I want to read the value of different stocks from websites. Therefore I wrote this tiny script, which reads the page source and then parses out the value:
stock_reader.py
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
from re import search
from urllib import request
def main():
links = [
[
'CSG',
'UBS',
],
[
'http://www.tradegate.de/orderbuch.php?isin=CH0012138530',
'http://www.tradegate.de/orderbuch.php?isin=CH0244767585',
],
]
for i in in range(len(links[0])):
url = links[1][i]
htmltext = request.urlopen(url).read().decode('utf-8')
source = htmltext.splitlines()
for line in source:
if 'id="bid"' in line:
m = search('\d+.\d+', line)
print('{}'.format(m.string[m.start():m.end()]))
if __name__ == '__main__':
main()
sometimes it works but sometimes this error gets raised:
error message
Traceback (most recent call last):
File "./aktien_reader.py", line 39, in <module>
main()
File "./aktien_reader.py", line 30, in main
htmltext = request.urlopen(url).read().decode('utf-8')
File "/usr/lib/python3.3/urllib/request.py", line 160, in urlopen
return opener.open(url, data, timeout)
File "/usr/lib/python3.3/urllib/request.py", line 479, in open
response = meth(req, response)
File "/usr/lib/python3.3/urllib/request.py", line 591, in http_response
'http', request, response, code, msg, hdrs)
File "/usr/lib/python3.3/urllib/request.py", line 511, in error
result = self._call_chain(*args)
File "/usr/lib/python3.3/urllib/request.py", line 451, in _call_chain
result = func(*args)
File "/usr/lib/python3.3/urllib/request.py", line 696, in http_error_302
return self.parent.open(new, timeout=req.timeout)
File "/usr/lib/python3.3/urllib/request.py", line 479, in open
response = meth(req, response)
File "/usr/lib/python3.3/urllib/request.py", line 591, in http_response
'http', request, response, code, msg, hdrs)
File "/usr/lib/python3.3/urllib/request.py", line 511, in error
result = self._call_chain(*args)
File "/usr/lib/python3.3/urllib/request.py", line 451, in _call_chain
result = func(*args)
File "/usr/lib/python3.3/urllib/request.py", line 696, in http_error_302
return self.parent.open(new, timeout=req.timeout)
File "/usr/lib/python3.3/urllib/request.py", line 479, in open
response = meth(req, response)
File "/usr/lib/python3.3/urllib/request.py", line 591, in http_response
'http', request, response, code, msg, hdrs)
File "/usr/lib/python3.3/urllib/request.py", line 511, in error
result = self._call_chain(*args)
File "/usr/lib/python3.3/urllib/request.py", line 451, in _call_chain
result = func(*args)
File "/usr/lib/python3.3/urllib/request.py", line 696, in http_error_302
return self.parent.open(new, timeout=req.timeout)
File "/usr/lib/python3.3/urllib/request.py", line 479, in open
response = meth(req, response)
File "/usr/lib/python3.3/urllib/request.py", line 591, in http_response
'http', request, response, code, msg, hdrs)
File "/usr/lib/python3.3/urllib/request.py", line 511, in error
result = self._call_chain(*args)
File "/usr/lib/python3.3/urllib/request.py", line 451, in _call_chain
result = func(*args)
File "/usr/lib/python3.3/urllib/request.py", line 696, in http_error_302
return self.parent.open(new, timeout=req.timeout)
File "/usr/lib/python3.3/urllib/request.py", line 479, in open
response = meth(req, response)
File "/usr/lib/python3.3/urllib/request.py", line 591, in http_response
'http', request, response, code, msg, hdrs)
File "/usr/lib/python3.3/urllib/request.py", line 511, in error
result = self._call_chain(*args)
File "/usr/lib/python3.3/urllib/request.py", line 451, in _call_chain
result = func(*args)
File "/usr/lib/python3.3/urllib/request.py", line 686, in http_error_302
self.inf_msg + msg, headers, fp)
urllib.error.HTTPError: HTTP Error 302: The HTTP server returned a redirect error that would lead to an infinite loop.
The last 30x error message was:
Found
My question is: why is it happening and how can I avoid it?
This happens probably because the destination site uses cookies and redirect you in case you don't send cookies.
What you can use is something like that :
from http.cookiejar import CookieJar
url = "http://www.tradegate.de/orderbuch.php?isin=CH0012138530"
req = urllib.request.Request(url, None, {'Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,*/*;q=0.8','Accept-Charset': 'ISO-8859-1,utf-8;q=0.7,*;q=0.3','Accept-Encoding': 'gzip, deflate, sdch','Accept-Language': 'en-US,en;q=0.8','Connection': 'keep-alive'})
cj = CookieJar()
opener = urllib.request.build_opener(urllib.request.HTTPCookieProcessor(cj))
response = opener.open(req)
response.read()
This way, you support Cookies and website will allow you to get the page :-)
Another way would be to use the requests package which is really simplest to use. In your case, it would lead to :
import requests
url = "http://www.tradegate.de/orderbuch.php?isin=CH0012138530"
r = requests.get(url, headers={'Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8'}, timeout=15)
print(r.content)
This answer is a simplification of the one by Cédric J. You don't really need to import CookieJar or set various Accept headers if you don't want to. You should however generally set a timeout. It is tested with Python 3.7. I would typically remember to use a new opener for each random URL that I want cookies for.
from urllib.request import build_opener, HTTPCookieProcessor, Request
url = 'https://www.cell.com/cell-metabolism/fulltext/S1550-4131(18)30630-2'
opener = build_opener(HTTPCookieProcessor())
Without a Request object:
response = opener.open(url, timeout=30)
content = response.read()
With a Request object:
request = Request(url)
response = opener.open(request, timeout=30)
content = response.read()
HTTP Status code 302 it's a kind of a redirect, it will have a header with a new URL for access (Not necessary a working URL..)
Location: http://www.example.com/x/y/
This is quite often used to block bots who make to many requests in too of a short time frame. So not an coding problem.
Related
A simple web scraping code I wrote few weeks back keeps coming up with the error of:
HTTP Error 429: Too Many Requests
The code is designed to get the input from an excel file and find and download pdfs online.
I'm not too familiar with requests but I've slowed down the number of requests to see how many it can handle. It seems to be an unrelated issue somehow. The code will go through similar number of inputs (around 30) no matter if the delays I sat are at 5 seconds or 20 seconds. Here is the error message that keeps coming up:
Traceback (most recent call last):
File "D:\Python\New folder\Web Scraper.py", line 17, in <module>
for url in search(searchquery, stop=1, pause=2):
File "D:\Python\lib\site-packages\google-2.0.2-py3.7.egg\googlesearch\__init__.py", line 288, in search
html = get_page(url, user_agent)
File "D:\Python\lib\site-packages\google-2.0.2-py3.7.egg\googlesearch\__init__.py", line 154, in get_page
response = urlopen(request)
File "D:\Python\lib\urllib\request.py", line 222, in urlopen
return opener.open(url, data, timeout)
File "D:\Python\lib\urllib\request.py", line 531, in open
response = meth(req, response)
File "D:\Python\lib\urllib\request.py", line 641, in http_response
'http', request, response, code, msg, hdrs)
File "D:\Python\lib\urllib\request.py", line 563, in error
result = self._call_chain(*args)
File "D:\Python\lib\urllib\request.py", line 503, in _call_chain
result = func(*args)
File "D:\Python\lib\urllib\request.py", line 755, in http_error_302
return self.parent.open(new, timeout=req.timeout)
File "D:\Python\lib\urllib\request.py", line 531, in open
response = meth(req, response)
File "D:\Python\lib\urllib\request.py", line 641, in http_response
'http', request, response, code, msg, hdrs)
File "D:\Python\lib\urllib\request.py", line 569, in error
return self._call_chain(*args)
File "D:\Python\lib\urllib\request.py", line 503, in _call_chain
result = func(*args)
File "D:\Python\lib\urllib\request.py", line 649, in http_error_default
raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 429: Too Many Requests
And here is the code that I wrote:
import xlrd, requests
from googlesearch import search
from time import sleep
xlloc = ("D:/VesselBase.xlsx")
#Excel location
ws = xlrd.open_workbook(xlloc)
sheet = ws.sheet_by_index(0)
#Sheet name/index
sheet.cell_value(0, 0)
for i in range(sheet.nrows):
vesselname = sheet.cell_value(i, 1)
vesselimo = sheet.cell_value(i,0)
#Which column/row to choose, 2nd column for vessels. 0=A/1.
