Parsing JSON Object in Java [duplicate]

Question

This question already has an answer here:

• How to parse JSON in Java 31 answers

I have JSON object as follows:

member = "{interests : [{interestKey:Dogs}, {interestKey:Cats}]}";

In Java I want to parse the above jsonobject and store the values in an arraylist.

Can any one please provide me some code snippet through which i can achieve this.

Thanks

Answer 1

public class JsonParsing {

public static Properties properties = null;

public static JSONObject jsonObject = null;

static {
    properties = new Properties();
}

public static void main(String[] args) {

    try {

        JSONParser jsonParser = new JSONParser();

        File file = new File("src/main/java/read.json");

        Object object = jsonParser.parse(new FileReader(file));

        jsonObject = (JSONObject) object;

        parseJson(jsonObject);

    } catch (Exception ex) {
        ex.printStackTrace();
    }
}

public static void getArray(Object object2) throws ParseException {

    JSONArray jsonArr = (JSONArray) object2;

    for (int k = 0; k < jsonArr.size(); k++) {

        if (jsonArr.get(k) instanceof JSONObject) {
            parseJson((JSONObject) jsonArr.get(k));
        } else {
            System.out.println(jsonArr.get(k));
        }

    }
}

public static void parseJson(JSONObject jsonObject) throws ParseException {

    Set<Object> set = jsonObject.keySet();
    Iterator<Object> iterator = set.iterator();
    while (iterator.hasNext()) {
        Object obj = iterator.next();
        if (jsonObject.get(obj) instanceof JSONArray) {
            System.out.println(obj.toString());
            getArray(jsonObject.get(obj));
        } else {
            if (jsonObject.get(obj) instanceof JSONObject) {
                parseJson((JSONObject) jsonObject.get(obj));
            } else {
                System.out.println(obj.toString() + "\t"
                        + jsonObject.get(obj));
            }
        }
    }
}}

Answer 2

I'm assuming you want to store the interestKeys in a list.

Using the org.json library:

JSONObject obj = new JSONObject("{interests : [{interestKey:Dogs}, {interestKey:Cats}]}");

List<String> list = new ArrayList<String>();
JSONArray array = obj.getJSONArray("interests");
for(int i = 0 ; i < array.length() ; i++){
    list.add(array.getJSONObject(i).getString("interestKey"));
}

Answer 3

There are many JSON libraries available in Java.

The most notorious ones are: Jackson, GSON, Genson, FastJson and org.json.

There are typically three things one should look at for choosing any library:

1. Performance
2. Ease of use (code is simple to write and legible) - that goes with features.
3. For mobile apps: dependency/jar size

Specifically for JSON libraries (and any serialization/deserialization libs), databinding is also usually of interest as it removes the need of writing boiler-plate code to pack/unpack the data.

For 1, see this benchmark: https://github.com/fabienrenaud/java-json-benchmark I did using JMH which compares (jackson, gson, genson, fastjson, org.json, jsonp) performance of serializers and deserializers using stream and databind APIs. For 2, you can find numerous examples on the Internet. The benchmark above can also be used as a source of examples...

Quick takeaway of the benchmark: Jackson performs 5 to 6 times better than org.json and more than twice better than GSON.

For your particular example, the following code decodes your json with jackson:

public class MyObj {

    private List<Interest> interests;

    static final class Interest {
        private String interestKey;
    }

    private static final ObjectMapper MAPPER = new ObjectMapper();
    public static void main(String[] args) throws IOException {
        MyObj o = JACKSON.readValue("{\"interests\": [{\"interestKey\": \"Dogs\"}, {\"interestKey\": \"Cats\" }]}", MyObj.class);
    }
}

Let me know if you have any questions.

Answer 4

Thank you soooooooo much to CODE I can read any JSON file thanks to your code :) Now, I'm trying to organize all the elements by levels, for could use them!

