如何使用 spring 来编组和解组 xml? [英] How to use spring to marshal and unmarshal xml?
问题描述
我有一个 Spring Boot 项目.我的项目中有几个 xsd.我已经使用 maven-jaxb2-plugin 生成了这些类.我用过 this运行示例 Spring Boot 应用程序的教程.
I have a spring boot project. I have a few xsds in my project. I have generated the classes using maven-jaxb2-plugin. I have used this tutorial to get a sample spring boot application running.
import org.kaushik.xsds.XOBJECT;
@SpringBootApplication
public class JaxbExample2Application {
public static void main(String[] args) {
//SpringApplication.run(JaxbExample2Application.class, args);
XOBJECT xObject = new XOBJECT('a',1,2);
try {
JAXBContext jc = JAXBContext.newInstance(User.class);
Marshaller marshaller = jc.createMarshaller();
marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
marshaller.marshal(xObject, System.out);
} catch (PropertyException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (JAXBException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
但我担心的是我需要映射架构的所有 jaxb 类.在 Spring 中还有什么东西可以用来让我的任务更容易.我看过 Spring OXM 项目,但它有在 xml 中配置的应用程序上下文.Spring Boot 是否有任何我可以开箱即用的东西.任何示例都会有所帮助.
But my concern is that I need to have all the jaxb classes of the schema mapped. Also is there something in Spring that I can use to make my task easier. I have looked at the Spring OXM project but it had application context configured in xml. Does spring boot have anything that I can use out of the box. Any examples will be helpful.
编辑
我尝试了 xerx593 的答案 我使用 main 方法运行了一个简单的测试
I tried xerx593's answer and I ran a simple test using main method
JaxbHelper jaxbHelper = new JaxbHelper();
Jaxb2Marshaller marshaller = new Jaxb2Marshaller();
marshaller.setClassesToBeBound(XOBJECT.class);
jaxbHelper.setMarshaller(marshaller);
XOBJECT xOBJECT= (PurchaseOrder)jaxbHelper.load(new StreamSource(new FileInputStream("src/main/resources/PurchaseOrder.xml")));
System.out.println(xOBJECT.getShipTo().getName());
它运行得非常好.现在我只需要使用 spring boot 将其插入即可.
It ran perfectly fine. Now I just need to plug it in using spring boot.
推荐答案
OXM 绝对适合您!
Jaxb2Marshaller 的简单 java 配置如下所示:
A simple java configuration of a Jaxb2Marshaller would look like:
//...
import java.util.HashMap;
import org.springframework.oxm.jaxb.Jaxb2Marshaller;
//...
@Configuration
public class MyConfigClass {
@Bean
public Jaxb2Marshaller jaxb2Marshaller() {
Jaxb2Marshaller marshaller = new Jaxb2Marshaller();
marshaller.setClassesToBeBound(new Class[]{
//all the classes the context needs to know about
org.kaushik.xsds.All.class,
org.kaushik.xsds.Of.class,
org.kaushik.xsds.Your.class,
org.kaushik.xsds.Classes.class
});
// "alternative/additiona - ly":
// marshaller.setContextPath(<jaxb.context-file>)
// marshaller.setPackagesToScan({"com.foo", "com.baz", "com.bar"});
marshaller.setMarshallerProperties(new HashMap<String, Object>() {{
put(javax.xml.bind.Marshaller.JAXB_FORMATTED_OUTPUT, true);
// set more properties here...
}});
return marshaller;
}
}
在您的应用程序/服务类中,您可以这样处理:
In your Application/Service class you could approach like this:
import java.io.InputStream;
import java.io.StringWriter;
import javax.xml.bind.JAXBException;
import javax.xml.transform.Result;
import javax.xml.transform.stream.StreamResult;
import javax.xml.transform.stream.StreamSource;
import org.springframework.oxm.jaxb.Jaxb2Marshaller;
@Component
public class MyMarshallerWrapper {
// you would rather:
@Autowired
private Jaxb2Marshaller marshaller;
// than:
// JAXBContext jc = JAXBContext.newInstance(User.class);
// Marshaller marshaller = jc.createMarshaller();
// marshalls one object (of your bound classes) into a String.
public <T> String marshallXml(final T obj) throws JAXBException {
StringWriter sw = new StringWriter();
Result result = new StreamResult(sw);
marshaller.marshal(obj, result);
return sw.toString();
}
// (tries to) unmarshall(s) an InputStream to the desired object.
@SuppressWarnings("unchecked")
public <T> T unmarshallXml(final InputStream xml) throws JAXBException {
return (T) marshaller.unmarshal(new StreamSource(xml));
}
}
参见 Jaxb2Marshaller-javadoc,以及相关的答案
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