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    lt  xml version  1 0  encoding  UTF-8   gt   lt o gt  lt array json class  array  gt  lt e json type  number  gt 1 lt  e gt  lt e json type  number  gt 2 lt  e gt  lt e json type  number  gt 3 lt  e gt  lt  array gt  lt boolean json type  boolean  gt true lt  boolean gt  lt double json type  number  gt 2 0 lt  double gt  lt integer json type  number  gt 1 lt  integer gt  lt name json type  string  gt JSON lt  name gt  lt nested json class  object  gt  lt id json type  number  gt 42 lt  id gt  lt  nested gt  lt  o gt    my biggest problem is how to write my own attributes instead of json type  number  and also writing my own sub elements like

User · Answer

Underscore-java library has static method U.jsonToXml(jsonstring). I am the maintainer of the project. Live example

import com.github.underscore.lodash.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]}";  
        System.out.println(json); 
        String xml = U.jsonToXml(json);  
        System.out.println(xml); 
    }
}

Output:

{"name":"JSON","integer":1,"double":2.0,"boolean":true,"nested":{"id":42},"array":[1,2,3]}
<?xml version="1.0" encoding="UTF-8"?>
<root>
  <name>JSON</name>
  <integer number="true">1</integer>
  <double number="true">2.0</double>
  <boolean boolean="true">true</boolean>
  <nested>
    <id number="true">42</id>
  </nested>
  <array number="true">1</array>
  <array number="true">2</array>
  <array number="true">3</array>
</root>

User · Answer

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

User · Answer

For json to xml use the following Jackson example  final String str    quot    quot name  quot    quot JSON  quot    quot integer  quot  1   quot double  quot  2 0   quot boolean  quot  true   quot nested  quot     quot id  quot  42    quot array  quot   1 2 3   quot   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   StringWriter w   new StringWriter    xmlMapper writeValue w  node   System out println w toString      Prints   lt  xml version  1 1  encoding  UTF-8   gt   lt ObjectNode gt     lt name gt JSON lt  name gt     lt integer gt 1 lt  integer gt     lt double gt 2 0 lt  double gt     lt boolean gt true lt  boolean gt     lt nested gt       lt id gt 42 lt  id gt     lt  nested gt     lt array gt 1 lt  array gt     lt array gt 2 lt  array gt     lt array gt 3 lt  array gt   lt  ObjectNode gt   To convert it back  xml to json  take a look at this answer https   stackoverflow com a 62468955 1485527

User · Answer

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

User · Answer

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

User · Answer

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   lt dependency gt       lt groupId gt org json lt  groupId gt       lt artifactId gt json lt  artifactId gt       lt version gt 20160212 lt  version gt   lt  dependency gt    original post updated with new links

[java] Converting JSON to XML in Java

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