CODEWITHSUNDEEP
javajsonxml
dina
dina

Reputation: 4289

Convert XML to JSON ignoring attributes

I'm trying to convert an XML to JSON in JAVA removing the tag attributes from the XML.

I tried using org.json.XML but it did not meet my needs.

Is there a library for doing what I want to do?

Example input:

<?xml version="1.0"?>
<company g="j">
    <staff id="1001">
        <firstname hi="5">jim</firstname>
        <lastname>fox</lastname>
    </staff>
    <staff id="2001">
        <firstname a="7">jay</firstname>
        <details tmp="0">
            <lastname>box</lastname>
            <nickname >fong fong</nickname>
            <salary id="99">200000</salary>
        </details>
    </staff>
</company>

Desired output:

{
    "company": {
        "staff": [
            {
                "firstname": "jim"
                "lastname": "fox",
            },
            {
                "firstname": "jay",
                "details": {
                    "lastname": "box",
                    "nickname": "fong fong",
                    "salary":"200000",
            }
        ]
    }
}

I tried the following but it convert the xml using the attributes:

package my.transform.data.utils;

import java.io.File;
import org.apache.commons.io.FileUtils;
import org.json.XML;
import org.json.JSONObject;

public class JSONObjectConverter {

    public static void main(String[] args) throws Exception {

        String xml = FileUtils.readFileToString(new File("src/main/resources/staff.xml"));
        JSONObject aJson = XML.toJSONObject(xml);
        System.out.println(aJson.toString());

    }

}

any suggestions?

Upvotes: 2

Views: 3625

Answers (3)

Valentyn Kolesnikov
Valentyn Kolesnikov

Reputation: 2097

Underscore-java library has static methods U.fromXmlWithoutAttributes(string) and U.toJson(object). Live example

import com.github.underscore.U;
import java.util.Map;

public class JsonConversion {
    @SuppressWarnings("unchecked")
    public static void main(String args[]) {
        String xmlString  = "<?xml version=\"1.0\"?>"
        + "<company g=\"j\">"
        + "    <staff id=\"1001\">"
        + "        <firstname hi=\"5\">jim</firstname>"
        + "        <lastname>fox</lastname>"
        + "    </staff>"
        + "    <staff id=\"2001\">"
        + "        <firstname a=\"7\">jay</firstname>"
        + "        <details tmp=\"0\">"
        + "            <lastname>box</lastname>"
        + "            <nickname >fong fong</nickname>"
        + "            <salary id=\"99\">200000</salary>"
        + "        </details>"
        + "    </staff>"
        + "</company>"; 

        Map<String, Object> map = (Map) U.fromXmlWithoutAttributes(xmlString);  
        System.out.println(U.toJson(map));  
    }
}

Output:

{
  "company": {
    "staff": [
      {
        "firstname": "jim",
        "lastname": "fox"
      },
      {
        "firstname": "jay",
        "details": {
          "lastname": "box",
          "nickname": "fong fong",
          "salary": "200000"
        }
      }
    ]
  }
}

Upvotes: 1

Michael Kay
Michael Kay

Reputation: 163262

Try doing an XSLT transformation to get the XML into the desired form before conversion. (You could also consider using the XSLT 3.0 xml-to-json() function).

I think it's very likely that any general-purpose converter will do exactly what you want without pre- or post- processing.

Upvotes: 1

Vasu
Vasu

Reputation: 22394

You need to use JAXB to unmarshal the xml content to a java object and then use that java object to prepare the JSON.

JAXB converts the given xml to a java object (this is called unmarshalling) and then that java object can be used to form the JSON

You can refer the below code snippet:

public class JAXBToJsonConverter {
     public static void main(String[] args) {
        try {  
            //save the company details content to a .xml file
            // and refer the path below
            File file = new File("C:\\myproject\\company.xml");  

            //create the jaxb context and unmarshall
            JAXBContext jaxbContext = JAXBContext.newInstance(Company.class);  

            Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();  
            Company company= (Company) jaxbUnmarshaller.unmarshal(file);  

            //create the JSON object
            JSONObject json = new JSONObject(company);
            System.out.println(json);
          } catch (JAXBException e) {  
            e.printStackTrace();  
          }  
    }
  }

Company class:

import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlRootElement;

@XmlRootElement
public class Company {

    private Staff staff;

    @XmlElement
    public Staff getStaff() {
        return staff;
    }

    public void setStaff(Staff staff) {
        this.staff = staff;
    }
  }

Staff class:

public class Staff {
    private String firstname;
    private String lastname;

    @XmlElement
    public String getFirstname() {
        return firstname;
    }
    public void setFirstname(String firstname) {
        this.firstname = firstname;
    }

    @XmlElement
    public String getLastname() {
        return lastname;
    }
    public void setLastname(String lastname) {
        this.lastname = lastname;
    }
}

Details class:

 public class Details {
     private String lastname;
     private String nickname;
     private int salary;

@XmlElement
public String getLastname() {
   return lastname;
}
public void setLastname(String lastname) {
    this.lastname = lastname;
}

 @XmlElement
 public String getNickname() {
     return nickname;
 }
 public void setNickname(String nickname) {
    this.nickname = nickname;
 }

 @XmlElement
 public int getSalary() {
    return salary;
}
public void setSalary(int salary) {
    this.salary = salary;
}
}

I need something more dynamic, since my xml is in a different structure every time.

You can have a look at here which uses staxon:

https://github.com/beckchr/staxon/wiki/Converting-XML-to-JSON

Upvotes: 2

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