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    <p>Pycurl is a pretty thin wrapper for libcurl. If you can do it with libcurl, you can do it with pycurl. (Mostly.)</p> <p>For instance:</p> <blockquote> <p>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.</p> </blockquote> <p>See: <a href="http://pycurl.sourceforge.net/doc/curlobject.html" rel="nofollow">http://pycurl.sourceforge.net/doc/curlobject.html</a></p> <p>That being said, the <code>curl -d</code> option is for sending HTTP POST requests... not the GET style your example shows.</p> <p>libcurl does expect that urls it recives already be URL encoded. Just use <a href="http://docs.python.org/library/urllib.html" rel="nofollow">http://docs.python.org/library/urllib.html</a> if needed.</p> <p>The sample URL in your question already has 2 parameters (userId and page). </p> <p>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.</p> <p>Just use the urlencode function:</p> <pre><code>&gt;&gt;&gt; import urllib &gt;&gt;&gt; params = [('name1','value1'), ('name2','value2 with spaces &amp; stuff')] &gt;&gt;&gt; pairs = urllib.urlencode(params) &gt;&gt;&gt; fullurl = 'http://status.dummy.com/status.json' + '?' + pairs &gt;&gt;&gt; print fullurl http://status.dummy.com/status.json?name1=value1&amp;name2=value2+with+spaces+%26+stuff &gt;&gt;&gt; </code></pre> <p>Also, see the <code>urllib.urlopen</code> function. Perhaps you do not need curl at all? (But I do not know your application...)</p> <p><strong>Hope this helps.</strong> If so, mark answered and let me know. :-)</p>
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    1. COThanks for info nonot1 but I am looking for a way to include multiple parameters in the URL. My guess is that I am doing wrong in the URL but I don't how to fix it.
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    2. CO@coolcajun See updated answer
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