Urllib error httperror http error 403 forbidden python

NSE website has changed and the older scripts are semi-optimum to current website. This snippet can gather daily details of security. Details include symbol, security type, previous close, open price, high price, low price, average price, traded quantity, turnover, number of trades, deliverable quantities and ratio of delivered vs traded in percentage. These conveniently presented as list of dictionary form.

Python 3.X version with requests and BeautifulSoup

from requests import get
from csv import DictReader
from bs4 import BeautifulSoup as Soup
from datetime import date
from io import StringIO 

SECURITY_NAME="3MINDIA" # Change this to get quote for another stock
START_DATE= date(2017, 1, 1) # Start date of stock quote data DD-MM-YYYY
END_DATE= date(2017, 9, 14)  # End date of stock quote data DD-MM-YYYY


BASE_URL = "https://www.nseindia.com/products/dynaContent/common/productsSymbolMapping.jsp?symbol={security}&segmentLink=3&symbolCount=1&series=ALL&dateRange=+&fromDate={start_date}&toDate={end_date}&dataType=PRICEVOLUMEDELIVERABLE"




def getquote(symbol, start, end):
    start = start.strftime("%-d-%-m-%Y")
    end = end.strftime("%-d-%-m-%Y")

    hdr = {'User-Agent': 'Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.11 (KHTML, like Gecko) Chrome/23.0.1271.64 Safari/537.11',
         'Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8',
         'Referer': 'https://cssspritegenerator.com',
         'Accept-Charset': 'ISO-8859-1,utf-8;q=0.7,*;q=0.3',
         'Accept-Encoding': 'none',
         'Accept-Language': 'en-US,en;q=0.8',
         'Connection': 'keep-alive'}

    url = BASE_URL.format(security=symbol, start_date=start, end_date=end)
    d = get(url, headers=hdr)
    soup = Soup(d.content, 'html.parser')
    payload = soup.find('div', {'id': 'csvContentDiv'}).text.replace(':', 'n')
    csv = DictReader(StringIO(payload))
    for row in csv:
        print({k:v.strip() for k, v in row.items()})


 if __name__ == '__main__':
     getquote(SECURITY_NAME, START_DATE, END_DATE)

Besides this is relatively modular and ready to use snippet.

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1. the urllib Module in Python
2. Check robots.txt to Prevent urllib HTTP Error 403 Forbidden Message
3. Adding Cookie to the Request Headers to Solve urllib HTTP Error 403 Forbidden Message
4. Use Session Object to Solve urllib HTTP Error 403 Forbidden Message

Today’s article explains how to deal with an error message (exception), urllib.error.HTTPError: HTTP Error 403: Forbidden, produced by the error class on behalf of the request classes when it faces a forbidden resource.

the urllib Module in Python

The urllib Python module handles URLs for python via different protocols. It is famous for web scrapers who want to obtain data from a particular website.

The urllib contains classes, methods, and functions that perform certain operations such as reading, parsing URLs, and robots.txt. There are four classes, request, error, parse, and robotparser.

Check robots.txt to Prevent urllib HTTP Error 403 Forbidden Message

When using the urllib module to interact with clients or servers via the request class, we might experience specific errors. One of those errors is the HTTP 403 error.

We get urllib.error.HTTPError: HTTP Error 403: Forbidden error message in urllib package while reading a URL. The HTTP 403, the Forbidden Error, is an HTTP status code that indicates that the client or server forbids access to a requested resource.

Therefore, when we see this kind of error message, urllib.error.HTTPError: HTTP Error 403: Forbidden, the server understands the request but decides not to process or authorize the request that we sent.

To understand why the website we are accessing is not processing our request, we need to check an important file, robots.txt. Before web scraping or interacting with a website, it is often advised to review this file to know what to expect and not face any further troubles.

To check it on any website, we can follow the format below.

https://<website.com>/robots.txt

For example, check YouTube, Amazon, and Google robots.txt files.

https://www.youtube.com/robots.txt
https://www.amazon.com/robots.txt
https://www.google.com/robots.txt

Checking YouTube robots.txt gives the following result.

# robots.txt file for YouTube
# Created in the distant future (the year 2000) after
# the robotic uprising of the mid-'90s wiped out all humans.

