Question

I want to make a curl call with a URL that has multiple parameters. I have listed the code below. Is there an equivalent option for “curl -d @filter” for instance or do I have URL encode the parameters?

SER = foobar
PASS = XXX

STREAM_URL = "http://status.dummy.com/status.json?userId=12&page=1"
class Client:   
          def __init__(self):
        self.buffer = ""
        self.conn = pycurl.Curl()
        self.conn.setopt(pycurl.USERPWD, "%s:%s" % (USER,PASS))
        self.conn.setopt(pycurl.URL, STREAM_URL)
        self.conn.setopt(pycurl.WRITEFUNCTION, self.on_receive)
        self.conn.perform()

    def on_receive(self,data):
        self.buffer += data

Answer 1

Pycurl is a pretty thin wrapper for libcurl. If you can do it with libcurl, you can do it with pycurl. (Mostly.)

For instance:

pycurl.setopt corresponds to curl_easy_setopt in libcurl, where option is specified with the CURLOPT_* constants in libcurl, except that the CURLOPT_ prefix has been removed.

See: http://pycurl.sourceforge.net/doc/curlobject.html

That being said, the curl -d option is for sending HTTP POST requests… not the GET style your example shows.

libcurl does expect that urls it recives already be URL encoded. Just use http://docs.python.org/library/urllib.html if needed.

The sample URL in your question already has 2 parameters (userId and page).

In general the format is: URL followed by a ‘question mark’, followed by name=value pairs joined by an ampersand symbol. If the name or value contain special chars, you will need to percent-encoded them.

Just use the urlencode function:

>>> import urllib
>>> params = [('name1','value1'), ('name2','value2 with spaces & stuff')]
>>> pairs = urllib.urlencode(params)
>>> fullurl = 'http://status.dummy.com/status.json' + '?' + pairs
>>> print fullurl
http://status.dummy.com/status.json?name1=value1&name2=value2+with+spaces+%26+stuff
>>> 

Also, see the urllib.urlopen function. Perhaps you do not need curl at all? (But I do not know your application…)

Hope this helps. If so, mark answered and let me know. 🙂