How to parse a JSON and turn its values into an Array?

Question

public static void parseProfilesJson(String the_json){
       try {
            JSONObject myjson = new JSONObject(the_json);

            JSONArray nameArray = myjson.names();
            JSONArray valArray = myjson.toJSONArray(nameArray);
            for(int i=0;i<valArray.length();i++)
            {
                String p = nameArray.getString(i) + "," + ValArray.getString(i);
                Log.i("p",p);
            }       

        } catch (JSONException e) {
                e.printStackTrace();
        }
    }

As you can see, this sample code will print out the KEY of the JSONs, followed by the VALUES of the JSONS.

It would print profiles, john if the json was like this:

{'profiles':'john'}

That's cool. That's fine, as I can work with those variables. However, what if the JSON was like this:

{'profiles': [{'name':'john', 'age': 44}, {'name':'Alex','age':11}]}

In this case, the entire value would be the array. Basically, I just want to grab that array (which is the "value" in this case)...and turn it into an actual array that JAVA could use. How can I do that? Thanks.

Be aware that this wouldn't pass a stricter parser. JSON requires double-quotes around names and strings (single-quote is not valid). — Jonathon Faust, Feb 12 '10 at 21:44
Don't worry Jonathon...JSONObject is a JSON.org code provided for JSON parsing in java. What alex put was just example.... — Buhake Sindi, Feb 12 '10 at 21:47
I gave you a point for providing simple code for me to parse my json string. — crafter, Jul 04 '11 at 00:39

score 58 · Accepted Answer · edited Dec 05 '11 at 01:58

for your example:

{'profiles': [{'name':'john', 'age': 44}, {'name':'Alex','age':11}]}

you will have to do something of this effect:

JSONObject myjson = new JSONObject(the_json);
JSONArray the_json_array = myjson.getJSONArray("profiles");

this returns the array object.

Then iterating will be as follows:

    int size = the_json_array.length();
    ArrayList<JSONObject> arrays = new ArrayList<JSONObject>();
    for (int i = 0; i < size; i++) {
        JSONObject another_json_object = the_json_array.getJSONObject(i);
            //Blah blah blah...
            arrays.add(another_json_object);
    }

//Finally
JSONObject[] jsons = new JSONObject[arrays.size()];
arrays.toArray(jsons);

//The end...

You will have to determine if the data is an array (simply checking that charAt(0) starts with [ character).

Hope this helps.

What JSON library is being used here? `http://javadox.com/net.minidev/json-smart/1.2/net/minidev/json/JSONObject.html` is it this one? — greg, Apr 10 '19 at 16:59
I've tried this but Im only able to get ` [{'name':'john', 'age': 44}, {'name':'Alex','age':11}]` is there a way to get only the "name" in an array? — Olufsen, Mar 23 '20 at 06:09

score 1 · Answer 2 · edited Nov 13 '13 at 07:13

You can prefer quick-json parser to meet your requirement...

quick-json parser is very straight forward, flexible, very fast and customizable. Try this out

[quick-json parser] (https://code.google.com/p/quick-json/) - quick-json features -

Compliant with JSON specification (RFC4627)
High-Performance JSON parser
Supports Flexible/Configurable parsing approach
Configurable validation of key/value pairs of any JSON Heirarchy
Easy to use # Very Less foot print
Raises developer friendly and easy to trace exceptions
Pluggable Custom Validation support - Keys/Values can be validated by configuring custom validators as and when encountered
Validating and Non-Validating parser support
Support for two types of configuration (JSON/XML) for using quick-json validating parser
Require JDK 1.5 # No dependency on external libraries
Support for Json Generation through object serialization
Support for collection type selection during parsing process

For e.g.

JsonParserFactory factory=JsonParserFactory.getInstance();
JSONParser parser=factory.newJsonParser();
Map jsonMap=parser.parseJson(jsonString);

How to parse a JSON and turn its values into an Array?

2 Answers2

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