json might not be the best choice for on-disk formats; The trouble it has with appending data is a good example of why this might be. Specifically, json objects have a syntax that means the whole object must be read and parsed in order to understand any part of it.

Fortunately, there are lots of other options. A particularly simple one is CSV; which is supported well by python's standard library. The biggest downside is that it only works well for text; it requires additional action on the part of the programmer to convert the values to numbers or other formats, if needed.

Another option which does not have this limitation is to use a sqlite database, which also has built-in support in python. This would probably be a bigger departure from the code you already have, but it more naturally supports the 'modify a little bit' model you are apparently trying to build.

Update

Oh seems I forgot to post the original json file. Pls put the student.json with below content: (Need to wrap with [ & ] to make it an array)

[{"name":"Mahesh","age":10}]

Because you can not just append objects to original object, it will violate JSON syntax, the way to make it correct is to put them into an array.

Original answer

You should change to use an array to store multiple objects. And use FileWriter to output.

Pls try this code, it works in my locl env:

import org.json.simple.JSONArray;
import org.json.simple.JSONObject;

import org.json.simple.parser.JSONParser;
import org.json.simple.parser.ParseException;

import java.io.*;

public class JsonWritingProject {

    public static void main(String[] args) {
        JSONParser jsonParser = new JSONParser();

        try {
            Object obj = jsonParser.parse(new FileReader("D:\\student.json"));
            JSONArray jsonArray = (JSONArray)obj;

            System.out.println(jsonArray);

            JSONObject student1 = new JSONObject();
            student1.put("name", "BROCK");
            student1.put("age", new Integer(3));

            JSONObject student2 = new JSONObject();
            student2.put("name", "Joe");
            student2.put("age", new Integer(4));

            jsonArray.add(student1);
            jsonArray.add(student2);

            System.out.println(jsonArray);

            FileWriter file = new FileWriter("D:\\student.json");
            file.write(jsonArray.toJSONString());
            file.flush();
            file.close();

        } catch (ParseException | IOException e) {
            e.printStackTrace();
        }

    }
}

And the library dependency is:

dependencies {
    compile group: 'com.googlecode.json-simple', name: 'json-simple', version: '1.1'
}

I found the solution, thanks to the comments by other users and to a "retired" answer, not present here anymore. Maybe it was my fault not being clear.

public void addEntryToJsonFile(Context ctx, String id, String name, String size) {

    // parse existing/init new JSON 
    File jsonFile = new File(ctx.getDir("my_data_dir", 0), "my_json_file.json");
    String previousJson = null;
    if (jsonFile.exists()) {
        try {
            previousJson = Utils.readFromFile(jsonFile);
        } catch (IOException e) {
            e.printStackTrace();
        }
    } else {
        previousJson = "{}";
    }

    // create new "complex" object
    JSONObject mO = null;
    JSONObject jO = new JSONObject();

    try {
        mO = new JSONObject(previousJson);
        jO.put("completed", true);
        jO.put("name", name);
        jO.put("size", size);
        mO.put(id, jO); //thanks "retired" answer
    } catch (JSONException e) {
        e.printStackTrace();
    }

    // generate string from the object
    String jsonString = null;
    try {
        jsonString = mO.toString(4);
    } catch (JSONException e) {
        e.printStackTrace();
    }

    // write back JSON file
    Utils.writeToFile(jsonFile, jsonString);

}

Try this. Don't forget to define anchors array.

var data = fs.readFileSync('testOutput.json');
var json = JSON.parse(data);
json.push(...anchors);

fs.writeFile("testOutput.json", JSON.stringify(json))

Consider closing files when you're finished (or use with-as). Also not sure why you calculate position to truncate (on 'a' mode) at the end of appending.

import csv
import json

fieldnames = ("CustomerNo", "SalesOrder", "Brand", "Item", "Qty", "Month", "time")

# Handle the case when there is no newline at the end. Format a file before appending.

with open('daily.csv', 'r') as csvfile1:
    with open('test3.json', 'a') as jsonfile1:
        for row in csv.DictReader(csvfile1, fieldnames):
            json.dump(row, jsonfile1)
            jsonfile1.write('\n')

If you want the file to be valid JSON, you have to open your file, parse the JSON, append your new result to the array, transform it back into a string and save it again.

var fs = require('fs')

var currentSearchResult = 'example'

fs.readFile('results.json', function (err, data) {
    var json = JSON.parse(data)
    json.push('search result: ' + currentSearchResult)

    fs.writeFile("results.json", JSON.stringify(json))
})

Use [] to access a dynamic key on the object

json.users[username] = {a: 1, b: 2}

Be careful naming your variable like that tho because json the way you're using it is not JSON. JSON is a string, not an object with keys.

See the below demo for distinction

var json = '{"users":{"test1":{},"test2":{}}}';
var obj  = JSON.parse(json);
var newuser = 'test3';
obj.users[newuser] = {};
console.log(JSON.stringify(obj));
//=> {"users":{"test1":{},"test2":{},"test3":{}}}

Assuming you have a test.json file with the following content:

{"67790": {"1": {"kwh": 319.4}}}

Then, the code below will load the json file, update the data inside using dict.update() and dump into the test.json file:

import json

a_dict = {'new_key': 'new_value'}

with open('test.json') as f:
    data = json.load(f)

data.update(a_dict)

with open('test.json', 'w') as f:
    json.dump(data, f)

Then, in test.json, you'll have:

{"new_key": "new_value", "67790": {"1": {"kwh": 319.4}}}

Hope this is what you wanted.