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File Upload Recipe

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Problem

File uploads can be a little tricky if you're not familiar with form uploads, or CGI in general.

Solution

import web

urls = ('/upload', 'Upload')

class Upload:
    def GET(self):
        return """<html><head></head><body>
<form method="POST" enctype="multipart/form-data" action="">
<input type="file" name="myfile" />
<br/>
<input type="submit" />
</form>
</body></html>"""

    def POST(self):
        x = web.input(myfile={})
        web.debug(x['myfile'].filename) # This is the filename
        web.debug(x['myfile'].value) # This is the file contents
        web.debug(x['myfile'].file.read()) # Or use a file(-like) object
        raise web.seeother('/upload')


if __name__ == "__main__":
   app = web.application(urls, globals()) 
   app.run()

Hang ups

A couple of things to watch out for:

  • The form needs an attribute enctype="multipart/form-data", or this won't work correctly.
  • In the webpy code, a default value is needed (the myfile={} part) if you want it to be imported as a CGI FieldStorage object. If you don't specify the default value, the file will be passed as a string -- this works, but you lose the filename attribute.