Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

Decoding JSON String in Java

I am new to using the json-simple library in Java and I've been through both the encoding and decoding samples. Duplicating the encoding examples was fine, but I have not been able to get the decoding ones to work with mixed type JSON.

One of my problems is that there are too many classes in the library which are not properly documented, and for which I do not have the source (in order to be able to read through and understand their purpose). Consequently, I am struggling to understand how to use a lot of these classes.

After reading this example:

String jsonText = "{\"first\": 123, \"second\": [4, 5, 6], \"third\": 789}";
JSONParser parser = new JSONParser();

ContainerFactory containerFactory = new ContainerFactory(){
    public List creatArrayContainer() {
        return new LinkedList();
    }

    public Map createObjectContainer() {
        return new LinkedHashMap();
    }                     
};

try {
    Map json = (Map)parser.parse(jsonText, containerFactory);
    Iterator iter = json.entrySet().iterator();
    System.out.println("==iterate result==");

    while(iter.hasNext()) {
        Map.Entry entry = (Map.Entry)iter.next();
        System.out.println(entry.getKey() + "=>" + entry.getValue());
    }

    System.out.println("==toJSONString()==");
    System.out.println(JSONValue.toJSONString(json));
} catch(ParseException pe) {
    System.out.println(pe);
}

from the json-simple official decoding tutorial, I tried to decode this JSON:

{
"stat":{
    "sdr": "MAC address of FLYPORT",
    "rcv": "ff:ff:ff:ff:ff:ff",
    "time": "0000000000000",
    "type": 0,
    "subt": 0,
    "argv": [
        {"type": "6","val": "NetbiosName"},
        {"type": "6","val": "MACaddrFlyport"},
        {"type": "6","val": "FlyportModel"},
        {"type": "1","val": id}
    ]
}
}

I am writing following code to decode:

    String jsonString = "{\"stat\":{\"sdr\": \"aa:bb:cc:dd:ee:ff\",\"rcv\": \"aa:bb:cc:dd:ee:ff\",\"time\": \"UTC in millis\",\"type\": 1,\"subt\": 1,\"argv\": [{1,2},{2,3}]}}";
    JSONObject jsonObject = new JSONObject(jsonString);
    JSONObject newJSON = jsonObject.getJSONObject("stat");
    System.out.println(newJSON);

But it doesn't work. Infact I was not able to get the unmodified example working either, and the original authors have not explained their code.

What is the easiest way to decode this JSON as shown?

like image 541
Sharda Singh Avatar asked May 15 '13 20:05

Sharda Singh


People also ask

What is JSON string in Java?

JSON (JavaScript Object Notation) is a lightweight data-interchange format, and we most commonly use it for client-server communication. It's both easy to read/write and language-independent. A JSON value can be another JSON object, array, number, string, boolean (true/false) or null.

How do you parse a JSON object in Java?

We can convert a JSON to Java Object using the readValue() method of ObjectMapper class, this method deserializes a JSON content from given JSON content String.

What is encode and decode of JSON in Java?

Encoding JSON with Java By default, no order is set. To set a strict element's ordering, the JSONValue. toString() method is used. Let us see a program that shows the basic encoding of JSON data through Java. For using this JSON with Java, you have to implement this jar file: json-simple-1.1.

How do you escape a JSON string in Java?

You can escape String in Java by putting a backslash in double quotes e.g. " can be escaped as \" if it occurs inside String itself. This is ok for a small JSON String but manually replacing each double quote with an escape character for even a medium-size JSON is time taking, boring, and error-prone.


2 Answers

This is the best and easiest code:

public class test
{
    public static void main(String str[])
    {
        String jsonString = "{\"stat\": { \"sdr\": \"aa:bb:cc:dd:ee:ff\", \"rcv\": \"aa:bb:cc:dd:ee:ff\", \"time\": \"UTC in millis\", \"type\": 1, \"subt\": 1, \"argv\": [{\"type\": 1, \"val\":\"stackoverflow\"}]}}";
        JSONObject jsonObject = new JSONObject(jsonString);
        JSONObject newJSON = jsonObject.getJSONObject("stat");
        System.out.println(newJSON.toString());
        jsonObject = new JSONObject(newJSON.toString());
        System.out.println(jsonObject.getString("rcv"));
       System.out.println(jsonObject.getJSONArray("argv"));
    }
}

The library definition of the json files are given here. And it is not same libraries as posted here, i.e. posted by you. What you had posted was simple json library I have used this library.

You can download the zip. And then create a package in your project with org.json as name. and paste all the downloaded codes there, and have fun.

I feel this to be the best and the most easiest JSON Decoding.

like image 78
Veer Shrivastav Avatar answered Oct 21 '22 15:10

Veer Shrivastav


Well your jsonString is wrong.

String jsonString = "{\"stat\":{\"sdr\": \"aa:bb:cc:dd:ee:ff\",\"rcv\": \"aa:bb:cc:dd:ee:ff\",\"time\": \"UTC in millis\",\"type\": 1,\"subt\": 1,\"argv\": [{\"1\":2},{\"2\":3}]}}";

use this jsonString and if you use the same JSONParser and ContainerFactory in the example you will see that it will be encoded/decoded.

Additionally if you want to print your string after stat here it goes:

     try{
        Map json = (Map)parser.parse(jsonString, containerFactory);
        Iterator iter = json.entrySet().iterator();
        System.out.println("==iterate result==");
        Object entry = json.get("stat");
        System.out.println(entry);
      }

And about the json libraries, there are a lot of them. Better you check this.

like image 24
anvarik Avatar answered Oct 21 '22 14:10

anvarik