.:: Koding::..

Kode berikut akan menjelaskan tentang kode pada halaman asp.Net untuk mengupload gambar pada server. Pada kode berikut digunakan 3 buah control yakni 2 buah html control biasa dan 1 buah asp web control. Untuk html control agar dapat dikenali dalam code behind, diwajibkan mengisi runat=”server” pada deklarasi html control tersebut.
Aspx file:
<p>
Upload Image File :
<input id="FileUpload1" type="file" name="File1" runat="server"/>
</p>
<p>
<input id="Upload" type="submit" value="Upload Image" name="Submit" runat="server"
onserverclick="btnUpload_Click"/>
</p>
<p>
<asp:Label ID="lbStatus" runat="server"></asp:Label></p>

Code Behind:

Protected Sub btnUpload_Click(ByVal sender As Object, ByVal e As System.EventArgs) Handles Upload.ServerClick
If Not (FileUpload1.PostedFile Is Nothing) Then
Dim strLongFilePath As String = FileUpload1.PostedFile.FileName
Dim intFileNameLength As Integer = InStr(1, StrReverse(strLongFilePath), "\")
Dim strFileName As String = Mid(strLongFilePath, (Len(strLongFilePath) - intFileNameLength) + 2)
Select Case FileUpload1.PostedFile.ContentType
'Memastikan file berextension JPEG
Case "image/pjpeg", "image/jpeg"
FileUpload1.PostedFile.SaveAs(Server.MapPath(strFileName))
lbStatus.Text = strFileName & " was uploaded successfully to: " & Server.MapPath(strFileName)
Case Else
'Bukan file gambar JPEG
lbStatus.Text = "Upload Failed! Invalid jpeg image"
End Select
End If
End Sub
Happy Programing.:D