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python发送post和get请求get请求:使用get方式时,请求数据直接放在url中。方法一、import urllibimport urllib2url = "

1 首先安装requests这个模块,方法是 输入pip install requests。当然你要装好pip这个包管理器。 2 然后 import requests url = " " # 这里是你的url your_param = {'Refer':"sina_index"} # 这里是你要发送的请求参数!, 它在url后边加上 ??Re...

http://blog.csdn.net/u012374229/article/details/46743877

你运气真好,我这刚好有现成的,下面大写的常量就是一些字符串,涉及我们自己的协议,就不发你了,你随便写个就行 def RequestCenter( data,url,web="127.0.0.1" ,port=9228 ): headDic = {} headDic[REQUEST_HEAD_LENGTH] = len( data ) tempLi...

使用requests库就可以了: pip install requests payload = {'key1': 'value1', 'key2': 'value2'} r = requests.post("ht tp:/ /httpbin.or g/post", data=payload) print(r.text) 然后参考这个网址: htt p:/ /docs.python-reques ts.org /en/...

import httplib, urllib from urlparse import urlparse def httppost(url, **kwgs): httpClient = None conn = urlparse(url) try: params = urllib.urlencode(dict(kwgs)) header = {"Content-type": "application/x-www-form-urlencoded", "A...

没用过3.3 不过看你提示 是 ascii_letters这个模块无法导入 那你改哈写法 from string import * 不去指定特定的模块 运行看看,还有其他的错误没

import urllib2import urllib#定义一个要提交的数据数组(字典)data = {}data['username'] = 'zgx030030'data['password'] = '123456'#定义post的地址url = 'http://www.test.com/post/'post_data = urllib.urlencode(data)#提交,发送数据req = ...

import httplib, urllibfrom urlparse import urlparsedef httppost(url, **kwgs): httpClient = None conn = urlparse(url) try: params = urllib.urlencode(dict(kwgs)) header = {"Content-type": "application/x-www-form-urlencoded", "Acc...

使用requests库就可以了: pip install requests payload = {'key1': 'value1', 'key2': 'value2'} r = requests.post("http://httpbin.org/post", data=payload) print(r.text) 然后参考这个网址: http://docs.python-requests.org/en/latest/...

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