I have tried every possible fix I can find online, unfortunately, I'm new to this and not sure if I'm getting closer or not.
Ultimately, all I am trying to do is print a JSON feed into a Google Sheet.
GSpread is working (I've appended just number values as a test), but I simply cannot get the JSON feed to print there.
I've gotten it printing to terminal, so I know it's accessible, but writing the loop to append the data becomes the issue.
This is my current script:
# import urllib library
import json
from urllib.request import urlopen
import gspread
gc = gspread.service_account(filename='creds.json')
sh = gc.open_by_key('1-1aiGMn2yUWRlh_jnIebcMNs-6phzUNxkktAFH7uY9o')
worksheet = sh.sheet1
# import json
# store the URL in url as
# parameter for urlopen
url = 'https://api.chucknorris.io/jokes/random'
# store the response of URL
response = urlopen(url)
# storing the JSON response
# from url in data
data_json = json.loads(response.read())
# print the json response
# print(data_json)
result = []
for key in data_json:
result.append([key, data_json[key]])
worksheet.update('a1', result)
I've hit a complete brick wall - any advice would be greatly appreciated
Update - suggested script with new error:
# import urllib library
import json
from urllib.request import urlopen
import gspread
gc = gspread.service_account(filename='creds.json')
sh = gc.open_by_key('1-1aiGMn2yUWRlh_jnIebcMNs-6phzUNxkktAFH7uY9o')
worksheet = sh.sheet1
url = 'https://api.chucknorris.io/jokes/random'
# store the response of URL
response = urlopen(url)
# storing the JSON response
# from url in data
data_json = json.loads(response.read())
# print the json response
# print(data_json)
result = []
for key in data_json:
result.append([key, data_json[key] if not isinstance(
data_json[key], list) else ",".join(map(str, data_json[key]))])
worksheet.update('a1', result)
Error:
Traceback (most recent call last):
File "c:\Users\AMadle\NBA-JSON-Fetch\PrintToSheetTest.py", line 17, in <module>
response = urlopen(url)
File "C:\Python\python3.10.5\lib\urllib\request.py", line 216, in urlopen
return opener.open(url, data, timeout)
File "C:\Python\python3.10.5\lib\urllib\request.py", line 525, in open
response = meth(req, response)
File "C:\Python\python3.10.5\lib\urllib\request.py", line 634, in http_response
response = self.parent.error(
File "C:\Python\python3.10.5\lib\urllib\request.py", line 563, in error
return self._call_chain(*args)
File "C:\Python\python3.10.5\lib\urllib\request.py", line 496, in _call_chain
result = func(*args)
File "C:\Python\python3.10.5\lib\urllib\request.py", line 643, in http_error_default
raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 403: Forbidden
Can confirm it is not a permissions issue, the script below prints the same URL to terminal no problem. Also have no problem writing other data to the sheet:
import requests as rq
from bs4 import BeautifulSoup
url = 'https://api.chucknorris.io/jokes/random'
req = rq.get(url, verify=False)
soup = BeautifulSoup(req.text, 'html.parser')
print(soup)
In your script, I thought that it is required to convert the JSON data to a 2-dimensional array. And, when I saw the value of data_json, I noticed that an array is included in the value. I think that it is required to be also considered. I thought that this might be the reason for your issue. When this is reflected in your script, how about the following modification?
From:
result.append([key, data_json[key]])
To:
result.append([key, data_json[key] if not isinstance(data_json[key], list) else ",".join(map(str, data_json[key]))])
In this modification, the array is converted to the string using join.
Related
I'm trying to import the Json data generated by an Impinj R420 reader.
