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更新：寻找 Jackson或JAXB 解决方案。

在对 Jackson 的行为进行了一些研究后，我发现 Jackson 将始终使用包装器进行收藏。所以我需要的可能与杰克逊无关。因此，将 JAXB 添加到标题中。

原始问题

我需要为以下 XML 模式创建 POJO。

<ABWrap>
    <A></A>
    <B></B>
    <A></A>
    <B></B>
    ...
    ... n times
</ABWrap>

我试过关注 POJO。但这些并没有产生预期的结果。

class AB {
    @JacksonXmlProperty(localName = "A")
    private String A;
    @JacksonXmlProperty(localName = "B")
    private String B;
}

@JacksonXmlRootElement(localName = "ABWrap")
class ABWrap {
    @JacksonXmlElementWrapper(useWrapping = false)
    private AB[] ab = new AB[n];
}

我需要保持那对<A></A>和<B></B>应该走到一起的状态。元素的顺序很重要。
以下模式不适用于我的情况：

<ABWrap>
    <A></A>
    <A></A>
    ...
    ... n times
    <B></B>
    <B></B>
    ...
    ... n times
</ABWrap>

我已经能够实现第二个。但我一直无法找到生成第一个模式的方法。

更新@mart 的回答：

我定义了ABWrap，如下：ABInterfaceA

@XmlAccessorType(XmlAccessType.FIELD)
@XmlRootElement(name = "ABWrap")
public class ABWrap {
    @XmlElements({@XmlElement(name = "A", type = A.class), @XmlElement(name = "B", type = B.class)})
    private List<ABInterface> ab;
}

public interface ABInterface { }

public class A implements ABInterface {
    @XmlValue
    private String a;
}

B定义类似于A。

主要方法如下：

public class Application {

    public static void main(final String[] args) throws JAXBException {

        JAXBContext jaxbContext = JAXBContext.newInstance(ABWrap.class);
        Marshaller marshaller = jaxbContext.createMarshaller();
        marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);

        A a = new A("a");
        B b = new B("b");
        ABWrap abWrap = new ABWrap(Arrays.asList(a, b));
        marshaller.marshal(abWrap, System.out);
    }
}

但是此解决方案失败并出现以下错误：（jaxbpoc是项目名称）

If a class has @XmlElement property, it cannot have @XmlValue property.
this problem is related to the following location:
    at private java.lang.String ...jaxbpoc.A.a
    at ...jaxbpoc.A
    at private java.util.List ...jaxbpoc.ABWrap.ab
    at ...jaxbpoc.ABWrap
this problem is related to the following location:
    at public java.lang.String ...A.getA()
    at ...jaxbpoc.A
    at private java.util.List ...jaxbpoc.ABWrap.ab
    at ...jaxbpoc.ABWrap
If a class has @XmlElement property, it cannot have @XmlValue property.
this problem is related to the following location:
    at private java.lang.String ...jaxbpoc.B.b
    at ...jaxbpoc.B
    at private java.util.List ...jaxbpoc.ABWrap.ab
    at ...jaxbpoc.ABWrap
    ....
    ....
Class has two properties of the same name "a"
this problem is related to the following location:
    at public java.lang.String ...jaxbpoc.A.getA()
    at ...jaxbpoc.A
    at private java.util.List ...jaxbpoc.ABWrap.ab
    at ...jaxbpoc.ABWrap
this problem is related to the following location:
    ....
    ....
4

1 回答 1

4

你可以这样做：

@XmlAccessorType(XmlAccessType.FIELD)
@XmlRootElement(name = "ABWrap")
public class ABWrap {
    @XmlElements({
            @XmlElement(name="A", type = A.class),
            @XmlElement(name="B", type = B.class),
    })
    private List<Letter> letters;
}

A, B 看起来像这样：

public class A implements Letter {
    @XmlValue
    private String a;
}

还有一个 A,B 的通用接口，作用不大：

public interface Letter { }

更新：

正如我在评论中提到的，我尝试将 XML 转换为 POJO，反之亦然，并且成功了。我把我用来测试的简单程序贴在这里，所以请告诉我它是如何为你工作的，以便我进一步探索。

XmlToPojo：

public static void main(String[] args) {
        try {
            File file = new File("AB.xml");
            JAXBContext jaxbContext = JAXBContext.newInstance(ABWrap.class);

            Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();

            ABWrap pojo = (ABWrap) jaxbUnmarshaller.unmarshal(file);
        } catch (JAXBException e) {
            e.printStackTrace();
        }

    }

和 POJO 到 xml：

public static void main(String[] args) {
        try {
            JAXBContext jaxbContext = JAXBContext.newInstance(ABWrap.class);
            Marshaller marshaller = jaxbContext.createMarshaller();
            marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);

            A a = new A("testA");
            B b = new B("testB");
            ABWrap abWrap = new ABWrap(Arrays.asList(a, b));
            marshaller.marshal(abWrap, System.out);

        } catch (JAXBException e) {
            e.printStackTrace();
        }
    }
于 2018-02-06T08:03:37.517 回答