Sharda Singh Sharda Singh - 1 month ago 16
JSON Question

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?

Answer

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.

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