Converting JSON to XML in Java

Question

I am new to json. I am having a program to generate xml from json object.

String str = "{'name':'JSON','integer':1,'double':2.0,'boolean':true,'nested':{'id':42},'array':[1,2,3]}";  
    JSON json = JSONSerializer.toJSON( str );  
    XMLSerializer xmlSerializer = new XMLSerializer();  
    xmlSerializer.setTypeHintsCompatibility( false );  
    String xml = xmlSerializer.write( json );  
    System.out.println(xml);

the output is:

<?xml version="1.0" encoding="UTF-8"?>
<o><array json_class="array"><e json_type="number">1</e><e json_type="number">2</e><e json_type="number">3</e></array><boolean json_type="boolean">true</boolean><double json_type="number">2.0</double><integer json_type="number">1</integer><name json_type="string">JSON</name><nested json_class="object"><id json_type="number">42</id></nested></o>

my biggest problem is how to write my own attributes instead of json_type="number" and also writing my own sub elements like .

vinod,Have you got the solution for the above question, "Writing your own attributes to XML tags generated from json".? I'm also looking for the solution, if you found that, please post that in this thread. Thanks in advance. — Malleswari
– Malleswari, Commented Jul 10, 2014 at 12:35
@Malleswari You may convert JSON to a map, modify it and then convert back. — Valentyn Kolesnikov
– Valentyn Kolesnikov, Commented Jan 11, 2020 at 5:29

Bruno Grieder · Accepted Answer · 2016-03-21 11:34:03Z

136

Use the (excellent) JSON-Java library from json.org then

JSONObject json = new JSONObject(str);
String xml = XML.toString(json);

toString can take a second argument to provide the name of the XML root node.

This library is also able to convert XML to JSON using XML.toJSONObject(java.lang.String string)

Check the Javadoc

Link to the the github repository

POM

<dependency>
    <groupId>org.json</groupId>
    <artifactId>json</artifactId>
    <version>20160212</version>
</dependency>

original post updated with new links

edited Mar 21, 2016 at 11:34

answered Nov 14, 2013 at 12:59

Bruno Grieder

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13 Comments

vinod Over a year ago

thanks for it. how to write json object to add attrubute to element like <Private Provider="AB"/>

Bruno Grieder Over a year ago

I guess the only way will be to deserialize the JSON to a (custom) Java Object. Then, using a framework like XMLBeans or XStream and with the help of annotations, specify which property goes as element and which goes as attribute

Bruno Grieder Over a year ago

Alternatively, use the XML from the simple two lines above and apply an XSLT to it

Christophe Roussy Over a year ago

Very convenient for simple one-to-one JSON to XML conversion.

Fabian Zeindl Over a year ago

This does not work for me, I get invalid xml out of valid json.

|

Valentyn Kolesnikov · Accepted Answer · 2024-11-14 13:22:07Z

9

Underscore-java library has static method U.jsonToXml(jsonstring). Live example

import com.github.underscore.U;

public class MyClass {
    public static void main(String[] args) {
        String json = "{\"name\":\"JSON\",\"integer\":1,\"double\":2.0,\"boolean\":true,\"nested\":{\"id\":42},\"array\":[1,2,3]}";
        String xml = U.jsonToXmlMinimum(json);
        System.out.println(xml);
    }
}

Output:

<root>
  <name>JSON</name>
  <integer>1</integer>
  <double>2.0</double>
  <boolean>true</boolean>
  <nested>
    <id>42</id>
  </nested>
  <array>1</array>
  <array>2</array>
  <array>3</array>
</root>

edited Nov 14, 2024 at 13:22

answered Jan 1, 2018 at 10:10

Valentyn Kolesnikov

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3 Comments

DubZ Over a year ago

thx for this. especially with mixed json object array structure its much easier than the top rated answer

PAA Over a year ago

@Valentyn - How to add

soapenv:Envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/" xmlns:soapenc="http://schemas.xmlsoap.org/soap/encoding/"

while converting Json to XML ?

Dinith Rukshan Kumara Over a year ago

Thanks. This is best solution in 2020. Some other libraries not work.

jschnasse · Accepted Answer · 2021-10-25 06:26:45Z

7

For json to xml use the following Jackson example:

final String str = "{\"name\":\"JSON\",\"integer\":1,\"double\":2.0,\"boolean\":true,\"nested\":{\"id\":42},\"array\":[1,2,3]}";
ObjectMapper jsonMapper = new ObjectMapper();
JsonNode node = jsonMapper.readValue(str, JsonNode.class);
XmlMapper xmlMapper = new XmlMapper();
                xmlMapper.configure(SerializationFeature.INDENT_OUTPUT, true);
                xmlMapper.configure(ToXmlGenerator.Feature.WRITE_XML_DECLARATION, true);
                xmlMapper.configure(ToXmlGenerator.Feature.WRITE_XML_1_1, true);
ObjectWriter ow = xmlMapper.writer().withRootName("root");
StringWriter w = new StringWriter();
ow.writeValue(w, node);
System.out.println(w.toString());

Prints:

<?xml version='1.1' encoding='UTF-8'?>
<root>
  <name>JSON</name>
  <integer>1</integer>
  <double>2.0</double>
  <boolean>true</boolean>
  <nested>
    <id>42</id>
  </nested>
  <array>1</array>
  <array>2</array>
  <array>3</array>
</root>

To convert it back (xml to json) take a look at this answer https://stackoverflow.com/a/62468955/1485527 .

edited Oct 25, 2021 at 6:26

answered Aug 24, 2020 at 12:28

jschnasse

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3 Comments

Simon Logic Over a year ago

You'd better show how to override "ObjectNode".

jschnasse Over a year ago

@SimonLogic Good hint! 'Feature' added.

Elliot Vargas Over a year ago

Is there a way to convert an array of objects in the JSON to a wrapped list in the XML? Say the JSON has "objects": [{"value": 1}, {"value": 2}], is it possible to get <objects><object><value>1</value></object><object><value>2</value></object></objects>?

Piyush Chordia · Accepted Answer · 2016-05-12 08:53:00Z

4

If you have a valid dtd file for the xml then you can easily transform json to xml and xml to json using the eclipselink jar binary.

Refer this: http://www.cubicrace.com/2015/06/How-to-convert-XML-to-JSON-format.html

The article also has a sample project (including the supporting third party jars) as a zip file which can be downloaded for reference purpose.

edited May 12, 2016 at 8:53

answered Jun 10, 2015 at 5:20

Piyush Chordia

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Comments

Martynas Jusevičius · Accepted Answer · 2017-12-08 14:37:46Z

0

Transforming with XSLT 3.0 is the only proper way to do it, as far as I can tell. It is guaranteed to produce valid XML, and a nice structure at that. https://www.w3.org/TR/xslt/#json

answered Dec 8, 2017 at 14:37

Martynas Jusevičius

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1 Comment

Martynas Jusevičius Over a year ago

This conversion now can also be done using our library: github.com/AtomGraph/JSON2XML

Saurabh · Accepted Answer · 2017-07-02 11:53:19Z

-2

If you want to replace any node value you can do like this

JSONObject json = new JSONObject(str);
String xml = XML.toString(json);
xml.replace("old value", "new value");

edited Jul 2, 2017 at 11:53

user3559349

answered Jul 2, 2017 at 11:30

Saurabh

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