How to retrieve jsonarray data sent from javascript by ajax in java

Question

My javascript code. I am creating a jsonarray with some elements. Now i want to retrieve that data retrieve and want to store in a java json object for further process.

var priceList = []; priceList.push({"itemId":itemId1,"idElem":idElem1,"isActive":isActive1,"description":description1}); priceList.push({"itemId":itemId2,"idElem":idElem2,"isActive":isActive2,"description":description2}); var param = {'objarray' : priceList, 'user' : user};

    $.ajax({
      url : 'cPayment,
      type: 'POST', 
      dataType: 'json',  
      data: param,
      success: function(result) {
          alert('SUCCESS');
      }
    });

In java code :

String objArray = request.getParameter("objarray");
JSONArray jsonArray = new JSONArray();
for(//for items in objArray) {
JSONObject jsonObj = new JSONObject(); 
jsonObj.put("itemId", itemId);
jsonObj.put("idElem", idElem);
jsonObj.put("isActive", isActive);
jsonObj.put("description", "description");
jsonArray.put( jsonObj);
}

Please help me in retrieving javascript json array data in java as java jsonobject.

You need to post more of your server-side code and be a little more specific as to what the actual problem is. — JoshG
– JoshG, Commented May 20, 2018 at 9:48
I am facing problem while converting javascript json array to java specific object and to iterate that in for loop. — Ram Thota
– Ram Thota, Commented May 20, 2018 at 10:06

kiranbirajdar · Accepted Answer · 2018-05-20 10:07:31Z

1

In Ajax call try JSON.parse(param);

answered May 20, 2018 at 10:07

kiranbirajdar

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2 Comments

Ram Thota Over a year ago

I am asking about how to retrieve at server side.

kiranbirajdar Over a year ago

$.ajax({ dataType: 'json', data: "objarray="+param, success: function(result) { } }); If needed we have to use JSON.stringify(param);

kiranbirajdar · Accepted Answer · 2018-05-21 05:43:29Z

0

Try following code in java:

JSONObject jsonObject = null;
try {
        jsonObject = new JSONObject(req.getParameter("objarray"));
    } catch(JSONException _instance) {
        // Exception Handle Message
    }
    System.out.println("Entered in to the controller ");
    String itemId ="" ;
    if(jsonObject.has("itemId")) {
       name = objarray.getString("itemId");
    }

answered May 21, 2018 at 5:43

kiranbirajdar

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1 Comment

Ram Thota Over a year ago

After catch block.What you are doing with jsonObject and where did you get objarray. I think you are missing something to add.

Ram Thota · Accepted Answer · 2018-05-21 17:06:56Z

I have used the following code it worked for me.

In Javascript code:

var param = {'objarray' : JSON.stringify(priceList), 'user' : user};

And on servlet side i used org.json.simple.JSONArray and org.json.simple.JSONObject to retrive my data.

My Java code :

JSONParser parser=new JSONParser();
    try {
        String objarray = URLDecoder.decode(request.getParameter("objarray"), "UTF-8");
        jsonArrayObj = (JSONArray) parser.parse(objarray);
    } catch (UnsupportedEncodingException e) {
        e.printStackTrace();
    } catch (ParseException e) {
        e.printStackTrace();
    }

    for(int i= 0; i <jsonArrayObj.size() ;i++) {
        JSONObject psItem = (JSONObject)jsonArrayObj.get(i);
        Integer itemId = Integer.valueOf((String) psItem.get("itemId "));
        String idElem = (String) psItem.get("idElem");
        String description= (String) psItem.get("description");
        String isActive= Boolean.valueOf((String)psItem.get("isActive"));
        Syste.out.println("Item Id --> " +itemId);
        Syste.out.println("Id Element --> "+idElem );
        Syste.out.println("Description --> "+description);
        Syste.out.println("Is Active --> "+isActive);
       }

It Worked for me.

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