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'
Related
I have a list of some links (which are pdf files). I'm running wget.download() for each link in the list. However, only some of them get downloaded and then I get:
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 urlretrieve
with contextlib.closing(urlopen(url, data)) as fp:
File "/usr/lib/python3.6/urllib/request.py", line 223, in urlopen
return opener.open(url, data, timeout)
File "/usr/lib/python3.6/urllib/request.py", line 532, in open
response = meth(req, response)
File "/usr/lib/python3.6/urllib/request.py", line 642, in http_response
'http', request, response, code, msg, hdrs)
File "/usr/lib/python3.6/urllib/request.py", line 570, in error
return self._call_chain(*args)
File "/usr/lib/python3.6/urllib/request.py", line 504, in _call_chain
result = func(*args)
File "/usr/lib/python3.6/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 also tried to use r = requests.get(link) but the issue is still the same and additionally, I get this error:
File "/usr/lib/python3.6/urllib/request.py", line 532, in open
response = meth(req, response)
File "/usr/lib/python3.6/urllib/request.py", line 642, in http_response
'http', request, response, code, msg, hdrs)
File "/usr/lib/python3.6/urllib/request.py", line 570, in error
return self._call_chain(*args)
File "/usr/lib/python3.6/urllib/request.py", line 504, in _call_chain
result = func(*args)
File "/usr/lib/python3.6/urllib/request.py", line 650, in http_error_default
raise HTTPError(req.full_url, code, msg, hdrs, fp
Example of link that does not get downloaded:
https://cnds.jacobs-university.de/courses/sads-2020/p6.pdf
If I open the link on my browser, I get the download. Also, this method was working until a few months back. Idk why it's not working anymore
I can't comment as of yet, but have you tried downloading the files with a browser manually? If that works then the issue must be that wget can't reach the target for some reason.
I am trying to parse some data from 'https://datausa.io/profile/geo/jacksonville-fl/#intro', but I am not sure how to access it from python. My code is:
adress, headers = urllib.request.urlretrieve(' https://datausa.io/profile/geo/jacksonville-fl/#intro')
handle = open(adress)
and it returns the error:
Traceback (most recent call last):
File "C:/Users/Jared/AppData/Local/Programs/Python/Python36-32/capstone1.py", line 16, in <module>
adress, headers = urllib.request.urlretrieve(' https://datausa.io/profile/geo/jacksonville-fl/#intro')
File "C:\Users\Jared\AppData\Local\Programs\Python\Python36-32\lib\urllib\request.py", line 248, in urlretrieve
with contextlib.closing(urlopen(url, data)) as fp:
File "C:\Users\Jared\AppData\Local\Programs\Python\Python36-32\lib\urllib\request.py", line 223, in urlopen
return opener.open(url, data, timeout)
File "C:\Users\Jared\AppData\Local\Programs\Python\Python36-32\lib\urllib\request.py", line 532, in open
response = meth(req, response)
File "C:\Users\Jared\AppData\Local\Programs\Python\Python36-32\lib\urllib\request.py", line 642, in http_response
'http', request, response, code, msg, hdrs)
File "C:\Users\Jared\AppData\Local\Programs\Python\Python36-32\lib\urllib\request.py", line 570, in error
return self._call_chain(*args)
File "C:\Users\Jared\AppData\Local\Programs\Python\Python36-32\lib\urllib\request.py", line 504, in _call_chain
result = func(*args)
File "C:\Users\Jared\AppData\Local\Programs\Python\Python36-32\lib\urllib\request.py", line 650, in http_error_default
raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 403: Forbidden
Please explain what is wrong or tell me a better way to access the page. Also, does the ' .io ' suffix affecthow python handles it?
Thanks.
This worked for me:
import requests
url = "https://datausa.io/profile/geo/jacksonville-fl/#intro"
req = requests.request("GET",url)
I seem to be getting this error with urllib.request and it gives me this url error that i cant seem to fix.
raceback (most recent call last):
File "C:\Users\Jarvis\Documents\Python Scripts\MultiCheck by Koala.py", line 133, in <module>
Migration()
File "C:\Users\Jarvis\Documents\Python Scripts\MultiCheck by Koala.py", line 116, in Migration
rawdata_uuid = urllib.request.urlopen(url)
File "C:\Python34\lib\urllib\request.py", line 161, in urlopen
return opener.open(url, data, timeout)
File "C:\Python34\lib\urllib\request.py", line 469, in open
response = meth(req, response)
File "C:\Python34\lib\urllib\request.py", line 579, in http_response
'http', request, response, code, msg, hdrs)
File "C:\Python34\lib\urllib\request.py", line 507, in error
return self._call_chain(*args)
File "C:\Python34\lib\urllib\request.py", line 441, in _call_chain
result = func(*args)
File "C:\Python34\lib\urllib\request.py", line 587, in http_error_default
raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 429: 42
The code im using is here is for a migration checker for a game:
def Migration():
url = "https://api.mojang.com/users/profiles/minecraft/" + einfos
rawdata = urllib.request.urlopen(url)
newrawdata = rawdata.read()
jsondata = json.loads(newrawdata.decode('utf-8'))
results = jsondata['id']
url = "https://sessionserver.mojang.com/session/minecraft/profile/" + results
rawdata_uuid = urllib.request.urlopen(url)
newrawdata_uuid = rawdata_uuid.read()
jsondata_uuid = json.loads(newrawdata_uuid.decode('utf-8'))
try:
results = jsondata_uuid['legacy']
print ("Unmigrated")
except:
print("Migrated")
Error 429 means: Too many requests. You seem to have hit a rate limit. The additional number gives are the seconds you have to wait for the limitation to be dropped. So, try again in 42s, or later.
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)