pycurl problem

2007-04-18 Thread pabloski 
Hi to all, I have a problem with a snippet of code that creates a Curl
object

the code is

c = pycurl.Curl()

c.key = keyCurrent
c.proxy = proxyCurrent
c.url = http://www.google.; + lg + /search?hl= + lg + q= + c.key +
meta=lr%3Dlang_ + lg + num=30

c.setopt(pycurl.URL, c.url)
c.setopt(pycurl.PROXY, c.proxy)

the last two lines raise a TypeError

obviously proxyCurrent, lg and keyCurrent are parameters

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
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Re: pycurl problem

2007-04-18 Thread Gabriel Genellina 
En Wed, 18 Apr 2007 11:48:06 -0300, pabloski [EMAIL PROTECTED] 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

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