Returning JSON response from Servlet to Javascript/JSP page

Question

I think (actually I KNOW!) I'm doing something wrong here I am trying to populate some values into HashMap and add each hasmap to a list which will be added to a JSON object:

JSONObject json = new JSONObject();
try
{
    Map address;
    List addresses = new ArrayList();

    int count = 15;

    for (int i=0 ; i<count ; i++)
    {
        address = new HashMap();
        address.put("CustomerName"     , "Decepticons" + i);
        address.put("AccountId"        , "1999" + i);
        address.put("SiteId"           , "1888" + i);
        address.put("Number"            , "7" + i);
        address.put("Building"          , "StarScream Skyscraper" + i);
        address.put("Street"            , "Devestator Avenue" + i);
        address.put("City"              , "Megatron City" + i);
        address.put("ZipCode"          , "ZZ00 XX1" + i);
        address.put("Country"           , "CyberTron" + i);
        addresses.add(address);
    }
    json.put("Addresses", addresses);
}
catch (JSONException jse)
{

}
response.setContentType("application/json");
response.getWriter().write(json.toString());

My problem is I know this is returning a string, which I cannot seem to parse (which is the problem). My question is how do I return the actual JSON encoded string (or even should I be doing this?) or what is the best method of attack for this type of problem. The JavaScript I am using for this is below:

function getReadyStateHandler(req)
{
    // Return an anonymous function that listens to the
    // XMLHttpRequest instance
    return function ()
    {
        // If the request's status is "complete"
        if (req.readyState == 4)
        {
            // Check that a successful server response was received
            if (req.status == 200)
            {
                msgBox("JSON Response recieved...");
                populateDatagrid(req.responseText.toJSON());
            }
            else
            {
                // An HTTP problem has occurred
                alert("HTTP error: " + req.status);
            }
        }
    }
}

Note the JSON Response comes back fine, but its a string. Any advice is greatly appreciated. I am also opening to using googles Gson, but don't have too much knowledge on that.

You shouldn't try to create a JSON encoded string yourself; I would be very surprised if your server side code doesn't provide a library function to properly encode an object into a JSON encoded string. — Marcel Korpel
– Marcel Korpel, Commented May 28, 2011 at 8:54
possible duplicate of Create JSON obect and convert it to String in Java — Marcel Korpel
– Marcel Korpel, Commented May 28, 2011 at 8:58

MByD · Accepted Answer · 2011-05-28 21:24:21Z

29

Got it working! I should have been building a JSONArray of JSONObjects and then add the array to a final "Addresses" JSONObject. Observe the following:

JSONObject json      = new JSONObject();
JSONArray  addresses = new JSONArray();
JSONObject address;
try
{
   int count = 15;

   for (int i=0 ; i<count ; i++)
   {
       address = new JSONObject();
       address.put("CustomerName"     , "Decepticons" + i);
       address.put("AccountId"        , "1999" + i);
       address.put("SiteId"           , "1888" + i);
       address.put("Number"            , "7" + i);
       address.put("Building"          , "StarScream Skyscraper" + i);
       address.put("Street"            , "Devestator Avenue" + i);
       address.put("City"              , "Megatron City" + i);
       address.put("ZipCode"          , "ZZ00 XX1" + i);
       address.put("Country"           , "CyberTron" + i);
       addresses.add(address);
   }
   json.put("Addresses", addresses);
}
catch (JSONException jse)
{ 

}
response.setContentType("application/json");
response.getWriter().write(json.toString());

This worked and returned valid and parse-able JSON. Hopefully this helps someone else in the future. Thanks for your help Marcel

edited May 28, 2011 at 21:24

MByD

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answered May 28, 2011 at 14:38

majestica

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2 Comments

Dan Over a year ago

How to read this in ajax , I tried alert(responsedata.Addresses.CustomerName);

Abhishek Ghosh Over a year ago

Amazing! I used this with GAE and worked just fine with the inbuilt import com.google.appengine.labs.repackaged.org.json.JSONObject package! Thank you so much.

Rahul Varadkar · Accepted Answer · 2017-05-20 07:09:19Z

I used JSONObject as shown below in Servlet.

    JSONObject jsonReturn = new JSONObject();

    NhAdminTree = AdminTasks.GetNeighborhoodTreeForNhAdministrator( connection, bwcon, userName);

    map = new HashMap<String, String>();
    map.put("Status", "Success");
    map.put("FailureReason", "None");
    map.put("DataElements", "2");

    jsonReturn = new JSONObject();
    jsonReturn.accumulate("Header", map);

    List<String> list = new ArrayList<String>();
    list.add(NhAdminTree);
    list.add(userName);

    jsonReturn.accumulate("Elements", list);

The Servlet returns this JSON object as shown below:

    response.setContentType("application/json");
    response.getWriter().write(jsonReturn.toString());

This Servlet is called from Browser using AngularJs as below

$scope.GetNeighborhoodTreeUsingPost = function(){
    alert("Clicked GetNeighborhoodTreeUsingPost : " + $scope.userName );

    $http({

        method: 'POST',
        url : 'http://localhost:8080/EPortal/xlEPortalService',
        headers: {
           'Content-Type': 'application/json'
         },
        data : {
            'action': 64,
            'userName' : $scope.userName
        }
    }).success(function(data, status, headers, config){
        alert("DATA.header.status : " + data.Header.Status);
        alert("DATA.header.FailureReason : " + data.Header.FailureReason);
        alert("DATA.header.DataElements : " + data.Header.DataElements);
        alert("DATA.elements : " + data.Elements);

    }).error(function(data, status, headers, config) {
        alert(data + " : " + status + " : " + headers + " : " + config);
    });

};

This code worked and it is showing correct data in alert dialog box:

Data.header.status : Success

Data.header.FailureReason : None

Data.header.DetailElements : 2

Data.Elements : Coma seperated string values i.e. NhAdminTree, userName

tinyd · Accepted Answer · 2011-05-27 16:41:41Z

1

I think that what you want to do is turn the JSON string back into an object when it arrives back in your XMLHttpRequest - correct?

If so, you need to eval the string to turn it into a JavaScript object - note that this can be unsafe as you're trusting that the JSON string isn't malicious and therefore executing it. Preferably you could use jQuery's parseJSON

answered May 27, 2011 at 16:41

tinyd

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2 Comments

majestica Over a year ago

Hi, I've tried using that and all I get is an object back - please see this thread: stackoverflow.com/questions/6153295/… This explains what I thought was valid JSON being returned. However, it appears that it was infact a string being passed back, so I am having difficulty parsing it - which is why am I here asking the question relating to trying to pass JSON back to my JavaScript function

Marcel Korpel Over a year ago

The question is about sending a JSON encoded string, not about how to process it in JavaScript.

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