pycurl problem

[Gabriel Genellina]

En Wed, 18 Apr 2007 11:48:06 -0300, pabloski <pabloski at gmail.com> escribió:

> I noted that if I define c.proxy and c.url as costants e.g. c.url =
> "http://www.google.com/search?blah blah...." and c.proxy =
> "127.0.0.1:8080" it works
>
> However if I define them as described before pycurl raises the TypeError

Python -with its long "batteries included" tradition- comes with two  
powerful tools for diagnosing problems: print and repr

print "c.url", type(c.url), repr(c.url)
print "c.proxy", type(c.proxy), repr(c.proxy)

-- 
Gabriel Genellina