Question

我有班车

Class Car{
    protected List<Attribute> attributes;
}

然后我们有类属性

Class Attribute{
    protected String name;
    protected int sortOrder;
}

然后我们有三种类型的属性

// single class represents Dropdown
class Single extends Attribute{
    protected List<AttributeOption> options;
}

// multiple class represents Checkbox
class Multiple extends Attribute{
    protected List<AttributeOption> options;
}

class Range extends Attribute{
    protected List<AttributeOption> startRange;
    protected List<AttributeOption> endRange;
}

class AttributeOption{
    protected String name;
    protected int sortOrder;
}

如何在hibernate中对上面的代码进行建模？

Answer 1

您没有提供足够的详细信息，因为Car和Attribute（一对多，或多对多）之间的确切关系是什么，您要将哪种继承策略用于{{ 1}}（单个表，每个类的表），但这应该让你开始

Attribute

@Entity
public class Car {
    ...
    @OneToMany(mappedBy = "car")
    private List<Attribute> attributes;
    ...
    // getters, setters
}

@Entity
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
public abstract class Attribute {
    ...
    private String name;
    private Integer sortOrder;
    @ManyToOne
    @JoinColumn(name = "car_id")
    private Car car;
    ...
    // getters, setters
}

@Entity
public class Single extends Attribute {
    ...
    @OneToMany(mappedBy = "attribute")
    private List<AttributeOption> options;
    // getters, setters
    ...
}

@Entity
public class Multiple extends Attribute {
    ...
    @OneToMany(mappedBy = "attribute")
    private List<AttributeOption> options;
    // getters, setters
    ...
}

@Entity
public class Range extends Attribute {
    ...
    @OneToMany(mappedBy = "attribute")
    @JoinTable(name = "startrange_option", 
               joinColumns = @JoinColumn(name = "range_id"), 
               inverseJoinColumns = @JoinColumn(name = "option_id")
    private List<AttributeOption> startRange;
    @OneToMany(mappedBy = "attribute")
    @JoinTable(name = "endrange_option", 
               joinColumns = @JoinColumn(name = "range_id"), 
               inverseJoinColumns = @JoinColumn(name = "option_id")
    private List<AttributeOption> endRange;
    ...
}

棘手的部分是与同一实体（@Entity public class AttributeOption { ... private String name; private Integer sortOrder @ManyToOne @JoinColumn(name = "attribute_id") private Attribute attribute; // getters, setters ... }和startRange）的两个关系，它们需要endRange实体中两个Attribute类型的字段，或者（在我的例子中）每个关系的单独连接表。

请注意，我直接在答案中输入了这个内容，因此可能存在错误。

存储不同类型的列表

1 个答案: