Esempio n. 1
0
        public string PostBlog()
        {
            string title = Request.Form["title"];
            string image = Request.Form["image"];
            string body  = Request.Form["body"];

            var dataAccess = new DatabaseAccess();

            if (dataAccess.AddBlogPost(title, image, body))
            {
                return("success");
            }
            else
            {
                return("fail");
            }
        }