Posted on Jul 5, 2011 by
Happy Camel
The quick start code sample is incorrect (at least for 0.2.2).
This line:
srv = pysftp.Connect('www.yoursever.com')
should read:
srv = pysftp.Connection('www.yoursever.com')
Comment #1
Posted on Feb 8, 2012 by Grumpy MonkeyI'll handle this one.
Comment #2
Posted on Feb 8, 2012 by Grumpy Monkeypage has been updated.
Status: Done
Labels:
Type-Defect
Priority-Medium