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Interacting with the web using WebRobot v1.0 - Uploading files to YouSendIt: Adding fields
Uploading files to YouSendIt: Adding fields
With Mozilla Firefox's Page Info dialog (right click on the page,
then click Page Info), we can see the form fields that have been added
via JavaScript to the form "tform". Of interest to us right now are the
"rcpt" field, "fname" field, and "rurl" field.
Let's add the required fields to the form:
Private Sub Form1_Load(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Load Dim wrobot As New foxtrot.xray.WebRobot wrobot.Base = "http://www.yousendit.com/" wrobot.LoadPage("/")
'getting file upload form from page Dim wform As foxtrot.xray.Form = wrobot.GetFormByName("tform") wform.AddField("text", "rcpt") wform.AddField("file", "fname") wform.AddField("hidden", "rurl", "http://www.yousendit.com/transfer.php?action=status") End Sub
Now we have the required fields to be able to post to the form,
so we are going to continue the upload process. But first, we shall add
a dialog to select the file to upload, and then we will read the file
into memory:
Private Sub Form1_Load(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Load Dim wrobot As New foxtrot.xray.WebRobot wrobot.Base = "http://www.yousendit.com/" wrobot.LoadPage("/") 'getting file upload form from page Dim wform As foxtrot.xray.Form = wrobot.GetFormByName("tform") wform.AddField("text", "rcpt") wform.AddField("file", "fname") wform.AddField("hidden", "rurl", "http://www.yousendit.com/transfer.php?action=status") 'creating an Open File Dialog to choose file to upload Dim fopendialog As New System.Windows.Forms.OpenFileDialog fopendialog.CheckFileExists = True If fopendialog.ShowDialog() = DialogResult.OK Then 'open file for reading Dim fstream As New System.IO.FileStream(fopendialog.FileName, IO.FileMode.Open) Dim bytFile(fstream.Length - 1) As Byte 'temporary buffer to store contents of file fstream.Read(bytFile, 0, fstream.Length) 'read in the file 'the contents of the file to be uploaded to the server Dim mstream As New System.IO.MemoryStream(bytFile) End If End Sub
Comments
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Posted by Alaa Jebran on 18 Jun 2007
Yes it'll save a plenty of work, thanks . -
Posted by fixx on 11 Apr 2007
[quote user="dinuX"]
This is a Very Good Example.I thin its very usefull to me.
but the problem is when i try this this not work properly error occured at this line " -
Posted by fixx on 11 Apr 2007
That looks very nice. It'll save me plenty of work when posting a form instead of hardcoding all the fields in a webrequest! http://www.fixx.be -
Posted by dinuX on 04 Oct 2006
This is a Very Good Example.I thin its very usefull to me.
but the problem is when i try this this not work properly error occured at this line "Dim
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