使用eclipse在java中将.xml文件转换为.json文件？

Question

This is my approach till now.

package com.xmltojson.parse_xml_to_json;

import java.io.IOException;
import java.io.InputStream;
import java.net.URL;
import net.sf.json.JSON;
import net.sf.json.JSONObject;
import net.sf.json.xml.XMLSerializer;
import org.apache.commons.io.IOUtils;

public class XMLtoJsonConverter 
{
    private URL url = null;
    private InputStream inputStream = null;

    public void getXMLfromJson() {
        try {
            url = XMLtoJsonConverter.class.getClassLoader().getResource("sampleParse.xml");
            inputStream = url.openStream();
            String xml = IOUtils.toString(inputStream);
            JSON objJson = new XMLSerializer().read(xml);


            System.out.println("JSON data : " + objJson);
        } catch (Exception e) {
            e.printStackTrace();
        } finally {
            try {
                if (inputStream != null) {
                    inputStream.close();
                }
                url = null;
            } catch (IOException ex) {
            }
        }
    }

    public static void main(String[] args) {
        new XMLtoJsonConverter().getXMLfromJson();
    }
}

Below is my sample xml file

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<terms>

<term id="108822" SC="CO" LC="EN" LQ="UK" Type="DES">
        <label>soyabeans</label>
        <hn>From 1983.</hn>
        <uf idref="108843"/>
        <bt idref="52990"/>
        <cpl idref="52335"/>
        <cpl idref="52340"/>
</term>

    <term id="108843" SC="CO" LC="EN" LQ="US" Type="NPT">
        <label>soybeans</label>
        <use idref="108822"/>
    </term>

    <term id="52990" SC="CO" LC="EN" LQ="NA" Type="DES">
        <label>grain legumes</label>
        <uf idref="98121"/>
        <bt idref="52982"/>
    </term>

    <term id="52335" SC="ON" LC="EN" LQ="NA" Type="DES">
        <label>Glycine max</label>
        <com>bhatta (Hindi) (India);
        bhatta (Nepali) (India);
        bhatta (Nepali) (Nepal);
        bhattamash (Nepali) (India);
        bhattamash (Nepali) (Nepal);</com>
        <uf idref="108125"/>
        <uf idref="52333"/>
    </term>

    <term id="52340" SC="ON" LC="EN" LQ="NA" Type="DES">
        <label>Glycine soja</label>
        <uf idref="52341"/>
        <uf idref="52344"/>
        <bt idref="52327"/>
        <hpr idref="108822"/>
    </term>

</terms>

I want to convert this xml to below shown jsonFormat

[
               {
                               "term": "soyabeans",
                               "termHelp": "uf: soybeans, bt: grain legumes",
               }
]

where term is label name and all other tags would come in termHElp. Some of the term help element contain reference to other XML tag reference from where there label value is picked up and put it in with reference to termhelp item

Answer 1

I see your JSON representation is not straight transformation of your input XML, so you can't achieve what you are asking using any available standard open source libraries like Jackson, Gson, Json etc. You will have to do this in 3 steps:

1) Deserialize XML to POJO(say class XMLPOJO) using JAXB

2) Transform XMLPOJO to JSONPOJO (class representing desired JSON format) using standard Java code

3) Serialize POJO (class JSONPOJO) to Json using any JSON parser libraries like Jackson etc.