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  • Status: Solved
  • Priority: Medium
  • Security: Public
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  • Last Modified:

getting ajax call to work

Here is my ajax call:

        $.ajax({
            type: "GET",
            url: "WebGlobalMethods.asmx/HelloWorld",
            contentType: "application/json; charset=utf-8",
            dataType: "json",
            success: TheTextFound,
            error: LoadFailed
        });

        function TheTextFound()
        {
            alert("success");
        }

        function LoadFailed()
        {
            alert("did not work");
        }

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Here is the web service code:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
using System.Web.Services;

namespace Campus_Webstore
{
    /// <summary>
    /// Summary description for WebGlobalMethods
    /// </summary>
    [WebService(Namespace = "http://tempuri.org/")]
    [WebServiceBinding(ConformsTo = WsiProfiles.BasicProfile1_1)]
    [System.ComponentModel.ToolboxItem(false)]
    // To allow this Web Service to be called from script, using ASP.NET AJAX, uncomment the following line. 
    // [System.Web.Script.Services.ScriptService]
    public class WebGlobalMethods : System.Web.Services.WebService
    {

        [WebMethod]
        public string HelloWorld()
        {
            return "Hello World";
        }
    }
}

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So far it is failing  ( I get the alert that says "did not work" )


I could use some help.  Thanks!

Tom
0
Tom Knowlton
Asked:
Tom Knowlton
1 Solution
 
leakim971PluritechnicianCommented:
Line 2, use POST instead GET

type: "POST",
0
 
richard_hughesCommented:
Hello knowlton

As leakim971 said, try the attached code.

Thanks,

Richard
$.ajax({
            type: "POST",
            url: "WebGlobalMethods.asmx/HelloWorld",
            contentType: "application/json; charset=utf-8",
            dataType: "json",
            success: TheTextFound,
            error: LoadFailed
        });

        function TheTextFound()
        {
            alert("success");
        }

        function LoadFailed()
        {
            alert("did not work");
        }

Open in new window

0
 
Tom KnowltonWeb developerAuthor Commented:
Yep, that was it.  It needed to be POST.
0

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