JAXB / Jackson：没有父标记的两个元素的序列 [英] JAXB/Jackson: Sequence of two elements without parent tag

问题描述

更新：寻找杰克逊 或 JAXB解决方案。

Update: Looking for a Jackson or JAXB solution.

之后研究一下杰克逊的行为，我发现杰克逊总是会用收藏品包装。所以我可能需要与杰克逊无关。因此，将JAXB添加到标题。

After researching a bit about behavior of Jackson, I found that Jackson will always use a wrapper for collections. So probably what I need is not possible to do with Jackson. Hence, added JAXB to title.

原始问题

我需要为以下XML模式创建POJO。

I need to create POJO for following XML pattern.

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

我尝试过关注POJO。但这些并没有产生预期的结果。

I've tried following POJOs. But these are not generating desired result.

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];
}

我需要维持一对<的条件;>< / A> 和< B>< / B> 应该结合在一起。元素序列很重要。

以下格式在我的情况下不起作用：

I need to maintain the condition that pair of <A></A> and <B></B> should come together. Sequence of elements is important.
Following pattern will not work in my case:

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

我已经能够获得第二名。但我无法找到生成第一种模式的方法。

I have been able to achieve second one. But I've not been able to figure out a way to generate the first pattern.

@更新mart的回答：

我定义了 ABWrap ， ABInterface 和 A 如下：

@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 已定义类似于 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);
    }
}

但此解决方案失败并出现以下错误：（ jaxbpoc 是项目名称）

But this solution failed with following error: (jaxbpoc is project name)

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:
    ....
    ....

推荐答案

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：

And A, B would look something like this:

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

A，B的通用接口没有做太多工作：

And a common interface for A,B that doesn't do much:

public interface Letter { }

更新：

正如我在评论中提到的，我尝试使用XML到POJO，反之亦然，它起作用了。我在这里粘贴了我用来测试的简单程序，所以请告诉我它是如何工作的，所以我可以进一步探索。

As I mentioned in the comment, I tried XML to POJO and vice versa and it worked. I paste here the simple programs I used to test, so please let me know how it works for you, so I can explore further.

XmlToPojo：

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();
        }
    }

这篇关于JAXB / Jackson：没有父标记的两个元素的序列的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！