标签: many-to-one

我有2个班:司机和汽车.汽车表在单独的过程中更新.我需要的是在驱动程序中拥有属性,允许我读取完整的汽车描述并只写入指向现有汽车的Id.这是一个例子:

@Entity(name = "DRIVER")
public class Driver {
... ID and other properties for Driver goes here .....

    @ManyToOne(fetch=FetchType.LAZY)
    @JoinColumn(name = "CAR_ID")
    private Car car;

    @JsonView({Views.Full.class})
    public Car getCar() {
      return car;
    }
    @JsonView({Views.Short.class})
    public long getCarId() {
      return car.getId();
    }
    public void setCarId(long carId) {
      this.car = new Car (carId);
    }

}

Car对象只是典型的JPA对象,没有对Driver的反向引用.

所以我试图通过这个实现:1)我可以使用详细的JSON View 2)阅读完整的汽车描述或者我只能阅读简短的JsonView 3中的汽车ID而且最重要的是,当创建新的驱动程序我只是想要传递汽车的JSON ID.这样我就不需要在持久化期间为汽车执行不必要的读取操作,而只是更新Id.

我得到以下错误:"对象引用未保存的瞬态实例 - 在刷新之前保存瞬态实例:com.Driver.car - > com.Car"

我不想在DB中更新Car的实例,而只是从Driver中引用它.知道怎么做到我想要的吗？

谢谢.

更新:忘记提及我在创建驱动程序期间传递的汽车ID是DB中现有汽车的有效ID.

我有一个Doctrine模型(Assignment),它与另一个模型(Region)有多对一的关系.分配由用户拥有(每个用户一次只能为每个区域分配一个分配),我试图用来indexBy分配用户的分配数组,并使用分配区域的ID.但是,我只获得标准的0..n数字键.

当我尝试运行DQL查询时SELECT am, reg, user FROM Assignment am INDEX BY [...] JOIN am.region reg JOIN am.user user WHERE user.id = ?1,INDEX BY的这些值都不起作用:

• region (错误:PathExpression无效.必须是StateFieldPathExpression.)
• region_id (错误:类... \赋值没有名为region_id的字段或关联)
• region.id (错误:字符串的预期结束,得到'.')

这可能吗？如果没有,那么User在没有区域的情况下访问某个区域的分配的便捷方式是indexBy什么？

我有一个将一个数据库转换为另一个数据库的项目 其中一个原始数据库列定义行的类别.此列应映射到新数据库中的新类别.

例如,我们假设原始类别是:parrot, spam, cheese_shop, Cleese, Gilliam, Palin

现在这对我来说有点冗长,而且我希望将这些行分类为sketch, actor- 也就是说,将所有草图和所有actor定义为两个等价类.

>>> monty={'parrot':'sketch', 'spam':'sketch', 'cheese_shop':'sketch', 
'Cleese':'actor', 'Gilliam':'actor', 'Palin':'actor'}
>>> monty
{'Gilliam': 'actor', 'Cleese': 'actor', 'parrot': 'sketch', 'spam': 'sketch', 
'Palin': 'actor', 'cheese_shop': 'sketch'}

这很尴尬 - 我更喜欢这样的东西:

monty={ ('parrot','spam','cheese_shop'): 'sketch', 
        ('Cleese', 'Gilliam', 'Palin') : 'actors'}

但是,这当然将整个元组设置为关键:

>>> monty['parrot']

Traceback (most recent call last):
  File "<pyshell#29>", line 1, in <module>
    monty['parrot']
KeyError: 'parrot'

如何在Python中创建优雅的多对一字典？

谢谢,

亚当

我试图通过Symfony2(2.3.0)使用Doctrine(2.2.3+)在我的数据库中的对象上建立一些ManyToOne/OneToMany关系,并得到一个奇怪的错误.以下是对象的相关部分(一个产品的许多属性):

/**
 * Product
 *
 * @ORM\Table(name="product")
 * @ORM\Entity
 */
class Product
{
    /**
     * @var integer
     *
     * @ORM\Column(name="id", type="integer")
     * @ORM\Id
     * @ORM\GeneratedValue(strategy="AUTO")
     */
    protected $id;

    ...

    /**
     *
     * @OneToMany(targetEntity="ProductAttributes", mappedBy="product")
     */
    protected $product_attributes;

    public function __construct() {
        $this->product_attributes = new \Doctrine\Common\Collections\ArrayCollection();
    }
}

/**
 * ProductAttributes
 *
 * @ORM\Table(name="product_attributes")
 * @ORM\Entity
 */
class ProductAttributes
{
    /**
     * @var integer
     *
     * @ORM\Column(name="pa_id", type="integer")
     * @ORM\Id
     * @ORM\GeneratedValue(strategy="AUTO")
     */
    protected $pa_id;

    /**
     * @var integer
     *
     * …

我正在使用Hibernate和spring.

我有这样的模型类.

@Entity
@Table(name = "forumtopic")
public final class Forumtopic extends AbstractUserTracking implements
    java.io.Serializable {

/**SNIP **/

    private Forumcategory forumcategory;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "FkForumcategoryId", nullable = false)
    public Forumcategory getForumcategory() {
        return this.forumcategory;
    }

    public void setForumcategory(final Forumcategory forumcategory) {
        this.forumcategory = forumcategory;
    }
}

它一般工作,但类不是懒惰加载,但在加载ForumEntry后急切.

