simpe code for file upload

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October 29th, 2003, 04:12 AM

Seth99

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Join Date: Aug 2003

Posts: 42

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simpe code for file upload

Hello,

I have search the web last day all day and i did not find how to make a simple upload in APS not classes and third parties and don't know what .dll's that must be copyed on the server just 3-4 lines of code just like in PHP:

PHP Code:


		
			
<input type="file" name="img">



$dir = "img/";

copy($img,$dir.$img) or die ("Error");

Can anyone help me with this becouse i have reach to the end of my search i don't know what else to do so i just asked you guys.

Thanks everyone

October 29th, 2003, 08:52 AM

don_sparko

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unfortunately there is no easy way to upload with asp. the best thing i found is here
its a real pain to upload with it since its not an inherent thing. my suggestion is: use something that makes it easy like cold fusion. one line of code there for file uploads. good luck.

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October 29th, 2003, 09:05 AM

Seth99

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Join Date: Aug 2003

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Hy,

i have seen that page i have search all day to get a simple upload but no chance probably you are right there in no easy way in ASP.

Thanks anyway for your help

October 29th, 2003, 09:22 AM

mopacfan

Junior Member

Join Date: Oct 2003

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File upload solution

If you can use a dll, visit Dundas . I have been using this free upload control for years and it works great. Very simple, very bulletproof.

If you can't use dll's, try this out:

Code:

<%
' Author Philippe Collignon
' Email PhCollignon@email.com

Sub BuildUploadRequest(RequestBin)
	'Get the boundary
	PosBeg = 1
	PosEnd = InstrB(PosBeg,RequestBin,getByteString(chr(13)))
	boundary = MidB(RequestBin,PosBeg,PosEnd-PosBeg)
	boundaryPos = InstrB(1,RequestBin,boundary)
	'Get all data inside the boundaries
	Do until (boundaryPos=InstrB(RequestBin,boundary & getByteString("--")))
		'Members variable of objects are put in a dictionary object
		Dim UploadControl
		Set UploadControl = CreateObject("Scripting.Dictionary")
		'Get an object name
		Pos = InstrB(BoundaryPos,RequestBin,getByteString("Content-Disposition"))
		Pos = InstrB(Pos,RequestBin,getByteString("name="))
		PosBeg = Pos+6
		PosEnd = InstrB(PosBeg,RequestBin,getByteString(chr(34)))
		Name = getString(MidB(RequestBin,PosBeg,PosEnd-PosBeg))
		PosFile = InstrB(BoundaryPos,RequestBin,getByteString("filename="))
		PosBound = InstrB(PosEnd,RequestBin,boundary)
		'Test if object is of file type
		If  PosFile<>0 AND (PosFile<PosBound) Then
			'Get Filename, content-type and content of file
			PosBeg = PosFile + 10
			PosEnd =  InstrB(PosBeg,RequestBin,getByteString(chr(34)))
			FileName = getString(MidB(RequestBin,PosBeg,PosEnd-PosBeg))
			'Add filename to dictionary object
			UploadControl.Add "FileName", FileName
			Pos = InstrB(PosEnd,RequestBin,getByteString("Content-Type:"))
			PosBeg = Pos+14
			PosEnd = InstrB(PosBeg,RequestBin,getByteString(chr(13)))
			'Add content-type to dictionary object
			ContentType = getString(MidB(RequestBin,PosBeg,PosEnd-PosBeg))
			UploadControl.Add "ContentType",ContentType
			'Get content of object
			PosBeg = PosEnd+4
			PosEnd = InstrB(PosBeg,RequestBin,boundary)-2
			Value = MidB(RequestBin,PosBeg,PosEnd-PosBeg)
			Else
			'Get content of object
			Pos = InstrB(Pos,RequestBin,getByteString(chr(13)))
			PosBeg = Pos+4
			PosEnd = InstrB(PosBeg,RequestBin,boundary)-2
			Value = getString(MidB(RequestBin,PosBeg,PosEnd-PosBeg))
		End If
		'Add content to dictionary object
	UploadControl.Add "Value" , Value	
		'Add dictionary object to main dictionary
	UploadRequest.Add name, UploadControl	
		 
		'Loop to next object
		BoundaryPos=InstrB(BoundaryPos+LenB(boundary),RequestBin,boundary)
	Loop

End Sub

'String to byte string conversion
Function getByteString(StringStr)
 For i = 1 to Len(StringStr)
 	char = Mid(StringStr,i,1)
	getByteString = getByteString & chrB(AscB(char))
 Next
End Function

'Byte string to string conversion
Function getString(StringBin)
 getString =""
 For intCount = 1 to LenB(StringBin)
	getString = getString & chr(AscB(MidB(StringBin,intCount,1))) 
 Next
End Function
%>

I did not write the code, but it works great and does not need a dll, only write permissions in the target directory.

I hope this helps,

Regards,

Michael

October 29th, 2003, 10:09 AM

Seth99

Contributing User

Join Date: Aug 2003

Posts: 42

Time spent in forums: 1 h 58 m 7 sec
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ok thanks for your reply but how i will use this will work if i code like this:

PHP Code:


		
			
<%

dim fs



if request.form("action") <> "add" then

%>

<form action="" method="post" name="form1">

<input type="hidden" name="action" value="add">

  <input type="file" name="img">

  <br>

  <input type="submit" name="Submit" value="Submit">

</form>

<%

else 



img = request.Form("img")



img = response.write("img")



set fs=Server.CreateObject("Scripting.FileSystemObject")

fs.CopyFile "e:/hosting/admin/upload.asp","e:/hosting/4real.co.uk/admin/img/"



response.write("yhahahahahaahahah")



set fs=nothing



end if

%>

i have made an page like the code above (is just some test) but i get this error message:

Cannot use Request.Form

/admin/tour.asp, line 131

Cannot use Request.Form collection after calling BinaryRead.

i want to be able to upload multiple images
i you know the answer please

Thanks to all

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