如何使用spring来编组和解组xml？

本文介绍了如何使用spring来编组和解组xml？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有一个春季启动项目。我的项目中有几个xsds。我使用maven-jaxb2-plugin生成了类。我使用了这个获取示例Spring启动应用程序运行的教程。

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。还有一些东西在Spring中可以用来使我的任务更容易。我查看了Spring OXM 项目，但它有在xml中配置的应用程序上下文弹簧靴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

我试过xerx593的回答我使用main运行了一个简单的测试方法

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());

它运行得很好。现在我只需要使用spring boot插入它。

It ran perfectly fine. Now I just need to plug it in using spring boot.

推荐答案

OXM绝对适合你！

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 }); //"alternatively" setContextPath(<jaxb.context>), marshaller.setMarshallerProperties(new HashMap<String, Object>() {{ put(javax.xml.bind.Marshaller.JAXB_FORMATTED_OUTPUT, true); }}); return marshaller; } }

在您的应用程序/服务类中，您可以这样接近：

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