我想通过解组将XML转换为Java

Question

这是我的XML，我想解组它。

<?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();
        }
    }
}

我尝试过解组但我没有得到java对象。我有xml格式。我正在做休息服务。为此，我需要解组XML文件，这样我才能得到Java对象。请你更新我先生。

Answer 1

我尝试了你的代码并最终出现了这个错误：

线程“main”中的异常javax.xml.bind.UnmarshalException：意外元素（uri：“”，local：“departments”）。预期要素是（无）

通过这种方式声明Departments来修复它：

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

    List<Department> deptname;

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

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

见that question

试试