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pycurl problem

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
Apr 18 '07 #1
1 1729
En Wed, 18 Apr 2007 11:48:06 -0300, pabloski <pa******@gmail.comescribió:
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

Apr 18 '07 #2

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