urllib

(Python 3.4.2) I've got a weird error when I run 'urllib.request.urlopen(url)' inside of a script. If I run it directly in the Python interpreter, it works fine, but not when I run it inside of a script through a bash shell (Linux). I'm guessing it has something to do with the 'url' string, maybe because I'm creating the string through the 'string.join' method. import urllib.request url = "".join((baseurl, other_string, midurl, query)) response = urllib.request.urlopen(url) The 'url' string prints perfectly, but when I try to create the 'response' string, I get this output: File "./script.py",

I am trying to convert working Python 2.7 code into Python 3 code and I am receiving a type error from the urllib request module. I used the inbuilt 2to3 Python tool to convert the below working urllib and urllib2 Python 2.7 code: import urllib2 import urllib url = "https://www.customdomain.com" d = dict(parameter1="value1", parameter2="value2") req = urllib2.Request(url, data=urllib.urlencode(d)) f = urllib2.urlopen(req) resp = f.read() The output from the 2to3 module was the below Python 3 code: import urllib.request, urllib.error, urllib.parse url = "https://www.customdomain.com" d = dict

I want to download image file from a url using python module "urllib.request", which works for some website (e.g. mangastream.com), but does not work for another (mangadoom.co) receiving error "HTTP Error 403: Forbidden". What could be the problem for the latter case and how to fix it? I am using python3.4 on OSX. import urllib.request # does not work img_url = 'http://mangadoom.co/wp-content/manga/5170/886/005.png' img_filename = 'my_img.png' urllib.request.urlretrieve(img_url, img_filename) At the end of error message it said: ... HTTPError: HTTP Error 403: Forbidden However, it works for

I am using following code to save webpage using Python: import urllib import sys from bs4 import BeautifulSoup url = 'http://www.vodafone.de/privat/tarife/red-smartphone-tarife.html' f = urllib.urlretrieve(url,'test.html') Problem : This code saves html as basic html without javascripts, images etc. I want to save webpage as complete (Like we have option in browser) Update : I am using following code now to save all the js/images/css files of webapge so that it can be saved as complete webpage but still my output html is getting saved like basic html: import pycurl import StringIO c = pycurl

艺赛旗 RPA9.0全新首发免费下载 点击下载 http://www.i-search.com.cn/index.html?from=line1 函数参数说明 urllib.request.urlretrieve(url, filename=None, reporthook=None, data=None) url: 文件下载链接 filename: 文件下载路径（如果参数未指定，urllib 会生成一个临时文件保存数据） reporthook: 回调函数，当连接上服务器、以及相应的数据块传输完毕时会触发该回调，我们可以利用这个回调函数来显示当前的下载进度 data: 指 post 到服务器的数据。该方法返回一个包含两个元素的元组 (filename, headers)，filename 表示保存到本地的路径，header 表示服务器的响应头 示例 import urllib.request def download(): #下载链接 down_url=r" https://av.sc.com/hk/zh/content/docs/hk-c-nddr-ff304m-ag-20190809.pdf " #储存路径 down_path=r"C:\Users\Administrator\Desktop\test\aa.pdf" #链接转义，防止链接中有中文或空格而报错 down_url

While porting code from python2 to 3 , I get this error when reading from a URL TypeError: initial_value must be str or None, not bytes. import urllib import json import gzip from urllib.parse import urlencode from urllib.request import Request service_url = 'https://babelfy.io/v1/disambiguate' text = 'BabelNet is both a multilingual encyclopedic dictionary and a semantic network' lang = 'EN' Key = 'KEY' params = { 'text' : text, 'key' : Key, 'lang' :'EN' } url = service_url + '?' + urllib.urlencode(params) request = Request(url) request.add_header('Accept-encoding', 'gzip') response = urllib

If I have a URL that, when submitted in a web browser, pops up a dialog box to save a zip file, how would I go about catching and downloading this zip file in Python? senderle Use urllib2.urlopen . The return value is a file-like object that you can read() , pass to zipfile and so on. yoavram As far as I can tell, the proper way to do this is: import requests, zipfile, StringIO r = requests.get(zip_file_url, stream=True) z = zipfile.ZipFile(StringIO.StringIO(r.content)) z.extractall() of course you'd want to check that the GET was successful with r.ok . For python 3+, sub the StringIO module