The code i use is:
# import urllib library
import urllib.request
from urllib.request import urlopen
# import json
import json
# store the URL in url as
# parameter for urlopen
url = "http://10.234.92.19:14150"
# store the response of URL
response = urllib.request.urlopen(url)
# storing the JSON response
# from url in data
data_json = json.loads(response())
# print the json response
print(data_json)
When i execute the programm it gives the following error:
Traceback (most recent call last):
File "C:\Users\V\PycharmProjects\Stapelcontrole\main.py", line 13, in <module>
response = urllib.request.urlopen(url)
File "C:\Users\V\AppData\Local\Programs\Python\Python310\lib\urllib\request.py", line 216, in urlopen
return opener.open(url, data, timeout)
File "C:\Users\V\AppData\Local\Programs\Python\Python310\lib\urllib\request.py", line 519, in open
response = self._open(req, data)
File "C:\Users\V\AppData\Local\Programs\Python\Python310\lib\urllib\request.py", line 536, in _open
result = self._call_chain(self.handle_open, protocol, protocol +
File "C:\Users\V\AppData\Local\Programs\Python\Python310\lib\urllib\request.py", line 496, in _call_chain
result = func(*args)
File "C:\Users\V\AppData\Local\Programs\Python\Python310\lib\urllib\request.py", line 1377, in http_open
return self.do_open(http.client.HTTPConnection, req)
File "C:\Users\V\AppData\Local\Programs\Python\Python310\lib\urllib\request.py", line 1352, in do_open
r = h.getresponse()
File "C:\Users\V\AppData\Local\Programs\Python\Python310\lib\http\client.py", line 1374, in getresponse
response.begin()
File "C:\Users\V\AppData\Local\Programs\Python\Python310\lib\http\client.py", line 318, in begin
version, status, reason = self._read_status()
File "C:\Users\V\AppData\Local\Programs\Python\Python310\lib\http\client.py", line 300, in _read_status
raise BadStatusLine(line)
http.client.BadStatusLine: {"epc":"3035307B2831B383E019E8EA","firstSeenTimestamp":"2022-04-11T11:24:23.592434Z","isHeartBeat":false}
Process finished with exit code 1
I know this is an error in the response where it gets a faulty HTTP status code.
Yet i don't know how to fix the error.
Could you advice me how to fix this?
The {"epc":"3035307B2831B383E019E8EA","firstSeenTimestamp":"2022-04-11T11:24:23.592434Z","isHeartBeat":false} is an answer i expect.
Thanks in advance
Edit:
Even with
with urllib.request.urlopen(url) as f:
data_json = json.load(f)`
i get the same BadStatusLine error.
I can't setup the reader any different, it can only sent a JSON response trough the IP-adress of the device. Is there a way to import the data without the HTTP Protocol?
# store the response of URL
response = urllib.request.urlopen(url)
# storing the JSON response
# from url in data
data_json = json.loads(response())
Here you are actually calling response, I do not know what you want to achieve by that, but examples in urllib.request docs suggest that urllib.request.urlopen should be treated akin to local file handle, thus please replace above using
with urllib.request.urlopen(url) as f:
data_json = json.load(f)
Observe that I used json.load not json.loads
EDIT: After Is there a way to import the data without the HTTP Protocol? I conclude more low-level solution is needed, hopefully socket will you allow to get what you want, using echo client example as starting point I prepared following code
import socket
with socket.socket(socket.AF_INET, socket.SOCK_STREAM) as s:
s.connect(("10.234.92.19",14150))
s.sendall(b'GET / HTTP/1.1\r\n')
data = s.recv(1024)
print(data)
If everything will work as intended you should get printed 1024 first bytes of answer. If it so change 1024 to value which will be always bigger than number of bytes of response and use json.dumps(data) to get parsed response
So recently I came across the Google Testing Tools API - Mobile Friendly Test (https://developers.google.com/webmaster-tools/search-console-api/reference/rest/v1/urlTestingTools.mobileFriendlyTest) but I couldn't work it even when I am trying on the site. I tried to use python for this app and followed the guide (https://developers.google.com/webmaster-tools/search-console-api/v1/samples) and I made some few changes to actually make it work (since urllib was merged into one library). So end of the day my code looked like this:
from __future__ import print_function
import urllib
import urllib.request as urllib2
api_key = 'API_KEY'
request_url = 'https://www.google.com/'
service_url = 'https://searchconsole.googleapis.com/v1/urlTestingTools/mobileFriendlyTest:run'
params = {
'url' : request_url,
'key' : api_key,
}
data = urllib.parse.urlencode(params)
content = urllib2.urlopen(url=service_url, data=str.encode(data)).read()
print(content)
And I got the error:
File ".\script2.py", line 14, in <module>
content = urllib2.urlopen(url=service_url, data=str.encode(data)).read()
File "C:\Python\lib\urllib\request.py", line 222, in urlopen
return opener.open(url, data, timeout)
File "C:\Python\lib\urllib\request.py", line 531, in open
response = meth(req, response)
File "C:\Python\lib\urllib\request.py", line 641, in http_response
'http', request, response, code, msg, hdrs)
File "C:\Python\lib\urllib\request.py", line 569, in error
return self._call_chain(*args)
File "C:\Python\lib\urllib\request.py", line 503, in _call_chain
result = func(*args)
File "C:\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 403: Forbidden
I also tried using curl command and the online tool (not https://search.google.com/test/mobile-friendly but Try This API section) but neither of them worked.
