So I`m trying to send a request to a webpage and read its response. I did a code that compares the request and the page, and I cant get the same page text. Am I using "requests" correctly?
I really think that I misunderstand how requests function works and what it does. Can someone help me please?
import requests
import urllib
def search():
pr = {'q':'pink'}
r = requests.get('http://stackoverflow.com/search',params=pr)
returntext = r.text
urllibtest(returntext)
def urllibtest(returntext):
connection = urllib.urlopen("http://stackoverflow.com/search?q=pink")
output = connection.read()
connection.close()
if output == returntext:
print("ITS THE SAME PAGE")
else:
print("ITS NOT THE SAME PAGE")
search()
First of all, there is no good reason to expect two different stack overflow searches to return the exact same response anyway.
There is one logical difference here too, requests automatically decodes the output for you:
>>> type(output)
str
>>> type(r.text)
unicode
You can use the content instead if you don't want it decoded, and use a more predictable source to see the same content returned - for example:
>>> r1 = urllib.urlopen('http://httpbin.org').read()
>>> r2 = requests.get('http://httpbin.org').content
>>> r1 == r2
True
Related
I am trying to pull the the number of followers from a list of Instagram accounts. I have tried using the "find" method within Requests, however, the string that I am looking for when I inspect the actual Instagram no longer appears when I print "r" from the code below.
Was able to get this code to run successfully find the past, however, will no longer run.
Webscraping Instagram follower count BeautifulSoup
import requests
user = "espn"
url = 'https://www.instagram.com/' + user
r = requests.get(url).text
start = '"edge_followed_by":{"count":'
end = '},"followed_by_viewer"'
print(r[r.find(start)+len(start):r.rfind(end)])
I receive a "-1" error, which means the substring from the find method was not found within the variable "r".
I think it's because of the last ' in start and first ' in end...this will work:
import requests
import re
user = "espn"
url = 'https://www.instagram.com/' + user
r = requests.get(url).text
followers = re.search('"edge_followed_by":{"count":([0-9]+)}',r).group(1)
print(followers)
'14061730'
I want to suggest an updated solution to this question, as the answer of Derek Eden above from 2019 does not work anymore, as stated in its comments.
The solution was to add the r' before the regular expression in the re.search like so:
follower_count = re.search(r'"edge_followed_by\\":{\\"count\\":([0-9]+)}', response).group(1)
This r'' is really important as without it, Python seems to treat the expression as regular string which leads to the query not giving any results.
Also the instagram page seems to have backslashes in the object we look for at least in my tests, so the code example i use is the following in Python 3.10 and working as of July 2022:
# get follower count of instagram profile
import os.path
import requests
import re
import urllib3
urllib3.disable_warnings(urllib3.exceptions.InsecureRequestWarning)
# get instagram follower count
def get_instagram_follower_count(instagram_username):
url = "https://www.instagram.com/" + instagram_username
filename = "instagram.html"
try:
if not os.path.isfile(filename):
r = requests.get(url, verify=False)
print(r.status_code)
print(r.text)
response = r.text
if not r.status_code == 200:
raise Exception("Error: " + str(r.status_code))
with open(filename, "w") as f:
f.write(response)
else:
with open(filename, "r") as f:
response = f.read()
# print(response)
follower_count = re.search(r'"edge_followed_by\\":{\\"count\\":([0-9]+)}', response).group(1)
return follower_count
except Exception as e:
print(e)
return 0
print(get_instagram_follower_count('your.instagram.profile'))
The method returns the follower count as expected. Please note that i added a few lines to not hammer Instagrams webserver and get blocked while testing by just saving the response in a file.
This is a slice of the original html content that contains the part we are looking for:
... mRL&s=1\",\"edge_followed_by\":{\"count\":110070},\"fbid\":\"1784 ...
I debugged the regex in regexr, it seems to work just fine at this point in time.
There are many posts about the regex r prefix like this one
Also the documentation of the re package shows clearly that this is the issue with the code above.
