Ignore a key in JSON to XML conversion

Question

I am using json library to convert json to xml but while converting I want to ignore a nested json object to be converted to xml tags.

eg.

Plane json is as :

{"id":"9568","name":"Customer Analysis","group":"demo","param":{"globalSettings":{"showLegends":false,"legendPosition":"bottom center"}}}

JSONObject json = new JSONObject("{\"id\":\"9568\",\"name\":\"Customer Analysis\",\"group\":\"demo\",\"param\":{\"globalSettings\":{\"showLegends\":false,\"legendPosition\":\"bottom center\"}}}");

    String xml = XML.toString(json);
    System.out.println(xml);

Now in above example, I want in xml with a json as it is inside. Whereas now various elements are created for showLegends and legendPosition inside globalSettings.

Current XML is as follows :

<name>Customer Analysis</name>
<id>9568</id>
<group>demo</group>
<param>
  <globalSettings>
     <showLegends>false</showLegends>
     <legendPosition>bottom center</legendPosition>
  </globalSettings>
</param>

Expected XML should be as follows :

<name>Customer Analysis</name>
<id>9568</id>
<group>demo</group>
<param>
  <globalSettings>
     {"showLegends":false,"legendPosition":"bottom center"}
  </globalSettings>
</param>

How can I handle this?

what is the expected XML output for the above JSON example? can you please add it to the post? there is a way to ignore some properties in the JSON if you are using POJO - https://www.baeldung.com/jackson-ignore-properties-on-serialization https://stackoverflow.com/questions/5455014/ignoring-new-fields-on-json-objects-using-jackson — Adi Ohana, Apr 30 '19 at 12:47
@PratikRawlekar since you have `var` local variable. Can't you just remove the key and value which you don't require before you convert it to json and then before converting to xml trying to ignore a key? — Yug Singh, Apr 30 '19 at 12:51
@PratikRawlekar Can you also post the **current output** you are getting? — Yug Singh, Apr 30 '19 at 12:52
@PratikRawlekar it seems that your expected o/p for the given `var` is not even a valid xml. — Yug Singh, Apr 30 '19 at 13:00
@YugSingh this is just an example and not the real life scenario which I am facing. — Pratik Rawlekar, Apr 30 '19 at 13:09
@PratikRawlekar IMO you need to edit the question and clearly mention what are your expectations. Even if you are doing above for learning but should use correct examples only. — Yug Singh, Apr 30 '19 at 13:13
@YugSingh updated with latest example in which I am facing an issue — Pratik Rawlekar, Apr 30 '19 at 13:34

Madplay · Accepted Answer · 2019-05-03T00:04:51.727

I think you need to tweak JSON before converting. Can you try this below?

String json = "{\n" +
        "   \"user\": \"gerry\",\n" +
        "   \"likes\": [1, 2, 4],\n" +
        "   \"followers\": {\n" +
        "      \"options\": {\n" +
        "        \"key1\": \"a\",\n" +
        "        \"key2\": \"b\"\n" +
        "      }        \n" +
        "   }\n" +
        "}";

JSONObject jsonObject = new JSONObject(json);
JSONObject followers = jsonObject.getJSONObject("followers");

String options = followers.optString("options");
followers.put("options", options);

String s = XML.toString(jsonObject);
System.out.println(XML.unescape(s));

result:

<followers><options>{"key1":"a","key2":"b"}</options></followers><user>gerry</user><likes>[1,2,4]</likes>

Extra Question:

What if I don't want options as an xml element and it should be part of json?

String json = "{\n" +
        "   \"user\": \"gerry\",\n" +
        "   \"likes\": [1, 2, 4],\n" +
        "   \"followers\": {\n" +
        "      \"options\": {\n" +
        "        \"key1\": \"a\",\n" +
        "        \"key2\": \"b\"\n" +
        "      }        \n" +
        "   }\n" +
        "}";

JSONObject jsonObject = new JSONObject(json);
jsonObject.put("followers", jsonObject.optString("followers"));

// org.json 20180813
String s = XML.toString(jsonObject);
System.out.println(XML.unescape(s));

result:

<followers>{"options":{"key1":"a","key2":"b"}}</followers><user>gerry</user><likes>1</likes><likes>2</likes><likes>4</likes>

Thanks for your response. I am not able to find unescape() on XML but we can achieve it with StringEscapeUtils.unescapeXml(s) — Pratik Rawlekar, Apr 30 '19 at 13:50
When I try to convert this xml to json using JSONObject obj = XML.toJSONObject(s) , \ will be printed in the options string. How can I overcome this? Also in above solution given by you, What if I don't want options as an xml element and it should be part of json? — Pratik Rawlekar, May 02 '19 at 06:49
@PratikRawlekar I tested it, but \ was not printed. I used org.json version ```20180813```. — Madplay, May 02 '19 at 23:55
My current version is 20150729 but I also tried with udpdated version 20180813. Still facing the same issue when I convert it to json from xml. — Pratik Rawlekar, May 03 '19 at 07:03

Adi Ohana · Answer 2 · 2019-04-30T13:14:33.487

0

You need to modify your json a bit:

    String json = "{\"user\":\"gerry\",\"likes\":[1,2,4],\"followers\":{\"options\":\"{key1:a,key2:b}\"}}";

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

    System.out.println(XML.unescape(xml));

note that the "options" should be as "String" (shaped like JSON), then the XML parser will treat it as a regular string.

edited Apr 30 '19 at 13:14

answered Apr 30 '19 at 13:08

Adi Ohana

927
2
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score 0 · Answer 3 · answered Apr 30 '19 at 13:17

0

You either need to tweak the JSON before conversion, or tweak the XML after conversion.

Since such tweaking is so often required, one approach is to do the conversion using XSLT 3.0 so you have transformation capability built-in to the tool.

answered Apr 30 '19 at 13:17

Michael Kay

156,231
11
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Valentyn Kolesnikov · Answer 4 · 2020-01-02T09:27:28.937

Underscore-java library has static methods U.fromJson(json) and U.toXml(map). You may modify map and generate xml. I am the maintainer of the project.

    Map<String, Object> map = U.fromJsonMap("{\"id\":\"9568\",\"name\":\"Customer Analysis\",\"group\":\"demo\",\"param\":{\"globalSettings\":{\"showLegends\":false,\"legendPosition\":\"bottom center\"}}}");
    U.set(map, "param.globalSettings", U.toJson((Map<String, Object>) U.get(map, "param.globalSettings")));
    System.out.println(U.toXml(map));

Output:

<?xml version="1.0" encoding="UTF-8"?>
<root>
  <id>9568</id>
  <name>Customer Analysis</name>
  <group>demo</group>
  <param>
    <globalSettings>{
  "showLegends": false,
  "legendPosition": "bottom center"
}</globalSettings>
  </param>
</root>

Ignore a key in JSON to XML conversion

4 Answers4