Question

我在JAXB unmarshalling中遇到了一个新问题

将此字符串作为输入

<ruleGroup id="1602" name="TestObject">
    <simple column="simple" type="value" operator="equals" value="1" not="true"/>
</ruleGroup>

该类不解析嵌套列表元素，只解析普通属性（表示为JSON）：

{"id":1602,"name":"TestObject"}

这是什么问题？我的课程是：

@XmlRootElement(name = "ruleGroup")
public class RuleGroup {

    @XmlAttribute(name = "id")
    public Long id;
    @XmlAttribute(name = "name")
    public String name;

    @XmlElement(name = "simple")
    public List<SimpleXML> simple;
}

和

public class SimpleXML {
    @XmlAttribute(name = "column")
    public String column = null;
    @XmlAttribute(name = "type")
    public String type = null;
    @XmlAttribute(name = "operator")
    public String operator = null;
    @XmlAttribute(name = "value")
    public String value = null;
    @XmlAttribute(name = "not")
}

我的解组非常标准：

RuleGroup rule;
JAXBContext jaxbContext = JAXBContext.newInstance(RuleGroup.class);
Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
StringReader reader = new StringReader(test);
JAXBElement<RuleGroup> root = jaxbUnmarshaller.unmarshal(new StreamSource(reader), RuleGroup.class);
rule = (RuleGroup) root.getValue();

所以我不知道问题是什么。关于我做错了什么想法？

Answer 1

在Java 8中运行代码解决了这个问题。如果有人想澄清为什么Marshaller会受到影响，我很想知道答案

JAXB Unmarshalling不解组嵌套列表

1 个答案: