我有一个简单的XML,我想要解组成一个模型类。我用JAXB注释注释了类,用于定义访问类型(FIELD):
import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
@XmlAccessorType(XmlAccessType.FIELD)
public class DtoTest {
private String name;
public DtoTest() {}
public DtoTest(String name) {
super();
this.name = name;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
@Override
public String toString() {
return "DtoTest [name=" + name + "]";
}
}
这是我的主类,我对一个保存在String变量中的简单XML运行一个unmarshal方法:
public class Test {
public static void main(String[] args) throws Exception {
Object obj = new DtoTest();
String testXML = "<dtoTest><name>example</name></dtoTest>";
obj = unmarshal(obj, testXML);
System.out.println(obj);
}
/* This is a generic unmarshall method which I've already used with success with other XML*/
public static <T> T unmarshal(T obj, String xml) throws Exception {
XMLInputFactory xif = XMLInputFactory.newFactory();
XMLStreamReader xsr = xif.createXMLStreamReader(new StringReader(xml));
Class<? extends Object> type = obj.getClass();
JAXBContext jc = JAXBContext.newInstance(type);
Unmarshaller unmarshaller = jc.createUnmarshaller();
obj = (T)unmarshaller.unmarshal(xsr, type).getValue();
xsr.close();
return obj;
}
}
每当我运行代码时,我都得到相同的输出:
DtoTest [name=null]
我不明白我做错了什么。
答案 0 :(得分:2)
我只是在jdk1.7.0_67上运行你的代码并且它可以运行。
DtoTest [name=example]
也许你对包含的库有些问题?我只使用普通的java运行它。
答案 1 :(得分:1)
你的问题对我来说完全没问题。您可以对其进行的一项优化是创建StreamSource
而不是XMLStreamReader
。
import javax.xml.bind.*;
import javax.xml.transform.stream.StreamSource;
import java.io.StringReader;
public class Test {
public static void main(String[] args) throws Exception {
Object obj = new DtoTest();
String testXML = "<dtoTest><name>example</name></dtoTest>";
obj = unmarshal(obj, testXML);
System.out.println(obj);
}
public static <T> T unmarshal(T obj, String xml) throws Exception {
StreamSource source = new StreamSource(new StringReader(xml));
Class<? extends Object> type = obj.getClass();
JAXBContext jc = JAXBContext.newInstance(type);
Unmarshaller unmarshaller = jc.createUnmarshaller();
obj = (T)unmarshaller.unmarshal(source, type).getValue();
return obj;
}
}
当解组不按预期工作时,填充JAXB模型并将其封送到XML以查看预期的XML是什么样的。