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.
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)
Here is the code:
quote_page = "https://www.timeanddate.com/holidays/fun/" + months [date.month].lower() + "/" + str(date.day)
page = urllib2.urlopen(quote_page)
soup = BeautifulSoup(page, 'html.parser')
event_box = soup.find('article', attrs={"class" : "fixed"})
event_box = event_box.find('h3')
event = event_box.text.strip()
print event
When I check my variable:
quote_page = https://www.timeanddate.com/holidays/fun/june/8
I've tried printing quote_page and the link works fine in my browser,
but when I run the code and print "event" I get this:
Traceback (most recent call last):
File "main.py", line 252, in <module>
page = urllib2.urlopen(req)
File "/usr/local/lib/python2.7/urllib2.py", line 154, in urlopen
return opener.open(url, data, timeout)
File "/usr/local/lib/python2.7/urllib2.py", line 435, in open
response = meth(req, response)
File "/usr/local/lib/python2.7/urllib2.py", line 548, in http_response
'http', request, response, code, msg, hdrs)
File "/usr/local/lib/python2.7/urllib2.py", line 473, in error
return self._call_chain(*args)
File "/usr/local/lib/python2.7/urllib2.py", line 407, in _call_chain
result = func(*args)
File "/usr/local/lib/python2.7/urllib2.py", line 556, in http_error_default
raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 403: ok
exit status 1
I think I remember that it worked earlier today but stopped when I tried again later.
Can anyone help?
(I'm fairly new to coding)
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)
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'
callurl = "http://vgintnh116:8001/master_data/"
params = urllib.urlencode({'res': 'arovit', 'qfields': 'prod' })
f = urllib2.urlopen(callurl, params)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/u/vgtools2/python-2.6.5/lib/python2.6/urllib2.py", line 126, in urlopen
return _opener.open(url, data, timeout)
File "/u/vgtools2/python-2.6.5/lib/python2.6/urllib2.py", line 397, in open
response = meth(req, response)
File "/u/vgtools2/python-2.6.5/lib/python2.6/urllib2.py", line 510, in http_response
'http', request, response, code, msg, hdrs)
File "/u/vgtools2/python-2.6.5/lib/python2.6/urllib2.py", line 435, in error
return self._call_chain(*args)
File "/u/vgtools2/python-2.6.5/lib/python2.6/urllib2.py", line 369, in _call_chain
result = func(*args)
File "/u/vgtools2/python-2.6.5/lib/python2.6/urllib2.py", line 518, in http_error_default
raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 403: FORBIDDEN
But it works with -
callurl = "http://vgintnh116:8001/master_data/res=arovit&qfields=prod"
f = urllib2.urlopen(callurl)
Please help. I want to use urlencode to avoid handling spaces and extra characters.
If you pass the second argument (data), request will be POST instead of GET.
Also, dictionaries in Python does not have order. To guarantee the order, you should use sequence.
callurl = "http://vgintnh116:8001/master_data/"
params = urllib.urlencode([('res', 'arovit'), ('qfields', 'prod')])
f = urllib2.urlopen(callurl + params)
From urllib2 documentation:
the HTTP request will be a POST instead of a GET when the data
parameter is provided
In your working example, you are making a GET request.