[Zope] Upload a file to the file system

[Dieter Maurer]

Gilles Lenfant writes:
 > I'm trying to upload a file to the file system (using an external method).
 > I got a :
 > 
 > <form action="uploadfs" method="post" enctype="multipart/form-data">
 > ....
 > <input type="file" name="file">
 > ....
 > <input type="submit" value=" Upload "></form>
Already quite good.

	def uploadfs(file):
  	    open(file.filename,"wb").write(file.read())
	    return 'uploaded'


ATTENTION:

  You *WILL* want to add checks that the client defined
  "filename" can not overwrite arbitrary files on the 
  server.

  Thus, in the production version, you will replace
  the "file.filename" above with something more intelligent
  and safer. You probably will only take the last component,
  prepend a standard prefix
  and add something to make the resulting name unique.


More enhancements expected....



Dieter