Question

我无法理解我在做什么，我需要解组如下所示的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>

我上下一课：

@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    

}

我的编组代码是：

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);
}
}

结果是我有appMetaData对象，它具有文件中的所有角色。角色具有属性，但角色内的所有列表均为空。不为空，但为空。我哪里出错了？

P.S。一切都与Java代码有关：）

Answer 1

我已经完成了：）

相反

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

在我使用过的角色类中

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

这里是功能类

@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;
}
}

使用标签解组XML仅包含列表的属性

1 个答案: