Getting Keys and Values of a JSON Array in the specific order

Question

I'm very new to Java, I'm using it to teach my Lego NXT Robot some ways out of a labyrinth. The algorithm parameters shall be outsourced and loaded in the code, so thats why I use JSON. MY JSON file is pretty simple (lefthand algorthm):

{"algorithm": 
    {
      "onGapLeft": "moveLeft",
      "onGapFront": "moveForward",
      "onGapRight": "moveRight",
      "default": "moveBackward"
    }
}

It's very important that this file is read in it's order. I.e. if you change Left and Right the algorithm would become a Right Hand Algorithm. This is the Java Code so far, I hope you understand what I'm trying to do. BTW: I'm using JSON.simple!

private static void loadAlgorithm() throws InterruptedException {

        JSONParser parser = new JSONParser();


            Object obj = parser.parse(new FileReader("lefthand.json"));             
            JSONObject jsonObject = (JSONObject) obj;
            JSONArray algorithm = (JSONArray) jsonObject.get("algorithm");
            int length = algorithm.size();

        for(int i = 0; i < length; i++)
        {
            switch (algorithm[i].key)
            {
                 case "onGapLeft" :  leftPos = i; break;
                 case "onGapFront": frontPos = i; break;
                 case "onGapRight": rightPos = i; break;
                 default: break;
            }

            switch (algorithm[i].value)
            {
                 case "moveLeft"    : directionAlgorithm[i] = direction.Left;     break;
                 case "moveFront"   : directionAlgorithm[i] = direction.Forward;  break;
                 case "moveRight"   : directionAlgorithm[i] = direction.Right;    break;
                 case "moveBackward": directionAlgorithm[3] = direction.Backward; break;
                 default: break;
            }
        }           
    }

I'll need to know now wether it is possible to get the key string (where I used algorithm[i].key actually) and the same for the value string (algorithm[i].value).

Thank you very much for your help!

I don't think it solves your problem completely, but you've misinterpreted "algorithm" as a JSONArray. It's actually a JSONObject, meaning just like how you got "algorithm" from the root object, you can get "onGapLeft"/"onGapFront" values from the algorithm object. Arrays have their values declared sequentially with [] brackets and commas, and do not have string labels (ex.: ["moveLeft", "moveRight"] — Katana314
– Katana314, Commented Sep 5, 2013 at 20:30
Yeah, there's no "array" in that JSON. And the order of entries in an "object" (map/dictionary) are "not defined" -- even if they arrive in one order the code that parses the JSON can legitimately rearrange them. — Hot Licks
– Hot Licks, Commented Sep 5, 2013 at 20:52
(One wonders why you have key names for the individual values, if they will always be in a given order. It would make more sense to just have an array of [moveLeft, moveForward, moveRight, moveBackward], if that's what you want. Or make the key names be "firstMove", "secondMove", "thirdMove", "fourthMove".) — Hot Licks
– Hot Licks, Commented Sep 5, 2013 at 20:56
(See json.org for the JSON spec. It takes only 5-10 minutes to learn.) — Hot Licks
– Hot Licks, Commented Sep 5, 2013 at 20:57
Thanks everyone! So if I change the JSON like zigdawgydawg stated below, it'll be sorted then? @HotLicks: This would be a solution, but we want to have it more custamizable, tehrefore we've added this two value solution! :) — Jonas Zimmer
– Jonas Zimmer, Commented Sep 5, 2013 at 21:08

zigdawgydawg · Accepted Answer · 2013-09-05 20:31:20Z

2

You should probably change your JSON so that it is ordered, something like this:

{"algorithm": 
    [
        { "key": "onGapLeft", "value" : "moveLeft" },
        { "key": "onGapFront", "value" : "moveForward" },
        { "key": "onGapRight", "value" : "moveRight" },
        { "key": "default", "value" : "moveBackward" }
    ]
}

And then modify your Java accordingly.

answered Sep 5, 2013 at 20:31

zigdawgydawg

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5 Comments

Jonas Zimmer Over a year ago

Thanks! Does this solve the problem Katana314 stated, that my JSON file is not an array at the moment? :) And is it then possible to iterate thorugh this via "algorithm[i].key"? i.e. algorithm[0].key would be OnGapLeft? :)

zigdawgydawg Over a year ago

Yes...it does solve the problem that @Katana314 mentioned. The [...] in JSON is a proper (and ordered) array. As you are iterating through the algorithm JSONArray, get the JSONObject at each index with JSONObject obj2 = (JSONObject) algorithm.get(i); and then to get that object's key obj2.get("key");. I hope that helps.

Jonas Zimmer Over a year ago

Okay! Now I have just one problem with the initialising. The line: Object obj = parser.parse(new FileReader("lefthand.json")); has an error saying: "Unhandled exception type ParseException" Any suggestions? (As I said: I'm completely new to Java and Eclipse and sometimes I don't get what it wants to tell me :)

zigdawgydawg Over a year ago

You'll either need to wrap that line in a try/catch block or explicitly specify that the loadAlgorithm method throws ParseException. Here is more information about the two options: here

Jonas Zimmer Over a year ago

Thank you very much! This helped, the program compiled, everythings fine. Big problem now: That damn Lego Robot doesn't understand simpleJSON. I now have to work myself through a non library but java file json lib.. I hope that's not too complicated. I think I'm going to cry a little bit now. Thanks anyways ;)

sr01853 · Accepted Answer · 2013-09-05 20:46:51Z

0

I am not familiar with JSON , but Since JSONObject is backed by a HashMap, you may be able to get the keys and values into a array in the same order like below,

Map<K, V> map = new HashMap<K, V>();
K[] keys = new K[map.size()];
V[] values = new V[map.size()];
int index = 0;
for (Map.Entry<K, V> mapEntry : map.entrySet()) {
    keys[index] = mapEntry.getKey();
    values[index] = mapEntry.getValue();
    index++;
}

answered Sep 5, 2013 at 20:46

sr01853

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