Well I actually solved my own problem, it is mainly caused by urllib I think. So here is what I did;
from __future__ import print_function
import urllib.parse as parser
import urllib.request as urllib2
import json
import base64
request_url = url
params = {
'url': request_url,
'key': api_key
}
data = bytes(parser.urlencode(params), encoding='utf-8')
content = urllib2.urlopen(url=service_url, data=data).read()
sContent = str(content, encoding='utf-8') #Shorthand for stringContent
I am trying to geocode a column of neighorhood names in a pandas Data Frame. I know that the first url generated works when I copy and paste it into a browser heading, but when I run the api query through the Python script I get an error.
Here's the function:
def geo_code(geo_vec):
partial_api = "https://maps.googleapis.com/maps/api/geocode/json?key="+API_KEY+"&address="
address_list = []
for g in geo_vec:
url = partial_api+g+", New York"
print(url)
response = urllib2.urlopen(url)
And here's the error I'm getting:
Traceback (most recent call last):
File "plot.py", line 33, in <module>
geo_code(geo['Descriptive Name'])
File "plot.py", line 22, in geo_code
response = urllib2.urlopen(url)
File "/usr/lib/python2.7/urllib2.py", line 154, in urlopen
return opener.open(url, data, timeout)
File "/usr/lib/python2.7/urllib2.py", line 437, in open
response = meth(req, response)
File "/usr/lib/python2.7/urllib2.py", line 550, in http_response
'http', request, response, code, msg, hdrs)
File "/usr/lib/python2.7/urllib2.py", line 475, in error
return self._call_chain(*args)
File "/usr/lib/python2.7/urllib2.py", line 409, in _call_chain
result = func(*args)
File "/usr/lib/python2.7/urllib2.py", line 558, in http_error_default
raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 400: Bad Request
I have literally copied the generated url from the console where I logged it and pasted it into the browser, and then it works fine. What might urllib2 be doing to make Google call this a bad request? Also it's the first request (first time through the loop) that I get this error, so it's not like it's an api limit error.
I think the problem is because I am using inputs to the url that include adding at the end names with spaces such as "New York City" and "East Harlem". How should I be processing these strings before adding them to the url?
Here's what I tried:
Using urllib.urlencode({''address' : g}) in the place of "address="+g in the url. This results in the same error even though now the spaces are all replaced with the appropriate + sign.
I'm quite baffled because the url seems fine, and other than url encoding I'm not sure what else I should be doing. Thanks for any suggestions.
To focus on your code and what is wrong with it. You are not handling your url parameters properly when you are formatting your string in your loop. You should be encoding your url parameters before making your call. Especially when you send "New York", the space will not be accepted.
To resolve this, you should be using urlencode.
When using urlencode, you just need to provide a dictionary of key value pairs and urlencode will convert it in to a properly encoded string in the format a=1&b=2.
