urllib

当url地址含有中文，或者参数有中文的时候，这个算是很正常了，但是把这样的url作为参数传递的时候（最常见的callback），需要把一些中文甚至'/'做一下编码转换。 一、urlencode urllib库里面有个urlencode函数，可以把key-value这样的键值对转换成我们想要的格式，返回的是a=1&b=2这样的字符串，比如： >>> from urllib import urlencode >>> data = { ... 'a': 'test', ... 'name': '魔兽' ... } >>> print urlencode(data) a=test&name=%C4%A7%CA%DE 如果只想对一个字符串进行urlencode转换，怎么办？urllib提供另外一个函数：quote() >>> from urllib import quote >>> quote('魔兽') '%C4%A7%CA%DE' 二、urldecode 当urlencode之后的字符串传递过来之后，接受完毕就要解码了——urldecode。urllib提供了unquote()这个函数，可没有urldecode()！ >>> from urllib import unquote >>> unquote('%C4%A7%CA%DE') '\xc4\xa7\xca\xde' >>>

I want to open and read https://yande.re/ with urllib.request , but I'm getting an SSL error. I can open and read the page just fine using http.client with this code: import http.client conn = http.client.HTTPSConnection('www.yande.re') conn.request('GET', 'https://yande.re/') resp = conn.getresponse() data = resp.read() However, the following code using urllib.request fails: import urllib.request opener = urllib.request.build_opener() resp = opener.open('https://yande.re/') data = resp.read() It gives me the following error: ssl.SSLError: [Errno 1] _ssl.c:392: error:1411809D:SSL routines:SSL

I'm slamming my head against the wall with this one. I've been trying every example, reading every last bit I can find online about basic http authorization with urllib2, but I can not figure out what is causing my specific error. Adding to the frustration is that the code works for one page, and yet not for another. logging into www.mysite.com/adm goes absolutely smooth. It authenticates no problem. Yet if I change the address to 'http://mysite.com/adm/items.php?n=201105&c=200' I receive this error: <h4 align="center" class="teal">Add/Edit Items</h4> <p><strong>Client:</strong> </p><p><strong

So, I'm messing around with urllib.request in Python 3 and am wondering how to write the result of getting an internet file to a file on the local machine. I tried this: g = urllib.request.urlopen('http://media-mcw.cursecdn.com/3/3f/Beta.png') with open('test.png', 'b+w') as f: f.write(g) But I got this error: TypeError: 'HTTPResponse' does not support the buffer interface What am I doing wrong? NOTE: I have seen this question , but it's related to Python 2's urllib2 which was overhauled in Python 3. change f.write(g) to f.write(g.read()) An easier way I think (also you can do it in two lines)

I have a program that uses urllib to periodically fetch a url, and I see intermittent errors like : I/O error(socket error): [Errno 111] Connection refused. It works 90% of the time, but the othe r10% it fails. If retry the fetch immediately after it fails, it succeeds. I'm unable to figure out why this is so. I tried to see if any ports are available, and they are. Any debugging ideas? For additional info, the stack trace is: File "/usr/lib/python2.6/urllib.py", line 203, in open return getattr(self, name)(url) File "/usr/lib/python2.6/urllib.py", line 342, in open_http h.endheaders() File "

　　python操作网络，即打开一个网站，或者请求一个http接口，可以通过使用python自带的标准模块urllib或第三方库requests实现 一、使用urllib模块操作网络 　　urllib模块是一个标准模块，直接import urllib即可，在python3里面只有urllib模块，在python2里面有urllib模块和urllib2模块。使用urlib模块发送请求实例如下： 1 from urllib import request 2 from urllib import parse 3 import json 4 5 #1、发送get请求 6 url = 'http://api.xxxx.cn/api/user/stu_info' 7 data = {'stu_name':'xiaohei'} 8 tmpData = parse.urlencode(data) #将数据格式变成Kv k=v 9 tmpUrl=url +'?'+tmpData #将接口url和参数拼接 10 res = request.urlopen(tmpUrl) #请求接口 11 resForRead=res.read() #通过read方法获取返回值结果，返回值结果是bytes 12 print(type(resForRead)) 13 resForString= resForRead

一、urllib模块 urllib模块是一个标准模块，直接import urllib即可，在python3里面只有urllib模块，在python2里面有urllib模块和urllib2模块。 urllib模块太麻烦了，传参数的话，都得是bytes类型，返回数据也是bytes类型，还得解码，想直接把返回结果拿出来使用的话，还得用json，发get请求和post请求，也不通，使用比较麻烦 1 import json 2 from urllib import request 3 from urllib import parse 4 5 #【get请求】 6 url = 'http://api.nnzhp.cn/api/user/stu_info' 7 8 data={"stu_name":"xiaohei"} 9 10 tmpData=parse.urlencode(data) #1、将数据变为k=v模式 11 print(tmpData) 12 # 接口+参数 13 tmpUrl=url+'?'+tmpData # 接口参数拼接 14 print(tmpUrl) 15 res = request.urlopen(tmpUrl) # 请求接口 16 resForRead = res.read() # 通过read安啊获取返回值结果，返回值结果为Bytes类型 17 print(res

I opened python code from github . I assumed it was python2.x and got the above error when I tried to run it. From the reading I've seen Python 3 has depreciated urllib itself and replaced it with a number of libraries including urllib.request . It looks like the code was written in python 3 (a confirmation from someone who knows would be appreciated.) At this point I don't want to move to Python 3 - I haven't researched what it would do to my existing code. Thinking there should be a urllib module for Python 2 , I searched Google (using "python2 urllib download") and did not find one. (It

一、使用python自带模块urllib 模拟页面请求服务端，python提供了一个urllib模块，作用是通过python代码调用接口进行参数传递并获取到接口的返回值信息 urllib模式是一个标准模块，直接import urllib即可 1、发送get请求 from urllib import request from urllib import parse import json #get请求 # url='http://www.nzhp.cn/api/user/stu_info' # # data={'stu_name':'xiaohei'} #传递参数 # # tmpData=parse.urlencode(data) #将数据变成kv形式，即k=v # #接口+参数 # tmpUrl=url+'?'+tmpData # 将接口和参数拼接 # res=request.urlopen(tmpUrl) #请求接口 # resForRead=res.read() #通过read方法获取返回值结果，返回值结果是二进制bytes的类型 # resForString=resForRead.decode() #通过decode将bytes 转换成str类型，反过来是用encode # resForDict=json.loads(resForString)