Question

这是“PersonType”类的生成代码。

package demo;

import java.util.ArrayList;
import java.util.List;
import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlType;

@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "PersonType", propOrder = {
"name",
"address"
})
public class PersonType {

@XmlElement(name = "Name", required = true)
protected String name;
@XmlElement(name = "Address", required = true)
protected List<AddressType> address;

public String getName() {
    return name;
}


public void setName(String value) {
    this.name = value;
}


public List<AddressType> getAddress() {
    if (address == null) {
        address = new ArrayList<AddressType>();
    }
    return this.address;
}

}

这是“AddressType”类的生成代码。

    package demo;

 import javax.xml.bind.annotation.XmlAccessType;
 import javax.xml.bind.annotation.XmlAccessorType;
 import javax.xml.bind.annotation.XmlElement;
 import javax.xml.bind.annotation.XmlSchemaType;
 import javax.xml.bind.annotation.XmlType;

 @XmlAccessorType(XmlAccessType.FIELD)
 @XmlType(name = "AddressType", propOrder = {
"number",
"street"
 })
public class AddressType {

@XmlElement(name = "Number")
@XmlSchemaType(name = "unsignedInt")
protected long number;
@XmlElement(name = "Street", required = true)
protected String street;


public long getNumber() {
    return number;
}


public void setNumber(long value) {
    this.number = value;
}


public String getStreet() {
    return street;
}

public void setStreet(String value) {
    this.street = value;
}

}

这是我要解组的xml文件

<Person xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:noNamespaceSchemaLocation="C:\JAXB Demo\demo.xsd">
  <Name>Sharon Krisher</Name>
  <Address>
    <Street>Iben Gevirol</Street>
    <Number>57</Number>
  </Address>
<Address>
  <Street>Moshe Sharet</Street>
   <Number>89</Number>
  </Address>
</Person>

代码敌人unarchall

JAXBContext context = JAXBContext.newInstance(PersonType.class);
    Unmarshaller unmarshaller =context.createUnmarshaller();
    //unmarshaller.setValidating(true);
    PersonType person =(PersonType) unmarshaller.unmarshal(new File("C:\\Users\\sithi\\workspace\\TestJAXB\\src\\data.xml") );

此代码提供了一个例外：

Exception in thread "main" javax.xml.bind.UnmarshalException: unexpected element (uri:"", local:"PersonType"). Expected elements are (none)

Answer 1

将类与根元素关联有两种方法。第一个是类@XmlRootElement，第二个是用@XmlElementDecl注释的类的@XmlRegistry注释（当模型从XML生成时，这个类称为模式{{1 }。

创建ObjectFactory

从XML Schema生成模型时，您应该在包名称上创建JAXBContext：

JAXBContext

或生成JAXBContext jc = JAXBContext.newInstance("demo");类：

ObjectFactory

这将确保处理JAXBContext jc = JAXBContext.newInstance(demo.ObjectFactory.class);类。

http://blog.bdoughan.com/2012/07/jaxb-and-root-elements.html

否ObjectFactory或@XmlRootElement？

如果没有@XmlElementDecl或@XmlRootElement将一个类与文档的根元素相关联，那么您将需要使用一个带有类参数的unmarshal方法。

@XmlElementDecl

Answer 2

问题是您的PersonType类没有@XmlRootElement。我想你想在这个页面中做“Unmarshalling into a known type”的例子，在PersonType上不需要@XmlRootElement。

JAXB中的解组错误

2 个答案: