How to add Fileupload file on Datagrid

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I want to add the selected file through Fileupload control on DataGrid while selecting Like Gmail Or Yahoo attachment.

How to do this while selecting the file.

Posted 5-Dec-11 21:42pm

Robymon

Updated 5-Dec-11 21:44pm

RaisKazi

v2

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Anuj Banka · Answer 1 · 2011-12-05T21:51:00

Solution 1

On Click on add attachment. Just save file on temp Loaction and Create a datatable and detail of the attachment in that datatable.

And finally give this datatable as datasource of the listview.

And on send mail you can clear the datatable and delete the attachment files

Posted 5-Dec-11 21:51pm

Anuj Banka

Updated 5-Dec-11 21:52pm

v2

Comments

Robymon 6-Dec-11 4:02am

Can i get some solution for this?

Anuj Banka 6-Dec-11 4:38am

Check the new solution that i have given. This is just a overview efforts you have to do.. If it help you mark as a answer

prashanth reddy pola · Answer 2 · 2011-12-05T21:55:00

Solution 2

Place file upload control in item template.

ASP.NET

<columns>
<asp:TemplateField>
<itemtemplate></itemtemplate></columns>

Here file upload control

HTML

<itemtemplate>
</itemtemplate>

-Prashanth

Posted 5-Dec-11 21:55pm

prashanth reddy pola

Updated 5-Dec-11 22:38pm

v2

Comments

Robymon 6-Dec-11 4:03am

I need to be like Gmail or Yahoo attachment. select the file, then after selecting it should add on to the Datagrid.

Anuj Banka · Answer 3 · 2011-12-05T22:36:00

Do something Like this: you can modify this according to your need

Protected Sub Button1_Click(ByVal sender As Object, ByVal e As System.EventArgs) Handles Button1.Click

       ListAttachment.Add(New Attachment With {.FileName = "YOr File Name"})
       ListView1.DataSource = ListAttachment
       ListView1.DataBind()
   End Sub



   Private _listAttachments As List(Of Attachment)
   Public Property ListAttachment() As List(Of Attachment)
       Get
           Return _listAttachments
       End Get
       Set(ByVal value As List(Of Attachment))
           _listAttachments = value
       End Set
   End Property
   Public Class Attachment
       Private _fileName As String
       Public Property FileName() As String
           Get
               Return _fileName
           End Get
           Set(ByVal value As String)
               _fileName = value
           End Set
       End Property
   End Class