java 如何在 JPA 列中使用自定义类型？
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How to use custom type in JPA column?
提问by yegor256
I have a class:
我有一堂课：
public class Email {
private String name;
private String domain;
public String toString() {
return name + "@" + domain;
}
}
I want to use it in JPA column:
我想在 JPA 列中使用它：
@Entity
public class User {
@Id private Integer id;
private Email email;
}
This is what Hibernate says:
这是 Hibernate 所说的：
org.hibernate.MappingException: Could not determine type for: com.XXX.Email
How to make it understand my custom type. I think that it's something very simple, but can't find in documentation.
如何让它理解我的自定义类型。我认为这很简单，但在文档中找不到。
回答by Bozho
回答by pakore
You can make email an entity and it will work...but it's pretty ineficcient.
您可以使电子邮件成为一个实体，它会起作用……但效率很低。
@Entity
public class Email {
...
}
Or you can swtich from Emailto Stringand it will work. (What's the point of wrapping a String anyway?)
或者你可以Email切换到String，它会起作用。（无论如何包装一个字符串有什么意义？）
You can read this tutorialabout custom user types in Hibernate (since you tagged it).
您可以阅读本教程，了解 Hibernate 中的自定义用户类型（因为您标记了它）。
Or you can use @Embebbedas Bozho says.
或者你可以@Embebbed像 Bozho 所说的那样使用。
