What's wrong with my code.

What's wrong with my code ?

i am uploading a file to image shack. i composted this code to my understanding by referring to http://www.vbforums.com/showpost.php...44&postcount=4

Code:

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Private Sub Command1_Click()
    Dim intFile    As Integer  '// Next available number for the Open statement
    Dim ImageData  As String  '// Image contents
    Dim Body        As String  '// Body contents
    Dim Header      As String  '// Header contents

    '// Get the image contents
    imagepath = App.Path & "\" & "test.jpg" ' max 3mb image size limited
    intFile = FreeFile
    Open imagepath For Binary As #intFile
        ImageData = String(LOF(intFile), Chr(0))
        Get #intFile, , ImageData
    Close #intFile

    Body = "Content-Disposition: form-data; name=""fileupload""; filename=""" & vbCrLf
    Body = Body & "Content-Type: multipart/form-data" & vbCrLf
    Body = Body & vbCrLf & ImageData


    Header = "Host: imageshack.us" & vbCrLf
    Header = Header & "Content-Type: multipart/form-data  & vbCrLf"
    Header = Header & "Content-Length: " & Len(ImageData) & vbCrLf & vbCrLf


    'upload now
    Inet.Execute "http://www.imageshack.us", "POST", Body, Header
   
      While Inet.StillExecuting
        DoEvents
      Wend
     
      Dim s$, ret$
     
      s = Inet.GetChunk(1024)
      Do Until s = ""
        ret = ret & s
        s = Inet.GetChunk(1024)
      Loop
     
      'check upload completed
      If InStr(1, ret, "Show Advanced Linking") > 0 Then MsgBox "upload completed"
  End Sub

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Could try giving us an idea of why you think something is wrong with your code...it'd help us help you

hai smUx
i dont undestand German language :D

Code:

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Could try giving us an idea of why you think something ....

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btw, ill be back in 30 min :)

http://translate.google.com/translate_t#

It's not rocket science here, if you have a problem there's a solution somewhere which gets you closer to understanding the bigger problem :-P

smUX. ok ok kool.

Quote:

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there's a solution somewhere

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yes, hope here i can find it :thumb:

Your Code:
imagepath = App.Path & "\" & "test.jpg"

My Code:
imagepath = App.Path & "\test.jpg"

Quote:

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Originally Posted by ThEiMp

Your Code:
imagepath = App.Path & "\" & "test.jpg"

My Code:
imagepath = App.Path & "\test.jpg"

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Those will both give the same result.

But it would be a good idea to make sure that the "\" is needed.
Sometimes App.Path provides it but most times it doesn't.

theEmp,
tx, but that is ok. no path problem in my code.
the file is open for binary access with out error.

Yes that is right, there is nothing more wrong with your code and I don't see why it won't run. Could you please post a working demo of your source code onto this thread, so that we can see if that is totally correct?

ofcause. give me a minute. ill attached it this this post it self.

Well I cannot get it to work, but with the source code that I could get to work. I added into it a message box, loading box. Which means that it will go on when it is loading, eg: visible = True and when it is not loading it will be visible = False.

ThEiMp,
is upload success and returns the results page in your test?
(which include the "Show Advanced Linking" keyword)