CODEWITHSUNDEEP
javaxmljson
Akshat
Akshat

Reputation: 575

Is it possible to maintain ordering in JSON while converting it from xml and vice versa

I need to convert XML to JSON and then after applying some business logic need to reconvert back in XML, But when i try to convert a XML to JSON and then reconvert back the JSON back to XML I am getting its attributes in Different order.

Eg Following XML

<breakfast_menu><food><name>Belgian Waffles</name><price>$5.95</price></food></breakfast_menu>

is converted to following JSON

{"breakfast_menu":{"food":{"price":"$5.95","name":"Belgian Waffles"}}}

and is reconverted to following XML

<breakfast_menu><food><price>$5.95</price><name>Belgian Waffles</name></food></breakfast_menu>

As there is name tag is replaced by price tag .

Is there any way so that we can maintain ordering so the conversion and reconversion produce same output . 

import org.json.JSONException;
import org.json.JSONObject;

import org.json.XML; 

public class XmlToJson {


 public static int PRETTY_PRINT_INDENT_FACTOR = 4;
    public static String TEST_XML_STRING =
            "<breakfast_menu>\n" +
                    "<food>\n" +
                    "<name>Belgian Waffles</name>\n" +
                    "<price>$5.95</price>\n" +
                    "</food>\n" +
                    "</breakfast_menu>";

    public static void main(String[] args) {
        try {
            JSONObject xmlJSONObj = XML.toJSONObject(TEST_XML_STRING);
            String jsonString = xmlJSONObj.toString();
            System.out.println(jsonString);
            System.out.println("================================");
            JSONObject jsonObj = new JSONObject(jsonString);
            String s1 = XML.toString(jsonObj);

            System.out.println(s1);
            s1 = s1.replace("\n", "").replace("\r", "");
            TEST_XML_STRING = TEST_XML_STRING.replace("\n", "").replace("\r", "");
            System.out.println("================================");
            System.out.println(TEST_XML_STRING);
            System.out.println("================================");

            System.out.println(s1.equals(TEST_XML_STRING));



        } catch (JSONException je) {
            System.out.println(je.toString());
        }
    }

}

Upvotes: 0

Views: 3778

Answers (2)

ErstwhileIII
ErstwhileIII

Reputation: 4853

While there are mechanisms that you might use to provide some sort of ordering, JSON does not guarantee a processor of any particular order. If you are interfacing with some process that requires a particular order (and that is under your control), I would recommend changing that process so it conforms to the normal JSON process to handle unordered input.

Upvotes: 1

Paul Lo
Paul Lo

Reputation: 6148

It might be naturally wrong because by default the definition of JSON is: 

An object is an unordered set of name/value pairs.

However, you might want to check JSON.simple library which read the JSON string and keep the order of keys

Upvotes: 1

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