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我已经跨越了新的问题来了,我JAXB解组JAXB解组不是解组嵌套列表
鉴于此字符串作为输入
<ruleGroup id="1602" name="TestObject">
<simple column="simple" type="value" operator="equals" value="1" not="true"/>
</ruleGroup>
类不解析嵌套列表元素,只有正常的属性(表示为JSON):
{"id":1602,"name":"TestObject"}
这里有什么问题?我的类别是:
@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();
所以我不知道这个问题可能是什么。关于我在做什么的任何想法都是错误的?