13

如何以有效的方式为以下xml解析和创建java pojo？请建议任何有效的解析器。

XML 格式是

<?xml version="1.0" encoding="utf-8"?>
<CCMainRootTag ID="12">
  <Header TableName="TableName"    TableVersion="12" TableID="12" CreatedDate="2013-02-09T15:35:33" CreatedByUserName="ABC" CreatedBySystem="ABC" />
  <ClassPrimary ID="12" Code="Y" DescriptionDK="DK language " DescriptionUK="" DefDK="" DefUK="" IFDGUID="">
    <ObjectClass ID="12" Code="YA" DescriptionDK="DK Language" DescriptionUK="" DefDK=""     DefUK="" IFDGUID="">
      <Synonym>
        <Concept Description="Description" Language="DK" />
        <Concept Description="" Language="UK" />
        <Concept Description="Description" Language="DK" />
        <Concept Description="" Language="UK" />
        <Concept Description="Description" Language="DK" />
        <Concept Description="" Language="UK" />
        <Concept Description="Description" Language="DK" />
        <Concept Description="" Language="UK" />
      </Synonym>
    </ObjectClass>
    <ObjectClass ID="12" Code="YB" DescriptionDK="DK Language" DescriptionUK="" DefDK="" DefUK="" IFDGUID=""> </ObjectClass>
    <ObjectClass ID="12" Code="YC" DescriptionDK="DK Language" DescriptionUK="" DefDK="" DefUK="" IFDGUID=""> </ObjectClass>
    <ObjectClass ID="12" Code="YD" DescriptionDK="DK language" DescriptionUK="" DefDK="" DefUK="" IFDGUID=""> </ObjectClass>
  </ClassPrimary>
</CCMainRootTag>

我已经使用了这个链接，但是它的性能很慢并且有问题没有有效的 pojo。

我想解析它以一种有效的方式为我提供直接的 java pojo。

4

4 回答 4

12

您可以使用JAXB将 XML 转换为 Java POJO。但在您最终确定解决方案之前，请检查此站点以进行性能比较。

于 2013-02-09T14:55:59.663 回答
8

对于那些寻找 JAXB 代码来将 xml 转换为 java 对象的人：

//Convert xml to String first
Element partyLoaderRequest; // your xml data
String xmlString = new XMLOutputter().outputString(partyLoaderRequest);   
InputStream is = new ByteArrayInputStream(xmlString.getBytes());
DocumentBuilder docBuilder = DocumentBuilderFactory.newInstance().newDocumentBuilder();
Document document = docBuilder.parse(is);
org.w3c.dom.Element varElement = document.getDocumentElement();
JAXBContext context = JAXBContext.newInstance(Person.class);
Unmarshaller unmarshaller = context.createUnmarshaller();
JAXBElement<Person> loader = unmarshaller.unmarshal(varElement, Person.class);
Person inputFromXml = loader.getValue();

而 Person 有适当的 XML 注释：

@XmlRootElement(name="Person")
public class CimbWlAdminUserAmendInput {
    @XmlElement(name="companyName",required=true,nillable=false) 
    private String companyName;
    ...
    //setters getters
    @XmlTransient
    public String getCompanyName() {
        return companyName;
    }

    public void setCompanyName(String companyName) {
        this.companyName = companyName;
    }
}
于 2015-05-21T02:15:40.830 回答
1

JacksonXmlModule 可以处理序列化和反序列化。

// Item.class - use lombok or create g/setters
@JsonPropertyOrder({"name", "description", "note"})
public class Item { 
   private String name;
   private String description;
   private String note;
}


// Test.class
package hello.service;

import com.fasterxml.jackson.core.type.TypeReference;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.SerializationFeature;
import com.fasterxml.jackson.dataformat.xml.JacksonXmlModule;
import com.fasterxml.jackson.dataformat.xml.XmlMapper;

import com.fasterxml.jackson.dataformat.xml.annotation.JacksonXmlRootElement;
import hello.entity.Item;
import org.junit.Before;

import org.junit.Test;

import java.util.ArrayList;
import java.util.List;

import static org.junit.Assert.*;

public class NameServiceImplTest {

    ObjectMapper objectMapper;

    @Before
    public void setup() {
        JacksonXmlModule xmlModule = new JacksonXmlModule();
        xmlModule.setDefaultUseWrapper(false);
        this.objectMapper = new XmlMapper(xmlModule);
        this.objectMapper.enable(SerializationFeature.INDENT_OUTPUT);
    }

    @Test
    public void serializeTest() {

        // Wrapper
        @JacksonXmlRootElement(localName = "names")
        class Names {
            public List<Item> item = new ArrayList<>();
        }

        Item item = new Item();
        item.setName("Vladimir");
        item.setDescription("Desc");
        item.setNote("Note");

        Item item2 = new Item();
        item2.setName("Iva");
        item2.setDescription("Desc2");
        item2.setNote("Note2");

        Names names = new Names();
        names.item.add(item);
        names.item.add(item2);

        try {
            String xml = objectMapper.writeValueAsString(names);
            assertNotNull(xml);
            System.out.println(xml);
        } catch (Exception e) { // IOException
            System.out.println(e.getMessage());
            fail();
        }
    }

    @Test
    public void deserializeTest() {
        String xml = "<names>" +
                "<item><name>name</name><description>desc</description><note>note</note></item>" +
                "<item><name>name</name><description>desc</description><note>note</note></item>" +
                "</names>";

        try {
            List<Item> names = objectMapper.readValue(xml, new TypeReference<List<Item>>() {});
            names.forEach(item -> {

                assertEquals("name", item.getName());
                assertEquals("desc", item.getDescription());
                assertEquals("note", item.getNote());

            } );
        } catch (Exception e) { // IOException

        }


    }
}
于 2018-08-04T22:54:17.903 回答
1 
public class XmlConvertUtil {

    public static void main(String[] args) {
        ResultDataSet resultDataSet = new ResultDataSet(new DtInformation("0", "Success"), new DtData("980000001"));
        //Method which uses JAXB to convert object to XML
        System.out.println(JaxbObjToXML(resultDataSet));
    }

    public static Object JaxbXmlToObj(String xmlString, Object obj) {

        JAXBContext jaxbContext;
        try {
            jaxbContext = JAXBContext.newInstance(obj.getClass());

            Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();

            Object resultDataSet = (Object) jaxbUnmarshaller.unmarshal(new StringReader(xmlString));

            //System.out.println(resultDataSet);
        }
        catch (JAXBException e) {
            e.printStackTrace();
        }
        return obj;
    }

    public static String JaxbObjToXML(Object object) {
        String xmlContent = null;
        try {
            //Create JAXB Context
            JAXBContext jaxbContext = JAXBContext.newInstance(object.getClass());

            //Create Marshaller
            Marshaller jaxbMarshaller = jaxbContext.createMarshaller();

            //Required formatting??
            jaxbMarshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, Boolean.TRUE);
            jaxbMarshaller.setProperty(Marshaller.JAXB_FRAGMENT, Boolean.TRUE);

            //Print XML String to Console
            StringWriter sw = new StringWriter();

            //Write XML to StringWriter
            jaxbMarshaller.marshal(object, sw);

            //Verify XML Content
            xmlContent = sw.toString();
            //System.out.println(xmlContent);
        }
        catch (JAXBException e) {
            e.printStackTrace();
        }
        return xmlContent;
    }
}
于 2020-06-25T18:22:57.643 回答