searchquery = 'Vessel specification information "%s" OR "%s" filetype:pdf' % (vesselname, vesselimo)
print('Searching "%s"' % searchquery)
for url in search(searchquery, stop=1, pause=20):
print('Searched for %s' % vesselname)
print('Found %s' % url)
open('D:/Newfolder/%s.pdf' % vesselname, 'wb').write(requests.get(url).content)
#Where to save
print('Saved %s' % vesselname)
My code is this:
import urllib.request
import re
http://www.weather-forecast.com/locations/Paris/forcasts/latest
city = input('Please enter a place: ')
url = 'http://www.weather-forecast.com/locations/'+city+'forcasts/latest'
data = urllib.request.urlopen(url).read()
data1 = data.decode('utf-8')
I'm having trouble with the url this is my output:
Traceback (most recent call last):
File "C:/Users/alext/AppData/Local/Programs/Python/Python36/Weather forecast.py", line 9, in
data = urllib.request.urlopen(url).read()
File "C:\Users\alext\AppData\Local\Programs\Python\Python36\lib\urllib\request.py", line 223, in urlopen
return opener.open(url, data, timeout)
File "C:\Users\alext\AppData\Local\Programs\Python\Python36\lib\urllib\request.py", line 532, in open
response = meth(req, response)
File "C:\Users\alext\AppData\Local\Programs\Python\Python36\lib\urllib\request.py", line 642, in http_response
'http', request, response, code, msg, hdrs)
File "C:\Users\alext\AppData\Local\Programs\Python\Python36\lib\urllib\request.py", line 564, in error
result = self._call_chain(*args)
File "C:\Users\alext\AppData\Local\Programs\Python\Python36\lib\urllib\request.py", line 504, in _call_chain
result = func(*args)
File "C:\Users\alext\AppData\Local\Programs\Python\Python36\lib\urllib\request.py", line 756, in http_error_302
return self.parent.open(new, timeout=req.timeout)
File "C:\Users\alext\AppData\Local\Programs\Python\Python36\lib\urllib\request.py", line 532, in open
response = meth(req, response)
File "C:\Users\alext\AppData\Local\Programs\Python\Python36\lib\urllib\request.py", line 642, in http_response
'http', request, response, code, msg, hdrs)
File "C:\Users\alext\AppData\Local\Programs\Python\Python36\lib\urllib\request.py", line 570, in error
return self._call_chain(*args)
File "C:\Users\alext\AppData\Local\Programs\Python\Python36\lib\urllib\request.py", line 504, in _call_chain
result = func(*args)
File "C:\Users\alext\AppData\Local\Programs\Python\Python36\lib\urllib \request.py", line 650, in http_error_default
raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 404: Not Found
I have checked the url and it is definitely correct. I have seen others with problems like this but am still unsure as to the solution.
you are missing a / after the city and a e in forecast. It should be
url = 'http://www.weather-forecast.com/locations/'+city+'/forecasts/latest'
I am writing a python script to use the urllib2 module as an equivalent to the command line utility wget. The only function I want for this is that it can be used to retrieve an arbitrary file based on URL and save it into a named file. I also only need to worry about two command line arguments, the URL from which the file is to be downloaded and the name of the file into which the content are to be saved.
Example:
python Prog7.py www.python.org pythonHomePage.html
This is my code:
import urllib
import urllib2
#import requests
url = 'http://www.python.org/pythonHomePage.html'
print "downloading with urllib"
urllib.urlretrieve(url, "code.txt")
print "downloading with urllib2"
f = urllib2.urlopen(url)
data = f.read()
with open("code2.txt", "wb") as code:
code.write(data)
urllib seems to work but urllib2 does not seem to work.
