Java XML to JSON using XMLSerializer making root element an Array

Question

Given that I have this XML:

<?xml version="1.0" encoding="UTF-8"?>
<Root>
    <op>update</op>
    <path>someString</path>
    <value>12345</value>
</Root>

and I want this output in JSON:

[{ "op":"update", "path":"someString", "value":"12345" }]

I have tried the following code:

package jsonconvertor;
import net.sf.json.JSON;
import net.sf.json.xml.XMLSerializer;

public class JSONConvertor {

    public static void main(String[] args) {

        String input = "<?xml version=\"1.0\" encoding=\"UTF-8\"?><Root><op>update</op><path>someString</path><value>12345</value></Root>";
        String output = "";

        XMLSerializer xml = new XMLSerializer();
        JSON jObj = xml.read( input );
        output = jObj.toString();

        System.out.println("My JSON:\n" + output);
    }
}

When I run that code I get the following response:

{"op":"update","path":"someString","value":"12345"}

However you will notice that the leading '[' and closing ']' are missing.

When I try changing the XML "input" string in the code to the following:

String input = "<?xml version=\"1.0\" encoding=\"UTF-8\"?><Root class=\"object\"><op>update</op><path>someString</path><value>12345</value></Root>";

which makes sense since itRoot was an object to start with, so I tried changing the "input" string to "array":

String input = "<?xml version=\"1.0\" encoding=\"UTF-8\"?><Root class=\"array\"><op>update</op><path>someString</path><value>12345</value></Root>";

however then I get the following:

["update","someString","12345"]

What am I missing? I want the output I get when class="object" however I want it enclosed in square brackets. I want the output to look like the example at the top of the post.

Try with this : String input2 = "<?xml version=\"1.0\" encoding=\"UTF-8\"?><Root class=\"array\"><SubRoot class=\"object\"><op>update</op><path>someString</path><value>12345</value></SubRoot></Root>"; — Abhishek
– Abhishek, Commented Nov 2, 2018 at 6:38
I can't thank you enough! .. This is the solution to my problem! I will certainly accept this as the solution I was looking for if you add it again as an answer! — AcidHawk
– AcidHawk, Commented Nov 2, 2018 at 6:44
Yes I've added as answer you can accept it . Happy to help! :) — Abhishek
– Abhishek, Commented Nov 2, 2018 at 6:45

Abhishek · Accepted Answer · 2018-11-02 06:43:17Z

1

Try with this input String it should work :

String input = "<?xml version=\"1.0\" encoding=\"UTF-8\"?><Root class=\"array\"><SubRoot class=\"object\"><op>update</op><path>someString</path><value>12345</value></SubRoot></Root>";

answered Nov 2, 2018 at 6:43

Abhishek

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Deepak Gunasekaran · Accepted Answer · 2018-11-02 06:34:37Z

0

In the given XML, you have only 1 ROOT element , so it's a single JSONObject and not a JSONArray. So, if you want a JSONArray as Output, you should provide XML Array as input to the xml.read( input ) method.

If your input looks like below, you will get your desired output.

<?xml version="1.0" encoding="UTF-8"?>
<Root>
    <SubRoot>
         <op>update</op>
         <path>someString</path>
         <value>12345</value>
    <SubRoot>
</Root>

Output :

[{"op":"update","path":"someString","value":"12345"}]

answered Nov 2, 2018 at 6:34

Deepak Gunasekaran

7575 silver badges13 bronze badges

4 Comments

AcidHawk Over a year ago

Thank you, however when adding String input = "<?xml version=\"1.0\" encoding=\"UTF-8\"?><Root><SubRoot><op>update</op><path>someString</path><value>12345</value></SubRoot></Root>"; I get the following result: {"SubRoot":["update","someString","12345"]}

Deepak Gunasekaran Over a year ago

OMG. But i am getting the output i pasted here which looks same as you. I think the difference is because of the dependency jar version's.Let me check again

samabcde Over a year ago

@Deepak Gunasekaran try to add class="array" to Root.

Deepak Gunasekaran Over a year ago

Okay @Samabcde . Thanks :)

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