Hibernate 中的枚举

发布于 2024-07-10 12:51:01 字数 786 浏览 2 评论 0原文

在 DAO 中拥有一个其值来自 Java 枚举的字段通常很有用。一个典型的示例是登录 DAO，其中通常有一个将用户特征描述为“NORMAL”或“ADMIN”的字段。在 Hibernate 中，我将使用以下 2 个对象以（半）类型安全的方式表示这种关系：

class User {
    String username;
    String passwd;
    UserType type;
}

class UserType {
    private enum Type {ADMIN, NORMAL};
    private String type;

    //Setters/Getters for Hibernate
    public void setType(String type);
    public String getType();

    //Setters/Getters for user
    public void setUserType(UserType.Type t);
    public UserType.Type getUserType();

    public static UserType fromType(UserType.Type t);
}

这可行，但我发现 UserType 类很笨拙，并且需要太多的官僚机构才能存储几个值。理想情况下，Hibernate 应该直接支持枚举字段，并创建一个额外的表来存储枚举值。

我的问题是：有没有办法在Hibernate中直接映射枚举类？如果没有，我表示枚举的模式是否足够好，还是我遗漏了一些东西？人们还使用哪些其他模式？

原文

It is often useful to have a field in a DAO whose value comes from a Java enumeration. A typical example is a login DAO where you usually have a field that characterises the user as "NORMAL" or "ADMIN". In Hibernate, I would use the following 2 objects to represent this relationship in a (semi-)typesafe way:

class User {
    String username;
    String passwd;
    UserType type;
}

class UserType {
    private enum Type {ADMIN, NORMAL};
    private String type;

    //Setters/Getters for Hibernate
    public void setType(String type);
    public String getType();

    //Setters/Getters for user
    public void setUserType(UserType.Type t);
    public UserType.Type getUserType();

    public static UserType fromType(UserType.Type t);
}

This works, but I find the UserType class ungly and requiring too much bureaucracy just to store a couple of values. Ideally, Hibernate should support enum fields directly and would create an extra table to store the enumeration values.

My question is: Is there any way to directly map an enumeration class in Hibernate? If not, is my pattern for representing enumerations good enough or am I missing something? What other patterns do people use?

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

凉风有信 2024-07-17 12:51:01

使用 hibernate 或 JPA 注释：

class User {
   @Enumerated(EnumType.STRING)
   UserType type
}

UserType 只是一个标准的 java 5 枚举。

我无法想象这仅限于注释，但我实际上不知道如何使用 hbm 文件来做到这一点。我猜它可能非常依赖于版本，但我很确定需要 hibernate 3.2+。

编辑：在 hbm 中是可能的，但有点混乱，看看这个论坛线程

using hibernate or JPA annotations:

class User {
   @Enumerated(EnumType.STRING)
   UserType type
}

UserType is just a standard java 5 enum.

I can't imagine this is just limited to just annotations but I don't actually know how to do this with hbm files. It may be very version dependant, I'm guessing but I'm pretty sure that hibernate 3.2+ is required.

edit: it is possible in a hbm, but is a little messy, have a look at this forum thread

回复收藏 0 原文

谁与争疯 2024-07-17 12:51:01

从 Hibernate 文档： http://www.hibernate.org/272.html

您可以为每个枚举创建一个新的 typedef 并在属性标记中引用 typedef。

示例映射 - 内联标记

  <property name='suit'>
    <type name="EnumUserType">
      <param name="enumClassName">com.company.project.Suit</param>
    </type>
  </property>

示例映射 - 使用

  <typedef name="suit" class='EnumUserType'>
      <param name="enumClassName">com.company.project.Suit</param>
  </typedef>

  <class ...>
    <property name='suit' type='suit'/>
  </class>

From the Hibernate documentation: http://www.hibernate.org/272.html

You can create a new typedef for each of your enums and reference the typedefs in the property tag.

Example Mapping - inline <type> tag

  <property name='suit'>
    <type name="EnumUserType">
      <param name="enumClassName">com.company.project.Suit</param>
    </type>
  </property>

Example Mapping - using <typedef>

  <typedef name="suit" class='EnumUserType'>
      <param name="enumClassName">com.company.project.Suit</param>
  </typedef>

  <class ...>
    <property name='suit' type='suit'/>
  </class>

回复收藏 0 原文

~没有更多了~