With that, you combine it with your URL and that should give you the properly structured URL you need to make your API call. Your adjusted code here ran for me but as expected received an error since I don't have a valid API key.
I've set comments in-line to provide more context.
def geo_code(geo_vec):
# Here simply use your base URL without query params. You can append the ?
partial_api = "https://maps.googleapis.com/maps/api/geocode/json?"
address_list = []
# Put together the initial structure of your query parameters here as a dictionary
d = {
"key": API_KEY,
"address": ""
}
for g in geo_vec:
# Add your new query param here
d["address"] = "g{}".format("New York")
# Call urlencode here to properly structure your query parameters
url_params = urlencode(d)
# Properly structure your URL string to make your API call
url = "{}{}".format(partial_api, url_params)
# Make API call
response = urlopen(url)
However, as an alternative, and it might make your life easier. It was mentioned in the comments. You should take a look at the requests module.
Hi my code won't work when actually running online, it returns None when i use Find how can i fix this?
This is my code;
import time
import sys
import urllib
import re
from bs4 import BeautifulSoup, NavigableString
print "Initializing Python Script"
print "The passed arguments are "
urls = ["http://tweakers.net/pricewatch/355474/gigabyte-gv-n78toc-3g/specificaties/", "http://tweakers.net/pricewatch/328943/sapphire-radeon-hd-7950-3gb-gddr5-with-boosts/specificaties/", "https://www.alternate.nl/GIGABYTE/GV-N78TOC-3GD-grafische-kaart/html/product/1115798", "http://tweakers.net/pricewatch/320116/raspberry-pi-model-b-(512mb)/specificaties/"]
i =0
regex = '<title>(.+?)</title>'
pattern = re.compile(regex)
word = "tweakers"
alternate = "alternate"
while i<len(urls):
dataraw = urllib.urlopen(urls[i])
data = dataraw.read()
soup = BeautifulSoup(data)
table = soup.find("table", {"class" : "spec-detail"})
print table
i+=1
Here is the outcome:
Initializing Python Script
The passed arguments are
None
None
None
None
Script finalized
i have tried using findAll and other methods.. But i don't seem to understand why it is working on my Command line but not on the server itself...
Any help?
Edit
Traceback (most recent call last):
File "python_script.py", line 35, in
soup = BeautifulSoup(urllib2.urlopen(url), 'html.parser')
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 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 403: Forbidden
I'm suspecting you are experiencing the differences between parsers.
Specifying the parser explicitly works for me:
import urllib2
from bs4 import BeautifulSoup
urls = ["http://tweakers.net/pricewatch/355474/gigabyte-gv-n78toc-3g/specificaties/",
"http://tweakers.net/pricewatch/328943/sapphire-radeon-hd-7950-3gb-gddr5-with-boosts/specificaties/",
"https://www.alternate.nl/GIGABYTE/GV-N78TOC-3GD-grafische-kaart/html/product/1115798",
"http://tweakers.net/pricewatch/320116/raspberry-pi-model-b-(512mb)/specificaties/"]
for url in urls:
soup = BeautifulSoup(urllib2.urlopen(url), 'html.parser')
table = soup.find("table", {"class": "spec-detail"})
print table
In this case, I'm using html.parser, but you can play around and specify lxml or html5lib, for example.
Note that the third url doesn't contain a table with class="spec-detail" and, therefore, it prints None for it.
I've also introduced few improvements:
removed unused imports
replaced a while loop with indexing with a nice for loop
removed unrelated code
replaced urllib with urllib2
You can also use requests module and set appropriate User-Agent header pretending to be a real browser:
from bs4 import BeautifulSoup
import requests
urls = ["http://tweakers.net/pricewatch/355474/gigabyte-gv-n78toc-3g/specificaties/",
"http://tweakers.net/pricewatch/328943/sapphire-radeon-hd-7950-3gb-gddr5-with-boosts/specificaties/",
"https://www.alternate.nl/GIGABYTE/GV-N78TOC-3GD-grafische-kaart/html/product/1115798",
"http://tweakers.net/pricewatch/320116/raspberry-pi-model-b-(512mb)/specificaties/"]
headers = {'User-Agent': 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_9_5) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/37.0.2062.124 Safari/537.36'}
for url in urls:
response = requests.get(url, headers=headers)
soup = BeautifulSoup(response.content, 'html.parser')
table = soup.find("table", {"class": "spec-detail"})
print table
Hope that helps.
i want to create a script that returns all the urls found in a page a google for example , so i create this script : (using BeautifulSoup)
import urllib2
from BeautifulSoup import BeautifulSoup
page = urllib2.urlopen("https://www.google.dz/search?q=see")
soup = BeautifulSoup(page.read())
links = soup.findAll("a")
for link in links:
print link["href"]
and it return this 403 forbidden result :
Traceback (most recent call last):
File "C:\Python27\sql\sql.py", line 3, in <module>
page = urllib2.urlopen("https://www.google.dz/search?q=see")
File "C:\Python27\lib\urllib2.py", line 126, in urlopen
return _opener.open(url, data, timeout)
File "C:\Python27\lib\urllib2.py", line 400, in open
response = meth(req, response)
File "C:\Python27\lib\urllib2.py", line 513, in http_response
'http', request, response, code, msg, hdrs)
File "C:\Python27\lib\urllib2.py", line 438, in error
return self._call_chain(*args)
File "C:\Python27\lib\urllib2.py", line 372, in _call_chain
result = func(*args)
File "C:\Python27\lib\urllib2.py", line 521, in http_error_default
raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 403: Forbidden
any idea to avoid this error or another methode to get the urls from of the search result ?
No problem using requests
import requests
from BeautifulSoup import BeautifulSoup
page = requests.get("https://www.google.dz/search?q=see")
soup = BeautifulSoup(page.content)
links = soup.findAll("a")
Some of the links have links are like search%:http:// where the end of one joins another so we need to split then using re
import requests
from bs4 import BeautifulSoup
page = requests.get("https://www.google.dz/search?q=see")
soup = BeautifulSoup(page.content)
import re
links = soup.findAll("a")
for link in soup.find_all("a",href=re.compile("(?<=/url\?q=)(htt.*://.*)")):
print re.split(":(?=http)",link["href"].replace("/url?q=",""))
['https://www.see.asso.fr/&sa=U&ei=ryv6U6PvEKzA7AaB4ICwCA&ved=0CBIQFjAA&usg=AFQjCNF2_I8jB98JwR3jcKniLZekSrRO7Q']
['http://webcache.googleusercontent.com/search%3Fq%3Dcache:f7M8NX1XmDsJ', 'https://www.see.asso.fr/%252Bsee%26hl%3Dfr%26%26ct%3Dclnk&sa=U&ei=ryv6U6PvEKzA7AaB4ICwCA&ved=0CBUQIDAA&usg=AFQjCNF8WJButjMNXQXvXBbtyXnF1SgiOg']
['https://www.see.asso.fr/3ei&sa=U&ei=ryv6U6PvEKzA7AaB4ICwCA&ved=0CBgQ0gIoADAA&usg=AFQjCNGnPL1RiX5TekI_yMUc-w_f2oVXtw']
['https://www.see.asso.fr/node/9587&sa=U&ei=ryv6U6PvEKzA7AaB4ICwCA&ved=0CBkQ0gIoATAA&usg=AFQjCNHX-6AzBgLQUF0s8TxFcZjIhxz_Hw']
['https://www.see.asso.fr/ree&sa=U&ei=ryv6U6PvEKzA7AaB4ICwCA&ved=0CBoQ0gIoAjAA&usg=AFQjCNGkkd8e1JjiNrhSM4HQYE-M6g6j-w']
['https://www.see.asso.fr/node/130&sa=U&ei=ryv6U6PvEKzA7AaB4ICwCA&ved=0CBsQ0gIoAzAA&usg=AFQjCNEkVdpcbXDz5-cV9u2NNYoV6aM8VA']
['http://www.wordreference.com/enfr/see&sa=U&ei=ryv6U6PvEKzA7AaB4ICwCA&ved=0CB0QFjAB&usg=AFQjCNHQGwcsGpro26dhxFP6q-fQvwbB0Q']
['http://webcache.googleusercontent.com/search%3Fq%3Dcache:ooK-I_HuCkwJ', 'http://www.wordreference.com/enfr/see%252Bsee%26hl%3Dfr%26%26ct%3Dclnk&sa=U&ei=ryv6U6PvEKzA7AaB4ICwCA&ved=0CCAQIDAB&usg=AFQjCNFRlV5Zv_n48Wivr4LeOkTQsA0D1Q']
['http://fr.wikipedia.org/wiki/S%25C3%25A9e&sa=U&ei=ryv6U6PvEKzA7AaB4ICwCA&ved=0CCMQFjAC&usg=AFQjCNGmtqmcXPqYZ_nwa0RWL0uYf5PMJw']
['http://webcache.googleusercontent.com/search%3Fq%3Dcache:GjcgkyzsUigJ', 'http://fr.wikipedia.org/wiki/S%2525C3%2525A9e%252Bsee%26hl%3Dfr%26%26ct%3Dclnk&sa=U&ei=ryv6U6PvEKzA7AaB4ICwCA&ved=0CCYQIDAC&usg=AFQjCNHesOIBU3OXBspARcONbK_k_8-gnw']
['http://fr.wikipedia.org/wiki/Camille_S%25C3%25A9e&sa=U&ei=ryv6U6PvEKzA7AaB4ICwCA&ved=0CCkQFjAD&usg=AFQjCNGO-WIDl4TrBeo88WY9QsopWmsMyQ']
['http://webcache.googleusercontent.com/search%3Fq%3Dcache:izhQjC85nOoJ', 'http://fr.wikipedia.org/wiki/Camille_S%2525C3%2525A9e%252Bsee%26hl%3Dfr%26%26ct%3Dclnk&sa=U&ei=ryv6U6PvEKzA7AaB4ICwCA&ved=0CCwQIDAD&usg=AFQjCNEfcIKsKbf026xgWT7NkrAueZvL0A']
['http://de.wikipedia.org/wiki/Zugersee&sa=U&ei=ryv6U6PvEKzA7AaB4ICwCA&ved=0CDEQ9QEwBA&usg=AFQjCNHpfJW5-XdsgpFUSP-jEmHjXQUWHQ']
['http://commons.wikimedia.org/wiki/File:Champex_See.jpg&sa=U&ei=ryv6U6PvEKzA7AaB4ICwCA&ved=0CDMQ9QEwBQ&usg=AFQjCNEordFWr2QIaob45WlR5Yi-ZvZSiA']
['http://www.all-free-photos.com/show/showphotop.php%3Fidtop%3D4%26lang%3Dfr&sa=U&ei=ryv6U6PvEKzA7AaB4ICwCA&ved=0CDUQ9QEwBg&usg=AFQjCNEC24FOIE5cvF4zmEDgq5-5xubM3w']
['http://www.allbestwallpapers.com/travel-zell_am_see,_kaprun,_austria_wallpapers.html&sa=U&ei=ryv6U6PvEKzA7AaB4ICwCA&ved=0CDcQ9QEwBw&usg=AFQjCNFkzMZDuthZHvnF-JvyksNUqjt1dQ']
['http://www.see-swe.org/&sa=U&ei=ryv6U6PvEKzA7AaB4ICwCA&ved=0CDkQFjAI&usg=AFQjCNF1zbcLfjanxgCXtHoOQXOdMgh_AQ']
['http://webcache.googleusercontent.com/search%3Fq%3Dcache:lzh6JxvKUTIJ', 'http://www.see-swe.org/%252Bsee%26hl%3Dfr%26%26ct%3Dclnk&sa=U&ei=ryv6U6PvEKzA7AaB4ICwCA&ved=0CDwQIDAI&usg=AFQjCNFYN6tzzVaHsAc5aOvYNql3Zy4m3A']
