I have a servlet which can accept the follow HTTP Form File operation, and save the file into the server's File System
<FORM METHOD="POST" ACTION="http://localhost:8
Select the file to upload : <INPUT TYPE="FILE" NAME="FILE1" SIZE="25"><BR>
<INPUT TYPE="SUBMIT" VALUE="Upload">
</FORM>
How can I write a VB Client that does the same thing? I got the VB as following, but it doesn't work at all.
I suspect I am sending a string, instead of the file object (sFile).
Any ideas?
Private Sub Command1_Click()
Dim xmlHttp As New MSXML2.xmlHttp
'Dim http As New WinHttp.WinHttpRequest
With CommonDialog1
.Flags = cdlOFNHideReadOnly Or cdlOFNFileMustExist
.Filter = "All Files|*.*"
.FileName = ""
.ShowOpen
sFile = .FileName
sTitle = .FileTitle
End With
'MsgBox sFile
xmlHttp.open "POST", "http://localhost:8080/ser
xmlHttp.setRequestHeader "ENCTYPE", "multipart/form-data"
xmlHttp.send sFile
If xmlHttp.Status = 200 Then
MsgBox "Upload Successful!!"
Else
MsgBox "Upload Unsuccessful!!"
End If
End Sub
Thanks!
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by: sinloongPosted on 2002-11-22 at 05:02:27ID: 7483179
hi mscproj,
you have to send the binary data of the file, instead of the file name.
xmlHttp.send <Binary data>
Use Open statement to get the binary data of the file.