Javascript和C＃处理程序-带有JSON的AJAX，发送数组

Question

When I send an array with AJAX (using JSON), my C# handler does not know how to handle the whole query (suddenly the querystrings combine with each other for some reason).

In this example I'm sending a very simple array to the server and the server says the querystring Name is null (but it is not null); Sending any request without the array works fine.

On that note, would appreciate if anyone could explain what the array looks like on the URL (if I wanted to send a request through the browser for example).

AJAX code:

    function btnClick() {
        var arr = new Array();
        arr[0] = "Hey";
        arr[1] = "Stackoverflow";
        arr[2] = "What's your name?";

        var jsonParam = { Name: "test", Pass: "123", Stuff: arr }
        $.ajax({
            url: "Test.ashx",
            type: "get",
            data: JSON.stringify(jsonParam),
            dataType: "json",
            contentType: 'application/json; charset=utf-8',
            async:false,
            success: function (response) { 
                alert(response.Name);
            }
        });
    }

Handler code:

public void ProcessRequest (HttpContext context) {
    context.Response.ContentType = "application/json";
    JavaScriptSerializer jss = new JavaScriptSerializer();



    string res = jss.Serialize(new UserInfo { Name = context.Request.QueryString["Name"], Pass = "pass" + context.Request.QueryString["Pass"], Stuff = new string[] { "1", "2" } });
    context.Response.Write(res);
}

Answer 1

You cannot get the json from querystring.

You can use this; You should install NewtonJson for JSonConvert from nuget. If you don't want that, you can use JavaScriptSerializer instead of that.

    protected object FromJsonToObject(Type t)
    {
        Context.Request.InputStream.Position = 0;
        string json;
        using (var reader = new StreamReader(Context.Request.InputStream))
        {
            json = reader.ReadToEnd();
        }

        return JsonConvert.DeserializeObject(json, t);
    }