如何将curl命令行转换为pycurl代码

这是我的原始代码

import urllib, urllib2
from xml.dom import minidom

base_url = 'test.com:8000'
username = 'MyId'
password = 'passwd'
search_query = 'search error | head 10'

# Login and get the session key
request = urllib2.Request(base_url + '/servicesNS/MyId/search/auth/login', 
data = urllib.urlencode({'username': username, 'password': password}))

server_content = urllib2.urlopen(request)

session_key = minidom.parseString(server_content.read()).\
getElementsByTagName('sessionKey')[0].childNodes[0].nodeValue
print "Session Key: %s" % session_key 

# Perform a search
request = urllib2.Request(base_url + '/servicesNS/MyId/search/search/jobs/export', 
data = urllib.urlencode({'search': search_query,'output_mode': 'json'}),
headers = { 'Authorization': ('Splunk %s' %session_key)})
search_results = urllib2.urlopen(request)
print search_results.read()

它导致404错误,所以我决定使用Curl.

我有这个卷曲的赞扬

  curl -k -u myID:Passwd -d 'search="google.com | head 10"' -d "output_mode=json" https://test.com:8089/servicesNS/myId/search/search/jobs/export

我正在尝试使用pycurl来调用Joson.

我如何将其转换为pycurl？

先感谢您.

将curl命令行转换为使用libcurl的任何东西(如pycurl或其他绑定)的最佳方法是运行命令行并在其后附加" --libcurl output.c",然后检查创建的胚胎为然后它显示了curl必须做什么.

将该C代码转换为pycurl代码通常是微不足道的.