我使用urllib2请求URL并读取它们的内容,但不幸的是,它并不适用于某些URL。请查看以下命令:
#No problem with this URL
urllib2.urlopen('http://www.huffingtonpost.com/2014/07/19/todd-akin-slavery_n_5602083.html')
#This one produced error
urllib2.urlopen('http://www.foxnews.com/us/2014/07/19/cartels-suspected-as-high-caliber-gunfi
嗨,我想要处理的错误如下(当wifi丢弃数据包时,这种情况时有发生):
Traceback (most recent call last):
File "twittersearch.py", line 40, in <module>
data = json.load(urllib2.urlopen(response))
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 127, in urlopen
return
我有一些链接的列表(这是pdf文件)。我对列表中的每个链接都运行了wget.download()。然而,只有其中的一部分被下载了,然后我得到了: File "/home/.local/lib/python3.6/site-packages/wget.py", line 526, in download
(tmpfile, headers) = ulib.urlretrieve(binurl, tmpfile, callback)
File "/usr/lib/python3.6/urllib/request.py", line 248, in ur
我已经在几个程序上得到这个错误,现在我已经尝试升级pytube,重新安装它,尝试一些修复,更改url和代码,但似乎都不起作用 from pytube import YouTube
#ask for the link from user
link = input("Enter the link of YouTube video you want to download: ")
yt = YouTube(link)
#Showing details
print("Title: ",yt.title)
print("Number of views: &
from selenium import webdriver
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
from selenium.webdriver.common.keys import Keys
import time
import os
import wget
PATH = "/Users/k
我需要这样做,但是用Python而不是Bash:
i=1
while [ $i <= 10 ] ; do
wget http://somewebsite.net/shared/fshared_$i.7z
$i = $i + 1
done
在Python中,我尝试使用以下方法:
import urllib, os
i = 0
while i <= 3:
os.system('wget http://somewebsite.net/shared/fshared_',i,'.7z')
i = i + 1
但是它不起作用,变量
self.driver.quit()
File "C:\Python27\lib\site-packages\selenium-2.40.0-py2.7.egg\selenium\webdriver\chrome\webdriver.py", line 82, in quit
self.service.stop()
File "C:\Python27\lib\site-packages\selenium-2.40.0-py2.7.egg\selenium\webdriver\chrome\service.py", line 97, in sto
我正在尝试从站点下载页面
像这样
url = 'http://statsheet.com'
urllib2.urlopen(url)
我尝试过使用Python模块urllib、urllib2和"reqests",但只收到了诸如“get a bad status line”、"BadStatusLine“或类似的错误消息
有什么办法可以绕过这个问题吗?
我只想简单地运行脚本
不管它给了
File "./sparsecov-example.py", line 92, in <module>
for symbol in symbols]
File "/usr/lib/pymodules/python2.7/matplotlib/finance.py", line 236, in quotes_historical_yahoo
fh = fetch_historical_yahoo(ticker, date1, date2, cachename)
File "/usr/l
不知道它是什么,但它一直让我发疯和我这么一个新手我尝试了这么多的东西,但我不知道这是什么意思,我尝试安装和重新安装钢管,但没有什么改变。
这是我的密码
from pytube import YouTube
link = input("Enter in a link: ")
video = YouTube(link)
print(video.title)
print(video.streams.filter(progressive=True))
stream = video.streams.get_by_itag(22)
stream.download()
print(
我使用下面的python程序,当我执行它时,我得到一个HTTP403:禁止的错误。
下面是代码:
import os
import re
from bs4 import BeautifulSoup
#import urllib.request
#from urllib.request import request, urlopen
from urllib import request
import pandas as pd
import numpy as np
import datetime
import time
import openpyxl
for a in range(0,len