I was working with Android reading a JSON from an URL and the only I had to change was the lines

Set<Object> set = jsonObject.keySet(); Iterator<Object> iterator = set.iterator();

for

Iterator<?> iterator = jsonObject.keys();

I share my implementation, to help someone:

public void parseJson(JSONObject jsonObject) throws ParseException, JSONException {

    Iterator<?> iterator = jsonObject.keys();
    while (iterator.hasNext()) {
        String obj = iterator.next().toString();

        if (jsonObject.get(obj) instanceof JSONArray) {
            //Toast.makeText(MainActivity.this, "Objeto: JSONArray", Toast.LENGTH_SHORT).show();
            //System.out.println(obj.toString());
            TextView txtView = new TextView(this);
            txtView.setText(obj.toString());
            layoutIzq.addView(txtView);
            getArray(jsonObject.get(obj));
        } else {
            if (jsonObject.get(obj) instanceof JSONObject) {
                //Toast.makeText(MainActivity.this, "Objeto: JSONObject", Toast.LENGTH_SHORT).show();
                parseJson((JSONObject) jsonObject.get(obj));
            } else {
                //Toast.makeText(MainActivity.this, "Objeto: Value", Toast.LENGTH_SHORT).show();
                //System.out.println(obj.toString() + "\t"+ jsonObject.get(obj));
                TextView txtView = new TextView(this);
                txtView.setText(obj.toString() + "\t"+ jsonObject.get(obj));
                layoutIzq.addView(txtView);
            }
        }
    }
}

Answer 5

1.) Create an arraylist of appropriate type, in this case i.e String

2.) Create a JSONObject while passing your string to JSONObject constructor as input

• As JSONObject notation is represented by braces i.e {}
• Where as JSONArray notation is represented by square brackets i.e []

3.) Retrieve JSONArray from JSONObject (created at 2nd step) using "interests" as index.

4.) Traverse JASONArray using loops upto the length of array provided by length() function

5.) Retrieve your JSONObjects from JSONArray using getJSONObject(index) function

6.) Fetch the data from JSONObject using index '"interestKey"'.

Note : JSON parsing uses the escape sequence for special nested characters if the json response (usually from other JSON response APIs) contains quotes (") like this

`"{"key":"value"}"`

should be like this

`"{\"key\":\"value\"}"`

so you can use JSONParser to achieve escaped sequence format for safety as

JSONParser parser = new JSONParser();
JSONObject json = (JSONObject) parser.parse(inputString);

---

Code :

JSONParser parser = new JSONParser();
String response = "{interests : [{interestKey:Dogs}, {interestKey:Cats}]}";

JSONObject jsonObj = (JSONObject) parser.parse(response);

or

JSONObject jsonObj = new JSONObject("{interests : [{interestKey:Dogs}, {interestKey:Cats}]}");

---

List<String> interestList = new ArrayList<String>();

JSONArray jsonArray = jsonObj.getJSONArray("interests");

for(int i = 0 ; i < jsonArray.length() ; i++){
    interestList.add(jsonArray.getJSONObject(i).optString("interestKey"));
}

---

Note : Sometime you may see some exceptions when the values are not available in appropriate type or is there is no mapping key so in those cases when you are not sure about the presence of value so use optString, optInt, optBoolean etc which will simply return the default value if it is not present and even try to convert value to int if it is of string type and vice-versa so Simply No null or NumberFormat exceptions at all in case of missing key or value

From docs

Get an optional string associated with a key. It returns the defaultValue if there is no such key.

 public String optString(String key, String defaultValue) {

  String missingKeyValue = json_data.optString("status","N/A");
  // note there is no such key as "status" in response
  // will return "N/A" if no key found 

or To get empty string i.e "" if no key found then simply use

  String missingKeyValue = json_data.optString("status");
  // will return "" if no key found where "" is an empty string

Further reference to study

• How to convert String to JSONObject in Java
• Convert one array list item into multiple Items