Response.Redirect with POST instead of Get?

Question

We have the requirement to take a form submission and save some data, then redirect the user to a page offsite, but in redirecting, we need to \"submit\" a form with POST, n

Answer 1

Copy-pasteable code based on Pavlo Neyman's method

RedirectPost(string url, T bodyPayload) and GetPostData() are for those who just want to dump some strongly typed data in the source page and fetch it back in the target one. The data must be serializeable by NewtonSoft Json.NET and you need to reference the library of course.

Just copy-paste into your page(s) or better yet base class for your pages and use it anywhere in you application.

My heart goes out to all of you who still have to use Web Forms in 2019 for whatever reason.

        protected void RedirectPost(string url, IEnumerable> fields)
        {
            Response.Clear();

            const string template =
@"


{1}


";

            var fieldsSection = string.Join(
                    Environment.NewLine,
                    fields.Select(x => $"")
                );

            var html = string.Format(template, HttpUtility.UrlEncode(url), fieldsSection);

            Response.Write(html);

            Response.End();
        }

        private const string JsonDataFieldName = "_jsonData";

        protected void RedirectPost(string url, T bodyPayload)
        {
            var json = JsonConvert.SerializeObject(bodyPayload, Formatting.Indented);
            //explicit type declaration to prevent recursion
            IEnumerable> postFields = new List>()
                {new KeyValuePair(JsonDataFieldName, json)};

            RedirectPost(url, postFields);

        }

        protected T GetPostData() where T: class 
        {
            var urlEncodedFieldData = Request.Params[JsonDataFieldName];
            if (string.IsNullOrEmpty(urlEncodedFieldData))
            {
                return null;// default(T);
            }

            var fieldData = HttpUtility.UrlDecode(urlEncodedFieldData);

            var result = JsonConvert.DeserializeObject(fieldData);
            return result;
        }