使用标记解组xml仅包含List的属性

问题描述 投票:0回答:1

我无法理解我做错了什么。我需要解组看起来像这样的xml文件:

<ApplicationMetadata xmlns="http://www.sas.com/xml/schema/namespace/ApplicationMetadata-9.4">
<Role Name="name1" Desc="desc1 " DisplayName="disp1 ">
    <Members/>
    <ContributingRoles/>
    <Capabilities>
        <Capability CapabilityId="1"/>
        <Capability CapabilityId="2"/>
        <Capability CapabilityId="3"/>
    </Capabilities>
</Role>
<Role Name="name2" Desc="desc2" DisplayName="disp2">
    <Members>
        <UserGroup Name="userGoup"/>
    </Members>
    <ContributingRoles/>
    <Capabilities>
        <Capability CapabilityId="1"/>
        <Capability CapabilityId="2"/>
    </Capabilities>
</Role>
</ApplicationMetadata>

我有下一堂课:

@XmlRootElement(name = "ApplicationMetadata", namespace = "http://www.sas.com/xml/schema/namespace/ApplicationMetadata-9.4")
@XmlAccessorType(XmlAccessType.FIELD)
public class ApplicationMetaData {

@XmlElement(name = "Role")
private List<Role> roles;

getters and setters

}

@XmlRootElement(name = "Role")
@XmlAccessorType(XmlAccessType.FIELD)
public class Role {

@XmlAttribute(name = "Name")
private String name;

@XmlAttribute(name = "Desc")
private String desc;

@XmlAttribute(name = "DisplayName")
private String displayName;

@XmlElementWrapper(name = "ContributingRoles")
@XmlElement(name = "UserGroup")
private List<UserGroup> contributionRoles;

@XmlElementWrapper(name = "Members")
@XmlElement(name = "UserGroup")
private List<UserGroup> members;

@XmlElementWrapper(name = "Capabilities")
@XmlElement(name = "Capability")
private List<Capability> capabilities;

getters and setters    

}

@XmlRootElement(name = "Capability")
@XmlAccessorType(XmlAccessType.FIELD)
public class Capability {

@XmlAttribute(name = "CapabilityId")
private String id;

getters and setters

}

@XmlRootElement (name = "UserGroup")
@XmlAccessorType(XmlAccessType.FIELD)
public class UserGroup {

@XmlAttribute(name = "Name")
private String name;

getters and setters    

}

我解密的代码是:

File file = new File(fileName);
ApplicationMetaData appMetaData = (ApplicationMetaData) 
WorkWithXml.unmarshalXml(file, ApplicationMetaData.class);

public static Object unmarshalXml(File file, Class unmarshallerClass) {
try {
    JAXBContext jaxbContext = JAXBContext.newInstance(unmarshallerClass);
    Unmarshaller unmarshaller = jaxbContext.createUnmarshaller();
    return unmarshaller.unmarshal(file);
} catch (JAXBException e) {
    LOG.error("", e);
    throw new AutotestError(e);
}
}

结果我有appMetaData对象与文件中的所有角色。角色具有属性,但角色内的所有列表都是空的。不为空,但为空。我哪里弄错了?

附:这是关于Java代码:)

java xml jaxb unmarshalling
1个回答
0
投票

我已经做了 :)

代替

@XmlElementWrapper(name = "Capabilities")
@XmlElement(name = "Capability")
private List<Capability> capabilities;

在我使用过的Role类中

@XmlElement (name = "Capabilities")
Capabilities capabilities;

这里是Capabilities类

@XmlRootElement(name = "Capabilities")
@XmlAccessorType(XmlAccessType.FIELD)
public class Capabilities {

@XmlElement(name = "Capability")
private List<Capability> capability;

public List<Capability> getCapability() {
    return capability;
}

public void setCapability(List<Capability> capability) {
    this.capability = capability;
}
}
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