![]() ![]() So, go for it and let me know if you have any problem. ![]() postieplacemedia: 1.8.10: This filter is called when replacing the img with an actual image reference and can be used to modify the html fragment. You can also select the file without using the FileUpload tool in an ASP.NET Web Application. postieemailnotifybody: 1.8.8: This filter is called before a success notification is sent and can be used to modify the message body. With this article you can send a HTML file body as an Email body to any person. Insert a breakpoint on rowcommand and check out the mail body: Step 2: Now at the Mail Body, we can check the HTML Body attached as an Email Body. NetworkCredential n = new NetworkCredential(FromEmailid, Pass).MailAddress fromMail = new MailAddress(FromEmailid).mystring = mystring.Replace( "$$Subject$$" , TxtSubject.Text).mystring = mystring.Replace( "$$Member$$" , Name).StreamReader reader = new StreamReader(Server.MapPath(Path.Combine( "~/UploadedFile/" , MyFile))) Viewing request errors from the console You will get an error message if Postman isnt able to send your request, or if it doesnt receive a response from.PostedFile.SaveAs(Server.MapPath(Path.Combine( "~/UploadedFile/" , MyFile))) MyFile = Path.GetFileName(PostedFile.FileName).if (PostedFile != null & PostedFile.ContentLength > 0).HttpPostedFile PostedFile = Request.Files.Label LblMail = ( "LblEmail" ) as Label.Label LblName = ( "LblName" ) as Label.int index = Convert.ToInt32(e.CommandArgument.ToString()).protected void GridView1_RowCommand( object sender, GridViewCommandEventArgs e).Paste the following code into the RowCommand Event of the Grid View: Now, we need to check determine whether the file was uploaded successfully. So, let us pproceed to understand it by the following points.Īt first we need to create a folder in the application for saving the HTML file uploaded by file upload We need to use the fileupload code in it. Step 3: Open the WebForm Code view to write the code in the Grid View Row Command. Step 2: Just double-click on the RowCommand event to generate its event. Step 1: Open the Grid View properties as shown below: Now, we have completed the designing part of our application and so let's proceed the code implementation with the procedure given below. ![]() Step 2: Let's check out the WebForm by debugging the application: Note: Please note that the Value of AutoGenerateColumns must be false, otherwise the source table may be displayed in the browser more then once. I changed the fields into the Template Field for getting the column values easily. ![]()
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