# 没有`@XmlRootElement`的 JAXB 编组 – 缺少`@XmlRootElement`错误 > 原文： [https://howtodoinjava.com/jaxb/marshal-without-xmlrootelement/](https://howtodoinjava.com/jaxb/marshal-without-xmlrootelement/) 很多时候，我们将需要**编组 Java 对象，而不使用 [JAXB 注解](https://howtodoinjava.com/jaxb/jaxb-annotations/)**，例如[`@XmlRootElement`](https://howtodoinjava.com/jaxb/xmlrootelement-annotation/)，并且我们不允许在源代码中进行任何更改。 当我们使用遗留代码或某些我们没有源代码的客户端 jar 时，可能会发生这种情况。 可能还有许多其他情况，但是想法是*我们无法使用 JAXB 注解*修改模型类。 这可能是**将 Java 对象转换为 xml 而没有 jaxb** 的示例。 ## 1.不带`@XmlRootElement`的编组问题 在这种情况下，如果我们尝试将[将 Java 对象直接编组为 XML](https://howtodoinjava.com/jaxb/write-object-to-xml/)，则将得到类似的错误。 ```java javax.xml.bind.MarshalException - with linked exception: [com.sun.istack.internal.SAXException2: unable to marshal type "com.howtodoinjava.demo.model.Employee" as an element because it is missing an @XmlRootElement annotation] at com.sun.xml.internal.bind.v2.runtime.MarshallerImpl.write(MarshallerImpl.java:311) at com.sun.xml.internal.bind.v2.runtime.MarshallerImpl.marshal(MarshallerImpl.java:236) at javax.xml.bind.helpers.AbstractMarshallerImpl.marshal(AbstractMarshallerImpl.java:95) at com.howtodoinjava.demo.JaxbExample.jaxbObjectToXML(JaxbExample.java:45) at com.howtodoinjava.demo.JaxbExample.main(JaxbExample.java:17) Caused by: com.sun.istack.internal.SAXException2: unable to marshal type "com.howtodoinjava.demo.model.Employee" as an element because it is missing an @XmlRootElement annotation at com.sun.xml.internal.bind.v2.runtime.XMLSerializer.reportError(XMLSerializer.java:234) at com.sun.xml.internal.bind.v2.runtime.ClassBeanInfoImpl.serializeRoot(ClassBeanInfoImpl.java:323) at com.sun.xml.internal.bind.v2.runtime.XMLSerializer.childAsRoot(XMLSerializer.java:479) at com.sun.xml.internal.bind.v2.runtime.MarshallerImpl.write(MarshallerImpl.java:308) ... 4 more ``` 其中`Employee.java`类如下。 它没有任何`@XmlRootElement`这样的 JAXB 注解。 ```java package com.howtodoinjava.demo.model; import java.io.Serializable; public class Employee implements Serializable { private static final long serialVersionUID = 1L; private Integer id; private String firstName; private String lastName; private Department department; public Employee() { super(); } public Employee(int id, String fName, String lName, Department department) { super(); this.id = id; this.firstName = fName; this.lastName = lName; this.department = department; } //Getters and Setters @Override public String toString() { return "Employee [id=" + id + ", firstName=" + firstName + ", lastName=" + lastName + ", department=" + department + "]"; } } ``` ## 2.不带`@XmlRootElement`注解的编组的解决方案 缺少`@XmlRootElement`注解时，JAXB 无法为`Employee`对象构建[`JAXBElement`](https://docs.oracle.com/javaee/7/api/javax/xml/bind/JAXBElement.html)实例。 这就是我们必须帮助 JAXB 手动构建它的地方。 #### 2.1 语法 ```java /** * @param name Java binding of xml element tag name * @param declaredType Java binding of xml element declaration's type * @param value Java instance representing xml element's value */ JAXBElement<T> elem = new JAXBElement(QName name, Class<T> declaredType, T value ); ``` 例如： ```java JAXBElement<Employee> jaxbElement = new JAXBElement<Employee>( new QName("", "employee"), Employee.class, employeeObj ); ``` #### 2.2 不带`@XmlRootElement`的 JAXB 编组示例 现在，让我们看看该编组代码的工作原理。 ```java package com.howtodoinjava.demo; import javax.xml.bind.JAXBContext; import javax.xml.bind.JAXBElement; import javax.xml.bind.JAXBException; import javax.xml.bind.Marshaller; import javax.xml.namespace.QName; import com.howtodoinjava.demo.model.Department; import com.howtodoinjava.demo.model.Employee; public class JaxbExample { public static void main(String[] args) { Employee employee = new Employee(1, "Lokesh", "Gupta", new Department(101, "IT")); jaxbObjectToXML( employee ); } private static void jaxbObjectToXML(Employee employeeObj) { try { JAXBContext jaxbContext = JAXBContext.newInstance(Employee.class); Marshaller jaxbMarshaller = jaxbContext.createMarshaller(); // To format XML jaxbMarshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, Boolean.TRUE); //If we DO NOT have JAXB annotated class JAXBElement<Employee> jaxbElement = new JAXBElement<Employee>( new QName("", "employee"), Employee.class, employeeObj); jaxbMarshaller.marshal(jaxbElement, System.out); //If we have JAXB annotated class //jaxbMarshaller.marshal(employeeObj, System.out); } catch (JAXBException e) { e.printStackTrace(); } } } ``` 程序输出： ```java <?xml version="1.0" encoding="UTF-8" standalone="yes"?> <employee> <department> <id>101</id> <name>IT</name> </department> <firstName>Lokesh</firstName> <id>1</id> <lastName>Gupta</lastName> </employee> ``` 请向我提供关于没有`@XmlRootElement`的 *JAXB 编组*示例的问题。 学习愉快！ 参考： [`XmlRootElement` Java 文档](https://docs.oracle.com/javase/7/docs/api/javax/xml/bind/annotation/XmlRootElement.html)