I am trying to download books from "http://www.gutenberg.org/". I want to know why my code gets nothing.
import requests
import re
import os
import urllib
def get_response(url):
response = requests.get(url).text
return response
def get_content(html):
reg = re.compile(r'(<span class="mw-headline".*?</span></h2><ul><li>.*</a></li></ul>)',re.S)
return re.findall(reg,html)
def get_book_url(response):
reg = r'a href="(.*?)"'
return re.findall(reg,response)
def get_book_name(response):
reg = re.compile('>.*</a>')
return re.findall(reg,response)
def download_book(book_url,path):
path = ''.join(path.split())
path = 'F:\\books\\{}.html'.format(path) #my local file path
if not os.path.exists(path):
urllib.request.urlretrieve(book_url,path)
print('ok!!!')
else:
print('no!!!')
def get_url_name(start_url):
content = get_content(get_response(start_url))
for i in content:
book_url = get_book_url(i)
if book_url:
book_name = get_book_name(i)
try:
download_book(book_url[0],book_name[0])
except:
continue
def main():
get_url_name(start_url)
if __name__ == '__main__':
start_url = 'http://www.gutenberg.org/wiki/Category:Classics_Bookshelf'
main()
I have run the code and get nothing, no tracebacks. How can I download the books automatically from the website?
I have run the code and get nothing,no tracebacks.
Well, there's no chance you get a traceback in the case of an exception in download_book() since you explicitely silent them:
try:
download_book(book_url[0],book_name[0])
except:
continue
So the very first thing you want to do is to at least print out errors:
try:
download_book(book_url[0],book_name[0])
except exception as e:
print("while downloading book {} : got error {}".format(book_url[0], e)
continue
or just don't catch exception at all (at least until you know what to expect and how to handle it).
I don't even know how to fix it
Learning how to debug is actually even more important than learning how to write code. For a general introduction, you want to read this first.
For something more python-specific, here are a couple ways to trace your program execution:
1/ add print() calls at the important places to inspect what you really get
2/ import your module in the interactive python shell and test your functions in isolation (this is easier when none of them depend on global variables)
3/ use the builtin step debugger
Now there are a few obvious issues with your code:
1/ you don't test the result of request.get() - an HTTP request can fail for quite a few reasons, and the fact you get a response doesn't mean you got the expected response (you could have a 400+ or 500+ response as well.
2/ you use regexps to parse html. DONT - regexps cannot reliably work on html, you want a proper HTML parser instead (BeautifulSoup is the canonical solution for web scraping as it's very tolerant). Also some of your regexps look quite wrong (greedy match-all etc).
start_url is not defined in main()
You need to use a global variable. Otherwise, a better (cleaner) approach is to pass in the variable that you are using. In any case, I would expect an error, start_url is not defined
def main(start_url):
get_url_name(start_url)
if __name__ == '__main__':
start_url = 'http://www.gutenberg.org/wiki/Category:Classics_Bookshelf'
main(start_url)
EDIT:
Nevermind, the problem is in this line: content = get_content(get_response(start_url))
The regex in get_content() does not seem to match anything. My suggestion would be to use BeautifulSoup, from bs4 import BeautifulSoup. For any information regarding why you shouldn't parse html with regex, see this answer RegEx match open tags except XHTML self-contained tags
Asking regexes to parse arbitrary HTML is like asking a beginner to write an operating system
As others have said, you get no output because your regex doesn't match anything. The text returned by the initial url has got a newline between </h2> and <ul>, try this instead:
r'(<span class="mw-headline".*?</span></h2>\n<ul><li>.*</a></li></ul>)'
When you fix that one, you will face another error, I suggest some debug printouts like this:
def get_url_name(start_url):
content = get_content(get_response(start_url))
for i in content:
print('[DEBUG] Handling:', i)
book_url = get_book_url(i)
print('[DEBUG] book_url:', book_url)
if book_url:
book_name = get_book_name(i)
try:
print('[DEBUG] book_url[0]:', book_url[0])
print('[DEBUG] book_name[0]:', book_name[0])
download_book(book_url[0],book_name[0])
except:
continue
I'm pretty new.
I wrote this python script to make an API call from blockr.io to check the balance of multiple bitcoin addresses.
The contents of btcaddy.txt are bitcoin addresses seperated by commas. For this example, let it parse this.
import urllib2
import json
btcaddy = open("btcaddy.txt","r")
urlRequest = urllib2.Request("http://btc.blockr.io/api/v1/address/info/" + btcaddy.read())
data = urllib2.urlopen(urlRequest).read()
json_data = json.loads(data)
balance = float(json_data['data''address'])
print balance
raw_input()
However, it gives me an error. What am I doing wrong? For now, how do I get it to print the balance of the addresses?
You've done multiple things wrong in your code. Here's my fix. I recommend a for loop.
import json
import urllib
addresses = open("btcaddy.txt", "r").read()
base_url = "http://btc.blockr.io/api/v1/address/info/"
request = urllib.urlopen(base_url+addresses)
result = json.loads(request.read())['data']
for balance in result:
print balance['address'], ":" , balance['balance'], "BTC"
You don't need an input at the end, too.
Your question is clear, but your tries not.
You said, you have a file, with at least, more than registry. So you need to retrieve the lines of this file.
with open("btcaddy.txt","r") as a:
addresses = a.readlines()
Now you could iterate over registries and make a request to this uri. The urllib module is enough for this task.
import json
import urllib
base_url = "http://btc.blockr.io/api/v1/address/info/%s"
for address in addresses:
request = urllib.request.urlopen(base_url % address)
result = json.loads(request.read().decode('utf8'))
print(result)
HTTP sends bytes as response, so you should to us decode('utf8') as approach to handle with data.
I have been having a persistent problem getting an rss feed from a particular website. I wound up writing a rather ugly procedure to perform this function, but I am curious why this happens and whether any higher level interfaces handle this problem properly. This problem isn't really a show stopper, since I don't need to retrieve the feed very often.
I have read a solution that traps the exception and returns the partial content, yet since the incomplete reads differ in the amount of bytes that are actually retrieved, I have no certainty that such solution will actually work.
#!/usr/bin/env python
import os
import sys
import feedparser
from mechanize import Browser
import requests
import urllib2
from httplib import IncompleteRead
url = 'http://hattiesburg.legistar.com/Feed.ashx?M=Calendar&ID=543375&GUID=83d4a09c-6b40-4300-a04b-f88884048d49&Mode=2013&Title=City+of+Hattiesburg%2c+MS+-+Calendar+(2013)'
content = feedparser.parse(url)
if 'bozo_exception' in content:
print content['bozo_exception']
else:
print "Success!!"
sys.exit(0)
print "If you see this, please tell me what happened."
# try using mechanize
b = Browser()
r = b.open(url)
try:
r.read()
except IncompleteRead, e:
print "IncompleteRead using mechanize", e
# try using urllib2
r = urllib2.urlopen(url)
try:
r.read()
except IncompleteRead, e:
print "IncompleteRead using urllib2", e
# try using requests
try:
r = requests.request('GET', url)
except IncompleteRead, e:
print "IncompleteRead using requests", e
# this function is old and I categorized it as ...
# "at least it works darnnit!", but I would really like to
# learn what's happening. Please help me put this function into
# eternal rest.
def get_rss_feed(url):
response = urllib2.urlopen(url)
read_it = True
content = ''
while read_it:
try:
content += response.read(1)
except IncompleteRead:
read_it = False
return content, response.info()
content, info = get_rss_feed(url)
feed = feedparser.parse(content)
As already stated, this isn't a mission critical problem, yet a curiosity, as even though I can expect urllib2 to have this problem, I am surprised that this error is encountered in mechanize and requests as well. The feedparser module doesn't even throw an error, so checking for errors depends on the presence of a 'bozo_exception' key.
Edit: I just wanted to mention that both wget and curl perform the function flawlessly, retrieving the full payload correctly every time. I have yet to find a pure python method to work, excepting my ugly hack, and I am very curious to know what is happening on the backend of httplib. On a lark, I decided to also try this with twill the other day and got the same httplib error.
P.S. There is one thing that also strikes me as very odd. The IncompleteRead happens consistently at one of two breakpoints in the payload. It seems that feedparser and requests fail after reading 926 bytes, yet mechanize and urllib2 fail after reading 1854 bytes. This behavior is consistend, and I am left without explanation or understanding.
At the end of the day, all of the other modules (feedparser, mechanize, and urllib2) call httplib which is where the exception is being thrown.
Now, first things first, I also downloaded this with wget and the resulting file was 1854 bytes. Next, I tried with urllib2:
>>> import urllib2
>>> url = 'http://hattiesburg.legistar.com/Feed.ashx?M=Calendar&ID=543375&GUID=83d4a09c-6b40-4300-a04b-f88884048d49&Mode=2013&Title=City+of+Hattiesburg%2c+MS+-+Calendar+(2013)'
>>> f = urllib2.urlopen(url)
>>> f.headers.headers
['Cache-Control: private\r\n',
'Content-Type: text/xml; charset=utf-8\r\n',
'Server: Microsoft-IIS/7.5\r\n',
'X-AspNet-Version: 4.0.30319\r\n',
'X-Powered-By: ASP.NET\r\n',
'Date: Mon, 07 Jan 2013 23:21:51 GMT\r\n',
'Via: 1.1 BC1-ACLD\r\n',
'Transfer-Encoding: chunked\r\n',
'Connection: close\r\n']
>>> f.read()
< Full traceback cut >
IncompleteRead: IncompleteRead(1854 bytes read)
So it is reading all 1854 bytes but then thinks there is more to come. If we explicitly tell it to read only 1854 bytes it works:
>>> f = urllib2.urlopen(url)
>>> f.read(1854)
'\xef\xbb\xbf<?xml version="1.0" encoding="utf-8"?><rss version="2.0" xmlns:atom="http://www.w3.org/2005/Atom">...snip...</rss>'
Obviously, this is only useful if we always know the exact length ahead of time. We can use the fact the partial read is returned as an attribute on the exception to capture the entire contents:
>>> try:
... contents = f.read()
... except httplib.IncompleteRead as e:
... contents = e.partial
...
>>> print contents
'\xef\xbb\xbf<?xml version="1.0" encoding="utf-8"?><rss version="2.0" xmlns:atom="http://www.w3.org/2005/Atom">...snip...</rss>'
This blog post suggests this is a fault of the server, and describes how to monkey-patch the httplib.HTTPResponse.read() method with the try..except block above to handle things behind the scenes:
import httplib
def patch_http_response_read(func):
def inner(*args):
try:
return func(*args)
except httplib.IncompleteRead, e:
return e.partial
return inner
httplib.HTTPResponse.read = patch_http_response_read(httplib.HTTPResponse.read)
I applied the patch and then feedparser worked:
>>> import feedparser
>>> url = 'http://hattiesburg.legistar.com/Feed.ashx?M=Calendar&ID=543375&GUID=83d4a09c-6b40-4300-a04b-f88884048d49&Mode=2013&Title=City+of+Hattiesburg%2c+MS+-+Calendar+(2013)'
>>> feedparser.parse(url)
{'bozo': 0,
'encoding': 'utf-8',
'entries': ...
'status': 200,
'version': 'rss20'}
This isn't the nicest way of doing things, but it seems to work. I'm not expert enough in the HTTP protocols to say for sure whether the server is doing things wrong, or whether httplib is mis-handling an edge case.
I find out in my case, send a HTTP/1.0 request , fix the problem, just adding this to the code:
import httplib
httplib.HTTPConnection._http_vsn = 10
httplib.HTTPConnection._http_vsn_str = 'HTTP/1.0'
after I do the request :
req = urllib2.Request(url, post, headers)
filedescriptor = urllib2.urlopen(req)
img = filedescriptor.read()
after I back to http 1.1 with (for connections that support 1.1) :
httplib.HTTPConnection._http_vsn = 11
httplib.HTTPConnection._http_vsn_str = 'HTTP/1.1'
I have fixed the issue by using HTTPS instead of HTTP and its working fine. No code change was required.
I'm perplexed as to why I'm not able to download the entire contents of some JSON responses from FriendFeed using urllib2.
>>> import urllib2
>>> stream = urllib2.urlopen('http://friendfeed.com/api/room/the-life-scientists/profile?format=json')
>>> stream.headers['content-length']
'168928'
>>> data = stream.read()
>>> len(data)
61058
>>> # We can see here that I did not retrieve the full JSON
... # given that the stream doesn't end with a closing }
...
>>> data[-40:]
'ce2-003048343a40","name":"Vincent Racani'
How can I retrieve the full response with urllib2?
Best way to get all of the data:
fp = urllib2.urlopen("http://www.example.com/index.cfm")
response = ""
while 1:
data = fp.read()
if not data: # This might need to be if data == "": -- can't remember
break
response += data
print response
The reason is that .read() isn't guaranteed to return the entire response, given the nature of sockets. I thought this was discussed in the documentation (maybe urllib) but I cannot find it.
Use tcpdump (or something like it) to monitor the actual network interactions - then you can analyze why the site is broken for some client libraries. Ensure that you repeat multiple times by scripting the test, so you can see if the problem is consistent:
import urllib2
url = 'http://friendfeed.com/api/room/friendfeed-feedback/profile?format=json'
stream = urllib2.urlopen(url)
expected = int(stream.headers['content-length'])
data = stream.read()
datalen = len(data)
print expected, datalen, expected == datalen
The site's working consistently for me so I can't give examples of finding failures :)
Keep calling stream.read() until it's done...
while data = stream.read() :
... do stuff with data
readlines()
also works