是否有一种让NHibernate处理这个问题的有效方法......

Answer 3

您需要确保为希望庄园拥有的每项功能加入MyFeatureValue一次。

一种方法是为每次迭代调用.CreateAlias，给它一个唯一的别名，然后添加表达式“aliasX.Id”


foreach (var id in idCollection)
{
   query = query.CreateAlias("MyFeatureValue", "feature" + id)
                .Add(Expression.Eq("feature" + id + ".Id",id);


}

我真的不记得语法是怎么回事了，写下这个，不确定你是否需要重新声明查询：）

但是，我认为这会让你开始。

编辑：由于Criteria API中的错误限制您使用CreateAlias或CreateCriteria多次关联集合，因此您需要求助于HQL。

http://derek-says.blogspot.com/2008/06/duplicate-association-path-bug-in.html

（Hibernate也遇到同样的问题）


select e   
FROM Estate AS e
INNER JOIN e.MyFeatureValue AS fv1
INNER JOIN e.MyFeatureValue AS fv2
WHERE fv1.Id = 3
   AND fv2.Id = 13

您需要动态构建HQL，以便您的别名变得唯一（fv1，fv2，fvX ......）

Answer 4

代码看起来像是在传递一个FeaturesValueIds列表，并且想要一个具有所有这些功能的List。如果是这种情况，我会查看正在生成的SQL，并针对数据库运行它以查看是否应该返回任何内容。

否则，如果您正在寻找具有您传入的任何功能的List，则应使用Disjunction而不是Conjunction。

Answer 5

    exec sp_executesql N'SELECT TOP 3 id11_1_, Address11_1_, Title11_1_, Descript4_11_1_, 
    Price11_1_, Discount11_1_, ForBankL7_11_1_, AddDate11_1_, LastUpdate11_1_, 
IsVisible11_1_, ViewCount11_1_, SaleOrRent11_1_, LocationId11_1_, StaffId11_1_, 
CategoryId11_1_, id27_0_, EstateId27_0_, FeatureV3_27_0_ FROM (SELECT ROW_NUMBER() 
OVER(ORDER BY __hibernate_sort_expr_0__) as row, query.id11_1_, query.Address11_1_, 
query.Title11_1_, query.Descript4_11_1_, query.Price11_1_, query.Discount11_1_, 
query.ForBankL7_11_1_, query.AddDate11_1_, query.LastUpdate11_1_, query.IsVisible11_1_, 
query.ViewCount11_1_, query.SaleOrRent11_1_, query.LocationId11_1_, query.StaffId11_1_, 
query.CategoryId11_1_, query.id27_0_, query.EstateId27_0_, query.FeatureV3_27_0_, 
query.__hibernate_sort_expr_0__ FROM (SELECT this_.id as id11_1_, this_.Address as 
Address11_1_, this_.Title as Title11_1_, this_.Description as Descript4_11_1_, this_.Price 
as Price11_1_, this_.Discount as Discount11_1_, this_.ForBankLoan as ForBankL7_11_1_, 
this_.AddDate as AddDate11_1_, this_.LastUpdate as LastUpdate11_1_, this_.IsVisible as 
IsVisible11_1_, this_.ViewCount as ViewCount11_1_, this_.SaleOrRent as SaleOrRent11_1_, 
this_.LocationId as LocationId11_1_, this_.StaffId as StaffId11_1_, this_.CategoryId as 
CategoryId11_1_, estatefeat1_.id as id27_0_, estatefeat1_.EstateId as EstateId27_0_, 
estatefeat1_.FeatureValueId as FeatureV3_27_0_, CURRENT_TIMESTAMP as 
__hibernate_sort_expr_0__ FROM Estate this_ inner join EstateFeature estatefeat1_ on 
this_.id=estatefeat1_.EstateId WHERE this_.CategoryId = @p0 and 
(estatefeat1_.FeatureValueId = @p1 and estatefeat1_.FeatureValueId = @p2 and 
estatefeat1_.FeatureValueId = @p3 and estatefeat1_.FeatureValueId = @p4 and 
estatefeat1_.FeatureValueId = @p5 and estatefeat1_.FeatureValueId = @p6 and 
estatefeat1_.FeatureValueId = @p7)) query ) page WHERE page.row > 0 ORDER BY 
__hibernate_sort_expr_0__',N'@p0 bigint,@p1 bigint,@p2 bigint,@p3 bigint,@p4 bigint,@p5 
bigint,@p6 bigint,@p7 bigint',@p0=3,@p1=7,@p2=8,@p3=9,@p4=10,@p5=11,@p6=12,@p7=16

Answer 6

我也试过这个，但结果是一样的：

DetachedCriteria features = DetachedCriteria.For<FeatureValue>();
features.SetProjection(Projections.Property("Id"));
features.Add(Property.ForName("Id").EqProperty("value.Id"));

var and = new Conjunction();

foreach (var l in FeatureIdCollection)
    and.Add(Expression.Eq("Id", l));

features.Add(and);

query.CreateCriteria("EstateFeatures")
     .CreateCriteria("MyFeatureValue","value")
     .Add(Subqueries.Exists(features));

Answer 7

看起来您需要or（Disjunction）而不是and（Conjunction）。现在，您正在搜索EstateFeature个对象，以便每个对象都有多个不同的Id s，这似乎不是您想要的。

var or = new Disjunction();
foreach(var id in idCollection)
    or.Add(Expression.Eq("MyFeatureValue.Id", id);

var query = DetachedCriteria.For<Estate>();
query
    .CreateCriteria("EstateFeatures")
    .Add(and);
var estates = query.GetExecutableCriteria(session).List<Estate>();

使用Nhibernate Criteria Api查询集合？

7 个答案: