CODEWITHSUNDEEP
asp.netajaxwebmethod
Deviland
Deviland

Reputation: 3374

ASP.NET AJAX returning XML when JSON specified

This is my first time using AJAX with ASP.NET so please allow m some rope.

I have a simple AJAX request that I wish to retuen a JSON string but ASP.NET keeps formatting the JSON string as XML

here is my code

[WebMethod]
[ScriptMethod(ResponseFormat = ResponseFormat.Json)] 
public string GetModels()
{
   string response;

   using (var context = new myDataModel())
   {
     var m = from model in context.ListModels select model;
     var serializer = new JavaScriptSerializer();
     response = serializer.Serialize(m);
   }

   return response;
}

this is called but returns an XML node containing my JSON string really confused

Upvotes: 1

Views: 1186

Answers (2)

kolin
kolin

Reputation: 2344

This is my method for returning JSON from a webservice using ASP.NET

[WebMethod]
    [ScriptMethod(ResponseFormat = ResponseFormat.Json)]
    public void GetJSON()
    {

        DataTable dtMarkers = new DataTable();
        dtMarkers.Load(<<STORED PROCEDURE>>.GetReader());

        string[][] myArray = new string[dtMarkers.Rows.Count][];
        int i = 0;
        foreach (DataRow marker in dtMarkers.Rows)
        {
            myArray[i] = new string[] { marker["ID"].ToString(), marker["Title"].ToString(), marker["Blurb"].ToString(), marker["Url"].ToString(), marker["LongLatPoint"].ToString() };
            i++;
        }

        JavaScriptSerializer js = new JavaScriptSerializer();
        string strJSON = js.Serialize(myArray);
        Context.Response.Clear();
        Context.Response.ContentType = "application/json";
        Context.Response.Flush();
        Context.Response.Write(strJSON);

    }

this particular example reads Googlemaps pointers from a database and returns them to the ASP.NET page for displaying on an instance of Google Maps.

Upvotes: 3

Simon Wang
Simon Wang

Reputation: 2943

Try to define it like:

public JsonResult GetModels()
{
    //.........

    return new JsonResult() { Data = new { result=response  } };
}

Upvotes: 0

Related Questions