['http://fr.wiktionary.org/wiki/see&sa=U&ei=ryv6U6PvEKzA7AaB4ICwCA&ved=0CD8QFjAJ&usg=AFQjCNFWYIGc1gj0prytowzqI-0LDFRvZA']
['http://webcache.googleusercontent.com/search%3Fq%3Dcache:G9v8lXWRCyQJ', 'http://fr.wiktionary.org/wiki/see%252Bsee%26hl%3Dfr%26%26ct%3Dclnk&sa=U&ei=ryv6U6PvEKzA7AaB4ICwCA&ved=0CEIQIDAJ&usg=AFQjCNENzi4E1n-9qHYsNahY6lQzaW5Xvg']
['http://en.wiktionary.org/wiki/see&sa=U&ei=ryv6U6PvEKzA7AaB4ICwCA&ved=0CEUQFjAK&usg=AFQjCNECGZjw-rBUALO43WaTh2yB9BUhDg']
['http://webcache.googleusercontent.com/search%3Fq%3Dcache:ywc4URuPdIQJ', 'http://en.wiktionary.org/wiki/see%252Bsee%26hl%3Dfr%26%26ct%3Dclnk&sa=U&ei=ryv6U6PvEKzA7AaB4ICwCA&ved=0CEgQIDAK&usg=AFQjCNE0pykIqXXRl08E-uTtoj03QEpnbg']
['http://see-concept.com/&sa=U&ei=ryv6U6PvEKzA7AaB4ICwCA&ved=0CEsQFjAL&usg=AFQjCNGFWjhiH7dEBhITJt01ob_JENlz1Q']
['http://webcache.googleusercontent.com/search%3Fq%3Dcache:jHTkOVEoRsAJ', 'http://see-concept.com/%252Bsee%26hl%3Dfr%26%26ct%3Dclnk&sa=U&ei=ryv6U6PvEKzA7AaB4ICwCA&ved=0CE4QIDAL&usg=AFQjCNECPgxt9ZSFmZzK_ker9Hw_FoCi_A']
['http://www.theconjugator.com/la/conjugaison/du/verbe/see.html&sa=U&ei=ryv6U6PvEKzA7AaB4ICwCA&ved=0CFEQFjAM&usg=AFQjCNETCTQ0vPDIdV_2Q57qq11dyN0d8Q']
['http://webcache.googleusercontent.com/search%3Fq%3Dcache:xD7_Qo7roS8J', 'http://www.theconjugator.com/la/conjugaison/du/verbe/see.html%252Bsee%26hl%3Dfr%26%26ct%3Dclnk&sa=U&ei=ryv6U6PvEKzA7AaB4ICwCA&ved=0CFQQIDAM&usg=AFQjCNF_hBCyDZncivYGnL7je5kYme9hEg']
['http://www.zellamsee-kaprun.com/fr&sa=U&ei=ryv6U6PvEKzA7AaB4ICwCA&ved=0CFcQFjAN&usg=AFQjCNFVDeBWrZMDSjK9jKYF4AQlIXa9lA']
['http://webcache.googleusercontent.com/search%3Fq%3Dcache:BFBEUp05w7YJ', 'http://www.zellamsee-kaprun.com/fr%252Bsee%26hl%3Dfr%26%26ct%3Dclnk&sa=U&ei=ryv6U6PvEKzA7AaB4ICwCA&ved=0CFoQIDAN&usg=AFQjCNHtrOeEpYWqvT3f0M1p-gxUkYT1IA']
The best way to do this is to use the google API (pip install google)
GeeksforGeeks writes about it here
from googlesearch import search
# to search
query = "see"
links = []
for j in search(query, tld="co.in", num=10, stop=10, pause=2):
links.append(j)
import urllib.request
from BeautifulSoup import BeautifulSoup
page = urllib.request.urlopen("https://www.google.dz/search?q=see")
soup = BeautifulSoup(page.read())
links = soup.findAll("a")
for link in links:
print link["href"]