Errors received:
File "Problem7.py", line 11, in <module>
f = urllib2.urlopen(url)
File "/usr/lib64/python2.6/urllib2.py", line 126, in urlopen
return _opener.open(url, data, timeout)
File "/usr/lib64/python2.6/urllib2.py", line 397, in open
response = meth(req, response)
File "/usr/lib64/python2.6/urllib2.py", line 510, in http_response
'http', request, response, code, msg, hdrs)
File "/usr/lib64/python2.6/urllib2.py", line 429, in error
result = self._call_chain(*args)
File "/usr/lib64/python2.6/urllib2.py", line 369, in _call_chain
result = func(*args)
File "/usr/lib64/python2.6/urllib2.py", line 616, in http_error_302
return self.parent.open(new, timeout=req.timeout)
File "/usr/lib64/python2.6/urllib2.py", line 397, in open
response = meth(req, response)
File "/usr/lib64/python2.6/urllib2.py", line 510, in http_response
'http', request, response, code, msg, hdrs)
File "/usr/lib64/python2.6/urllib2.py", line 435, in error
return self._call_chain(*args)
File "/usr/lib64/python2.6/urllib2.py", line 369, in _call_chain
result = func(*args)
File "/usr/lib64/python2.6/urllib2.py", line 518, in http_error_default
raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 404: NOT FOUND
And the URL is doesn't exist at all; https://www.python.org/pythonHomePage.html is indeed a 404 Not Found page.
The difference between urllib and urllib2 then is that the latter automatically raises an exception when a 404 page is returned, while urllib.urlretrieve() just saves the error page for you:
>>> import urllib
>>> urllib.urlopen('https://www.python.org/pythonHomePage.html').getcode()
404
>>> import urllib2
>>> urllib2.urlopen('https://www.python.org/pythonHomePage.html')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/Users/mj/Development/Library/buildout.python/parts/opt/lib/python2.7/urllib2.py", line 127, in urlopen
return _opener.open(url, data, timeout)
File "/Users/mj/Development/Library/buildout.python/parts/opt/lib/python2.7/urllib2.py", line 410, in open
response = meth(req, response)
File "/Users/mj/Development/Library/buildout.python/parts/opt/lib/python2.7/urllib2.py", line 523, in http_response
'http', request, response, code, msg, hdrs)
File "/Users/mj/Development/Library/buildout.python/parts/opt/lib/python2.7/urllib2.py", line 448, in error
return self._call_chain(*args)
File "/Users/mj/Development/Library/buildout.python/parts/opt/lib/python2.7/urllib2.py", line 382, in _call_chain
result = func(*args)
File "/Users/mj/Development/Library/buildout.python/parts/opt/lib/python2.7/urllib2.py", line 531, in http_error_default
raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 404: NOT FOUND
If you wanted to save the error page, you can catch the urllib2.HTTPError exception:
try:
f = urllib2.urlopen(url)
data = f.read()
except urllib2.HTTPError as err:
data = err.read()
It is due to the different behavior by urllib and urllib2.
Since the web page returns a 404 error (webpage not found) urllib2 "catches" it while urllib downloads the html of the returned page regardless of the error.
If you want to print the html to the text file you can print the error:
import urllib2
try:
data = urllib2.urlopen('http://www.python.org/pythonHomePage.html').read()
except urllib2.HTTPError, e:
print e.code
print e.msg
print e.headers
print e.fp.read()
with open("code2.txt", "wb") as code:
code.write(e.fp.read())
req will be a Request object, fp will be a file-like object with the
HTTP error body, code will be the three-digit code of the error, msg
will be the user-visible explanation of the code and hdrs will be a
mapping object with the headers of the error.
More data about HTTP error: urllib2 documentation
I know we can handle redirection with requests and urllib2 packages. I do not want to use requests as it is not a prebuilt package. Please help me with urllib2 to handle a URL which takes a 302 and then moves to 404. I am not concerned about the 404 and I would like to track whether it is a 301 or 302.
