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pythonsyntax-errorpycurl

Invalid syntax on ".perform()" with Python

I am trying to download a zip file to my Ubuntu 10.04 workstation and limit the transmit limit to 100 kB/s. When running the script I get the following:

  File "./iso.py", line 7
    iso.perform()
      ^
SyntaxError: invalid syntax

Here is the code I am using. I am not sure where the actual syntax error is. I have searched Google for a while now before asking here. Any help would be appreciated.

#!/usr/bin/env python
import pycurl
iso = pycurl.Curl()
iso.setopt(iso.URL, "http://downloads.sourceforge.net/sevenzip/7za920.zip")
iso.setopt(iso.MAX_RECV_SPEED_LARGE, 100000)
iso.setopt(iso.WRITEDATA, file("7za920.zip")
iso.perform()

FYI I am running Python Version 2.6.5

Solution

  • You forgot an extra parenthesis after your previous line.

    Change:

    iso.setopt(iso.WRITEDATA, file("7za920.zip")

    To:

    iso.setopt(iso.WRITEDATA, file("7za920.zip"))

    Python is interpretting this as if you are continuing to add to the function (eg, add more parameters). There's a SyntaxError because there's no comma.