Hibernate: 
    select
        forumtopic0_.PkId as PkId19_0_,
        forumtopic0_.CreateDate as CreateDate19_0_,
        forumtopic0_.FkCreateUserId as FkCreate3_19_0_,
        forumtopic0_.FkLastUserId as FkLastUs4_19_0_,
        forumtopic0_.LastChange as LastChange19_0_,
        forumtopic0_.FkForumcategoryId as FkForum10_19_0_,
        forumtopic0_.PublishCategory as PublishC6_19_0_,
        forumtopic0_.State as State19_0_,
        forumtopic0_.Text as Text19_0_,
        forumtopic0_.Topic as Topic19_0_, …

请指教我如何从many2one字段中删除"创建和编辑...".该项目显示在我使用域选项过滤的many2one字段中.

OpenERP第7版

我有一个复合键ContractServiceLocationPK做出来的三个ID.( ,,contractId 在嵌入类型长).使用此复合键的类使用注释将这些ID映射到其对象.这是它的样子(删除了setter/getters和不相关的属性):locationIdserviceIdContractServiceLocation@MapsId

合同

@Entity
@Table(name = "Contract")
public class Contract implements Serializable {

    public Contract() {
    }

    @Id
    @GeneratedValue
    private long id;
    @OneToMany(mappedBy = "contract", cascade = CascadeType.ALL, fetch= FetchType.EAGER)
    Collection<ContractServiceLocation> contractServiceLocation;
}

ContractServiceLocationPK

@Embeddable
public class ContractServiceLocationPK implements Serializable {

    private long contractId;
    private long locationId;
    private long serviceId;
}

ContractServiceLocation

@Entity
@Table(name="Contract_Service_Location")
public class ContractServiceLocation implements Serializable {

    @EmbeddedId
    ContractServiceLocationPK id;

    @ManyToOne(cascade = CascadeType.ALL)
    @MapsId("contractId")
    Contract contract;

    @ManyToOne(cascade = CascadeType.ALL) …

我正在研究JPA项目.我需要在具有三个主键的类上使用@OneToMany映射.您可以在此之后找到错误和类.

    javax.persistence.PersistenceException: No Persistence provider for EntityManager named JTA_pacePersistence: Provider named oracle.toplink.essentials.PersistenceProvider threw unexpected exception at create EntityManagerFactory: 
javax.persistence.PersistenceException
javax.persistence.PersistenceException: Exception [TOPLINK-28018] (Oracle TopLink Essentials - 2.0.1 (Build b09d-fcs (12/06/2007))): oracle.toplink.essentials.exceptions.EntityManagerSetupException
Exception Description: predeploy for PersistenceUnit [JTA_pacePersistence] failed.
Internal Exception: Exception [TOPLINK-7220] (Oracle TopLink Essentials - 2.0.1 (Build b09d-fcs (12/06/2007))): oracle.toplink.essentials.exceptions.ValidationException
Exception Description: The @JoinColumns on the annotated element [private java.util.Set isd.pacepersistence.common.Action.permissions] from the entity class [class isd.pacepersistence.common.Action] is incomplete. When the source entity class uses a composite primary key, …

我正在尝试根据product_item表中的productId和toolboxItemId选择product_item_sortorder表中的匹配行.

在适用于给定productId的普通SQL中:

SELECT pi.*, pis.* FROM product_item pi
LEFT JOIN product_item_sortorder pis
    ON pi.productId = pis.productId
    AND pi.toolboxItemId = pis.toolboxItemId
WHERE pi.productId = 6

我为它写了DQL如下:

$this->_em->createQuery(
    'SELECT pi
    FROM Entities\ProductItem pi
    LEFT JOIN pi.sequence s
    WHERE pi.product = ?1'
);

如果输出$ query-> getSQL(),我会得到以下SQL:

SELECT p0_.id AS id0, p0_.productId AS productId1, p0_.priceGroupId AS priceGroupId2, p0_.toolboxItemId AS toolboxItemId3, p0_.levelId AS levelId4, p0_.parentId AS parentId5, p0_.productId AS productId6, p0_.toolboxItemId AS toolboxItemId7 FROM product_item p0_ LEFT JOIN product_item_sortorder p1_ ON p0_.productId = p1_. AND …

我有两个具有外键关系的表。我试过搜索如何做到这一点，它总是导致 OneToMany 和 ManyToOne 映射。我有这两张桌子。

用户角色

用户

我试图通过将位置列的值获取到 user_role 表来显示所有用户并显示他们的位置。

这些是我的课

用户.java

@Entity
@Table(name="user")
public class User {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private long id;

    ..........

    private UserRole userRole;

    @ManyToOne()
    @JoinColumn(name="role_id")
    public UserRole getUserRole() {
        return userRole;
    }

    public void setUserRole(UserRole userRole) {
        this.userRole = userRole;
    }

    .......
}

用户角色.java

@Entity
@Table(name="user_role")
public class UserRole {

    .......

    private List<User> user;

    @OneToMany(targetEntity=User.class, mappedBy="userRole",cascade=CascadeType.ALL, fetch = FetchType.LAZY)
    public List<User> getUser() {
        return user;
    }

    public void setUser(List<User> user) {
        this.user = user; …