How to insert data using ASP.Net generic handler file ?

Question

How to insert data (file, textbox) in html control. how insert both of data using generic handler file.

Rupesh N · Answer

code behind ashx file :: public void ProcessRequest(HttpContext context) { //textinsert(_text); using (SqlConnection con = new SqlConnection(ConfigurationManager.Co nnectionStrings["dbconnection"].ConnectionString)) { context.Response.ContentType = "text/plain"; int uploadCount = 0; if (context.Request.Files.Count > 0) { HttpFileCollection files = context.Request.Files; for (int i = 0; i { HttpPostedFile file = files[i]; string fileName = new FileInfo(file.FileName).Name; // fileName = Path.Combine(context.Server.MapPath("/img/products/"), fileName); string sPath = HttpContext.Current.Server.MapPath("/img/products/"); //file.SaveAs(fileName); if (HttpContext.Current.Request.Browser.Browser.ToUpper() == "IE" || HttpContext.Current.Request.Browser.Browser.ToUpper() == "INTERNETEXPLORER" || file.ContentLength > 0) { Guid id = Guid.NewGuid(); string modifiedFileName = id.ToString() + "_" + fileName; file.SaveAs(sPath + Path.GetFileName(modifiedFileName)); string[] testfiles = file.FileName.Split(new char[] { '\' }); sPath = testfiles[testfiles.Length - 1]; string text = context.Request["text"]; if (!File.Exists(sPath + Path.GetFileName(modifiedFileName))) { uploadCount++; if (uploadCount > 0) { SqlCommand cmd = new SqlCommand("insert into tbl_test values(@id,@filename,@title,@text)", con); cmd.Parameters.AddWithValue("@id", id); cmd.Parameters.AddWithValue("@filename", "img/products/" + modifiedFileName.ToString()); cmd.Parameters.AddWithValue("@title", fileName); cmd.Parameters.AddWithValue("@text", text); con.Open(); cmd.ExecuteNonQuery(); con.Close(); } else { context.Response.Write("File Upload"); } } } else { fileName = file.FileName; } } } context.Response.ContentType = "text/plain"; context.Response.Write("File Upload"); } } Angularjs:; $scope.uploadFile = () => { var fd = new FormData(); angular.forEach($scope.files, (file) => { fd.append("file", file); }); var obj = JSON.stringify($scope.product); fd.append("form1", obj); $http({ method: 'POST', url: "../image.ashx", data: fd, headers: { 'Content-Type': undefined } }).then(res => { console.log(res.data); });

Sreekanth Reddy · Answer

@Rupesh N $("#submit").click(function () { var file = $("#file")[0].Files[0]; var fd = new FormData(); fd.append("firstname", $("#firstname").val()); fd.append("lastname", $("#lastname").val()); fd.append("avatar", file); $.ajax({ type: "POST", contentType: false, processData: false, url: " ", data: fd }).success(function (data) { }); }); Do as like above. If it helps, mark as answer. Thank you.