I referred this doc but it still throws the 404.
Here is my code
import urllib2
class My_HTTPRedirectHandler(urllib2.HTTPRedirectHandler):
def http_error_302(self, req, fp, code, msg, headers):
return urllib2.HTTPRedirectHandler.http_error_302(self, req, fp, code, msg, headers)
my_opener = urllib2.build_opener(My_HTTPRedirectHandler)
urllib2.install_opener(my_opener)
response =urllib2.urlopen("MY URL")
print response.read()
Here is my response
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/lib/python2.7/urllib2.py", line 126, in urlopen
return _opener.open(url, data, timeout)
File "/usr/lib/python2.7/urllib2.py", line 406, in open
response = meth(req, response)
File "/usr/lib/python2.7/urllib2.py", line 519, in http_response
'http', request, response, code, msg, hdrs)
File "/usr/lib/python2.7/urllib2.py", line 438, in error
result = self._call_chain(*args)
File "/usr/lib/python2.7/urllib2.py", line 378, in _call_chain
result = func(*args)
File "<stdin>", line 3, in http_error_302
File "/usr/lib/python2.7/urllib2.py", line 625, in http_error_302
return self.parent.open(new, timeout=req.timeout)
File "/usr/lib/python2.7/urllib2.py", line 406, in open
response = meth(req, response)
File "/usr/lib/python2.7/urllib2.py", line 519, in http_response
'http', request, response, code, msg, hdrs)
File "/usr/lib/python2.7/urllib2.py", line 444, in error
return self._call_chain(*args)
File "/usr/lib/python2.7/urllib2.py", line 378, in _call_chain
result = func(*args)
File "/usr/lib/python2.7/urllib2.py", line 527, in http_error_default
raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 404: Not Found
It's better to use httplib and HTTP/1.1 HEAD method. That way no response body is received.
What’s the best way to get an HTTP response code from a URL?
I am getting the following error while trying to parse a link using beautiful soup,looks like the authentication is failing....how to authenticate the link?
from bs4 import BeautifulSoup
import argparse
import urllib
import urllib2
import getpass
import re
def update (searchurl):
page = urllib2.urlopen(searchurl)
soup = BeautifulSoup(page)
#print(soup.textarea.string)
print(soup.get_text())
def main ():
#For logging
print "test"
parser = argparse.ArgumentParser(description='This is the update.py script created by test')
parser.add_argument('-u','--url',action='store',dest='url',default=None,help='<Required> url link',required=True)
results = parser.parse_args()# collect cmd line args
url = results.url
#print url
update(url)
if __name__ == '__main__':
main()
Following is the error while trying to open the link
Traceback (most recent call last):
File "announce_update2.py", line 24, in <module>
main()
File "announce_update2.py", line 22, in main
update(url)
File "announce_update2.py", line 9, in update
page = urllib2.urlopen(searchurl)
File "C:\Python27\lib\urllib2.py", line 126, in urlopen
return _opener.open(url, data, timeout)
File "C:\Python27\lib\urllib2.py", line 406, in open
response = meth(req, response)
File "C:\Python27\lib\urllib2.py", line 519, in http_response
'http', request, response, code, msg, hdrs)
File "C:\Python27\lib\urllib2.py", line 438, in error
result = self._call_chain(*args)
File "C:\Python27\lib\urllib2.py", line 378, in _call_chain
result = func(*args)
File "C:\Python27\lib\urllib2.py", line 625, in http_error_302
return self.parent.open(new, timeout=req.timeout)
File "C:\Python27\lib\urllib2.py", line 406, in open
response = meth(req, response)
File "C:\Python27\lib\urllib2.py", line 519, in http_response
'http', request, response, code, msg, hdrs)
File "C:\Python27\lib\urllib2.py", line 444, in error
return self._call_chain(*args)
File "C:\Python27\lib\urllib2.py", line 378, in _call_chain
result = func(*args)
File "C:\Python27\lib\urllib2.py", line 527, in http_error_default
raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)