Question

这是我的XML，我想取消编组。

<?xml version="1.0" encoding="UTF-8"?>
<departments>
    <deptname name="Research">
        <employee>
            <eid>r-001</eid>
            <ename>Dinesh R</ename>
            <age>35</age>
            <deptcode>d1</deptcode>
            <deptname>Research</deptname>
            <salary>20000</salary>
        </employee>
    </deptname>
    <deptname name="Sales">
        <employee>
            <eid>s-001</eid>
            <ename>Kanmani S</ename>
            <age>35</age>
            <deptcode>d2</deptcode>
            <deptname>Sales</deptname>
            <salary>30000</salary>
        </employee>
    </deptname>
</departments>

Department.java

public class Department 
{ 
    @XmlAttribute(name = "deptname")
    private String name;

    @XmlElement(name = "employee") 
    private List<Employee> employee = new ArrayList<>();//getter and setter//
}

这是我的Departments.java

public class Departments {

    List<Department> deptname;

    public List<Department> getDeptname() {
        return deptname;
    }

    public void setDeptname(List<Department> deptname) {
        this.deptname = deptname;
    }
}

这是我的Unmarshalling.java

public class Unmarshalling {
    public void testXML() {
        try {
            File file = new File(
                "/home/scrunch/work/workspace/sts/default/EmployeeUnmarshall/src/main/java/OutputXml.xml");
            JAXBContext jaxbContext = JAXBContext.newInstance(Departments.class);
            Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
            Departments departments = (Departments) jaxbUnmarshaller.unmarshal(file);
            System.out.println(departments);
        } catch (JAXBException e) {
            e.printStackTrace();
        }
    }
}

我尝试了编组，但是没有得到java对象。我得到了xml格式。我在做休息服务。为此，我需要解组 XML文件，这样我就可以获取Java对象。先生，请给我更新一下。

Answer 1

我尝试了您的代码并最终出现此错误：

线程“ main”中的异常javax.xml.bind.UnmarshalException：意外元素（uri：“”，local：“部门”）。预期元素为（无）

通过声明Departments来解决此问题：

@XmlRootElement(name="departments")
public class Departments {

    List<Department> deptname;

    public List<Department> getDeptname() {
        return deptname;
    }

    public void setDeptname(List<Department> deptname) {
        this.deptname = deptname;
    }
}

请参见that question

尝试一下

我想通过解组将XML转换为Java

1 个答案: