urllib

Mimic curl in python

前提是你 提交于 2020-05-15 04:52:36
问题 The following curl code works: curl --form addressFile=@t.csv --form benchmark=Public_AR_Census2010 http://geocoding.geo.census.gov/geocoder/locations/addressbatch t.csv is simply 1, 800 Wilshire Blvd, Los Angeles, CA, 90017 How do I mimic this in python. So far all of my attempts have resulted in 'Bad Requests'. I am also trying to keep everything in memory--no writing to file. One attempt: import requests url = "http://geocoding.geo.census.gov/geocoder/json/addressbatch" # data is csv like

scraping multiple page with urllib get response 403

一世执手 提交于 2020-04-18 05:48:26
问题 iam new to python, im trying to scraping page that not change the url when change the page, with this code : from time import sleep from selenium import webdriver from bs4 import BeautifulSoup as bs from selenium.webdriver.chrome.options import Options chrome_options = Options() chrome_options.add_argument("disable-extensions") chrome_options.add_argument("disable-gpu") chrome_options.add_argument("headless") path =r'F:\python latian\webdriver\chromedriver.exe' driver = webdriver.Chrome

Converting string to date in numpy unpack

喜夏-厌秋 提交于 2020-04-17 20:27:40
问题 I'm learning how to extract data from links and then proceeding to graph them. For this tutorial, I was using the yahoo dataset of a stock. The code is as follows import matplotlib.pyplot as plt import numpy as np import urllib import matplotlib.dates as mdates import datetime def bytespdate2num(fmt, encoding='utf-8'): strconverter = mdates.strpdate2num(fmt) def bytesconverter(b): s = b.decode(encoding) return strconverter(s) return bytesconverter def graph_data(stock): stock_price_url =

Save HTML of some website in a txt file with python

冷暖自知 提交于 2020-04-05 12:06:25
问题 I need save the HTML code of any website in a txt file, is a very easy exercise but I have doubts with this because a have a function that do this: import urllib.request def get_html(url): f=open('htmlcode.txt','w') page=urllib.request.urlopen(url) pagetext=page.read() ## Save the html and later save in the file f.write(pagetext) f.close() But this doesn't work. 回答1: Easiest way would be to use urlretrieve: import urllib urllib.urlretrieve("http://www.example.com/test.html", "test.txt") For

Python爬虫入门之三urllib库的基本使用

半腔热情 提交于 2020-03-25 20:54:35
前言 所谓网页抓取,就是把URL地址中指定的网络资源从网络流中读取出来,保存到本地。在Python中有很多库可以用来抓取网页,我们先学习urllib。 注:此博客开发环境为python3 urlopen 我们先来段代码: # urllib_urlopen.py # 导入urllib.request import urllib.request # 向指定的url发送请求,并返回服务器响应的类文件对象 response = urllib.request.urlopen("http://www.baidu.com") # 类文件对象支持 文件对象的操作方法,如read()方法读取文件全部内容,返回字符串 html = response.read() # 打印字符串 print(html) 执行写的python代码,将打印结果: python3 urllib_urlopen.py ** 实际上,如果我们在浏览器上打开百度主页, 右键选择“查看源代码”,你会发现,跟我们刚才打印出来的是一模一样。也就是说,上面的4行代码就已经帮我们把百度的首页的全部代码爬了下来。** 一个基本的url请求对应的python代码真的非常简单。 Request 在我们第一个例子里,urlopen()的参数就是一个url地址;但是如果需要执行更复杂的操作,比如增加HTTP报头,必须创建一个 Request

Using urlencode for Devanagari text

限于喜欢 提交于 2020-03-24 00:23:10
问题 The following code: import simplejson,urllib,urllib2 query=[u'नेपाल'] urlbase="http://search.twitter.com/search.json" values={'q':query[0]} data=urllib.urlencode(values) req=urllib2.Request(urlbase,data) response=urllib2.urlopen(req) json=simplejson.load(response) print json throws exception: SyntaxError: Non-ASCII character '\xe0' in file ques.py on line 3, but no encoding declared; see http://www.python.org/peps/pep-0263.html for details The code works if query contains standard ASCII

Using urlencode for Devanagari text

被刻印的时光 ゝ 提交于 2020-03-24 00:21:39
问题 The following code: import simplejson,urllib,urllib2 query=[u'नेपाल'] urlbase="http://search.twitter.com/search.json" values={'q':query[0]} data=urllib.urlencode(values) req=urllib2.Request(urlbase,data) response=urllib2.urlopen(req) json=simplejson.load(response) print json throws exception: SyntaxError: Non-ASCII character '\xe0' in file ques.py on line 3, but no encoding declared; see http://www.python.org/peps/pep-0263.html for details The code works if query contains standard ASCII

urllib运用(1)

筅森魡賤 提交于 2020-03-23 14:31:55
# 1.导入urllib中的request # 2.定义爬取的url(统一资源定位符) # 3.定义一个请求对象request # request的参数有url:访问的网址;data:发起请求时带的数据请求方式为post; # headers:包括发送HTTP报文的键值对(例如user-agent请求头)可以利用random中的choice方法随机一个user-agent # 4.定义一个响应对象接收访问的网页信息(此时返回的是一个response对象) # 5.使用read()方法和decode()方法对返回的数据进行处理返回成utf-8的格式 from urllib import request url_1 = 'https://www.baidu.com/' header = { 'User-Agent':'Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:68.0) Gecko/20100101 Firefox/68.0' } req = request.Request(url=url_1,headers=header) response = request.urlopen(req).read().decode('utf-8') print(response) 来源: https://www.cnblogs.com

Python Web编程

岁酱吖の 提交于 2020-03-21 07:43:32
1.统一资源定位符(URL) URL用来在Web上定位一个文档。浏览器只是Web客户端的一种,任何一个向服务器端发送请求来获取数据的应用程序都被认为是客户端 URL格式:port_sch://net_loc/path;params?query#frag port_sch  网络协议或者下载规划,如http /net_loc  服务器位置,如www.baidu.com path  斜杠/限定文件或者CGI应用程序的路径 params  可选参数 query  连接符&连接键值对 frag  拆分文档中的特殊锚 2.urllib模块 1 urlopen(urlstr,postQueryData=None) #打开一个给定URL字符串与Web连接,并返回了文件类的对象 2 f.read([bytes]) #从f中读出所有或bytes个字节 3 f.readline() #从f中读出一行 4 f.readlines() #从f中读出所有行并返回一个列表 5 f.close() #关闭f的URL的连接 6 f.fileno() #返回f文件的句柄 7 f.info() #获得f的MIME头文件,文件类型可以用哪类应用程序打开 8 f.geturl() #返回f所打开的真正的URL 9 10 urlretrieve(urlstr,localfile=None

爬虫入门学习 贴吧小案例

不打扰是莪最后的温柔 提交于 2020-03-10 19:54:12
1 import urllib.request 2 import urllib.parse 3 import random 4 5 #目标地址 6 url="http://tieba.baidu.com/f" 7 8 #伪造客户端 http请求头 9 ua_list = [ 10 "User-Agent: Mozilla/5.0 (Macintosh; Intel Mac OS X 10.6; rv:2.0.1) Gecko/20100101 Firefox/4.0.1", 11 "User-Agent: Mozilla/5.0 (Windows NT 6.1; rv:2.0.1) Gecko/20100101 Firefox/4.0.1", 12 "User-Agent: Opera/9.80 (Macintosh; Intel Mac OS X 10.6.8; U; en) Presto/2.8.131 Version/11.11", 13 "User-Agent: Opera/9.80 (Windows NT 6.1; U; en) Presto/2.8.131 Version/11.11", 14 "User-Agent: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_7_0) AppleWebKit/535.11 (KHTML,