NHibernate - CreateCriteria по сравнению с CreateAlias

Question

NHibernate - CreateCriteria по сравнению с CreateAlias

Swift 4 UIImage расширение, которое позволяет легко объединять / накладывать изображения.

extension UIImage { 

  func overlayWith(image: UIImage, posX: CGFloat, posY: CGFloat) -> UIImage {
    let newWidth = size.width < posX + image.size.width ? posX + image.size.width : size.width
    let newHeight = size.height < posY + image.size.height ? posY + image.size.height : size.height
    let newSize = CGSize(width: newWidth, height: newHeight)

    UIGraphicsBeginImageContextWithOptions(newSize, false, 0.0)
    draw(in: CGRect(origin: CGPoint.zero, size: size))
    image.draw(in: CGRect(origin: CGPoint(x: posX, y: posY), size: image.size))
    let newImage = UIGraphicsGetImageFromCurrentImageContext()!
    UIGraphicsEndImageContext()

    return newImage
  }

}

27

performance nhibernate criteria icriteria

задан Community 27 June 2012 в 17:06

1 ответ

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performance nhibernate criteria icriteria

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score 38 · Accepted Answer

с учетом этих требований не будет никакой разницы, сгенерированный SQL будет таким же: для сопоставлений:

    <?xml version="1.0" encoding="utf-8" ?>
<hibernate-mapping xmlns="urn:nhibernate-mapping-2.2">
    <class name="Project" table="Project">
        <id name="Id" type="Int32" unsaved-value="0">
            <column name="Id" sql-type="int" not-null="true" unique="true"/>
            <generator class="native" />
        </id>
        <many-to-one name="Job" column="FK_JobId" cascade="save-update" not-null="true" />
    </class>
</hibernate-mapping>
<?xml version="1.0" encoding="utf-8" ?>
<hibernate-mapping xmlns="urn:nhibernate-mapping-2.2">
    <class name="Job" table="Job">
        <id name="Id" type="Int32" unsaved-value="0">
            <column name="Id" sql-type="int" not-null="true" unique="true"/>
            <generator class="native" />
        </id>
        <property name="Name" type="String">
            <column name="Name" sql-type="nvarchar" length="50" not-null="true"/>
        </property>
    </class>
</hibernate-mapping>

и классы

public class Project
    {
        public Project() { }

        public virtual int Id { get; set; }

        public virtual Job Job { get; set; }
    }
public class Job
    {
        public Job() { }

        public virtual int Id { get; set; }

        public virtual String Name { get; set; }
    }

эти определения критериев

ICriteria criteriacrit = session
  .CreateCriteria(typeof (Project))
  .CreateCriteria("Job", "job")
  .Add(Restrictions.Eq("job.Name", "sometextA"));

ICriteria aliascrit = session
  .CreateCriteria(typeof (Project))
  .CreateAlias("Job", "job")
  .Add(Restrictions.Eq("job.Name", "sometextB"));

генерируют тот же SQL

SELECT 
  this_.Id as Id2_1_, 
  this_.FK_JobId as FK2_2_1_, 
  job1_.Id as Id1_0_, 
  job1_.Name as Name1_0_ 
FROM 
  Project this_ 
  inner join Job job1_ 
    on this_.FK_JobId=job1_.Id 
WHERE job1_.Name = @p0; @p0 = 'sometextA'

SELECT 
  this_.Id as Id2_1_, 
  this_.FK_JobId as FK2_2_1_, 
  job1_.Id as Id1_0_, 
  job1_.Name as Name1_0_ 
FROM
  Project this_ 
  inner join Job job1_ 
    on this_.FK_JobId=job1_.Id 
WHERE job1_.Name = @p0; @p0 = 'sometextB'

, однако обратите внимание, что CreateAlias полагается на сопоставления для генерации ассоциаций, тогда как CreateCriteria ] позволяет указать JoinType .

Итак, эти вызовы

ICriteria criteriacrit = session
  .CreateCriteria(typeof(Project))
  .CreateCriteria("Job",JoinType.LeftOuterJoin)
  .Add(Restrictions.Eq("Name", "sometextA"));

ICriteria aliascrit = session
  .CreateCriteria(typeof (Project))
  .CreateAlias("Job", "job")
  .Add(Restrictions.Eq("job.Name", "sometextB"));

генерируют эти операторы SQL

SELECT 
  this_.Id as Id2_1_, 
  this_.FK_JobId as FK2_2_1_, 
  job1_.Id as Id1_0_, 
  job1_.Name as Name1_0_ 
FROM 
  Project this_ 
  **left outer** join Job job1_
    on this_.FK_JobId=job1_.Id 
WHERE job1_.Name = @p0; @p0 = 'sometextA'

SELECT 
  this_.Id as Id2_1_, 
  this_.FK_JobId as FK2_2_1_, 
  job1_.Id as Id1_0_, 
  job1_.Name as Name1_0_ 
FROM Project this_ 
  **inner join** Job job1_ 
    on this_.FK_JobId=job1_.Id 
WHERE job1_.Name = @p0; @p0 = 'sometextB'