User-agent: Mediapartners-Google*
Disallow:

User-agent: *
Disallow: /channel/*/community
Disallow: /comment
Disallow: /get_video
Disallow: /get_video_info
Disallow: /get_midroll_info
Disallow: /live_chat
Disallow: /login
Disallow: /results
Disallow: /signup
Disallow: /t/terms
Disallow: /timedtext_video
Disallow: /user/*/community
Disallow: /verify_age
Disallow: /watch_ajax
Disallow: /watch_fragments_ajax
Disallow: /watch_popup
Disallow: /watch_queue_ajax

Sitemap: https://www.youtube.com/sitemaps/sitemap.xml
Sitemap: https://www.youtube.com/product/sitemap.xml

We can notice a lot of Disallow tags there. This Disallow tag shows the website’s area, which is not accessible. Therefore, any request to those areas will not be processed and is forbidden.

In other robots.txt files, we might see an Allow tag. For example, http://youtube.com/comment is forbidden to any external request, even with the urllib module.

Let’s write code to scrape data from a website that returns an HTTP 403 error when accessed.

Example Code:

import urllib.request
import re

webpage = urllib.request.urlopen('https://www.cmegroup.com/markets/products.html?redirect=/trading/products/#cleared=Options&sortField=oi').read()
findrows = re.compile('<tr class="- banding(?:On|Off)>(.*?)</tr>')
findlink = re.compile('<a href =">(.*)</a>')

row_array = re.findall(findrows, webpage)
links = re.findall(findlink, webpage)

print(len(row_array))

Output:

Traceback (most recent call last):
  File "c:UsersakinlDocumentsPythonindex.py", line 7, in <module>
    webpage = urllib.request.urlopen('https://www.cmegroup.com/markets/products.html?redirect=/trading/products/#cleared=Options&sortField=oi').read()
  File "C:Python310liburllibrequest.py", line 216, in urlopen
    return opener.open(url, data, timeout)
  File "C:Python310liburllibrequest.py", line 525, in open
    response = meth(req, response)
  File "C:Python310liburllibrequest.py", line 634, in http_response
    response = self.parent.error(
  File "C:Python310liburllibrequest.py", line 563, in error
    return self._call_chain(*args)
  File "C:Python310liburllibrequest.py", line 496, in _call_chain
    result = func(*args)
  File "C:Python310liburllibrequest.py", line 643, in http_error_default
    raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 403: Forbidden

The reason is that we are forbidden from accessing the website. However, if we check the robots.txt file, we will notice that https://www.cmegroup.com/markets/ is not with a Disallow tag. However, if we go down the robots.txt file for the website we wanted to scrape, we will find the below.

User-agent: Python-urllib/1.17
Disallow: /

The above text means that the user agent named Python-urllib is not allowed to crawl any URL within the site. That means using the Python urllib module is not allowed to crawl the site.

Therefore, check or parse the robots.txt to know what resources we have access to. we can parse robots.txt file using the robotparser class. These can prevent our code from experiencing an urllib.error.HTTPError: HTTP Error 403: Forbidden error message.

Passing a valid user agent as a header parameter will quickly fix the problem. The website may use cookies as an anti-scraping measure.

The website may set and ask for cookies to be echoed back to prevent scraping, which is maybe against its policy.

from urllib.request import Request, urlopen

def get_page_content(url, head):

  req = Request(url, headers=head)
  return urlopen(req)

url = 'https://example.com'
head = {
  'User-Agent': 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_14_6) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/99.0.4844.84 Safari/537.36',
  'Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,image/avif,image/webp,image/apng,*/*;q=0.8,application/signed-exchange;v=b3;q=0.9',
  'Accept-Charset': 'ISO-8859-1,utf-8;q=0.7,*;q=0.3',
  'Accept-Encoding': 'none',
  'Accept-Language': 'en-US,en;q=0.8',
  'Connection': 'keep-alive',
  'refere': 'https://example.com',
  'cookie': """your cookie value ( you can get that from your web page) """
}

data = get_page_content(url, head).read()
print(data)

Output:

<!doctype html>n<html>n<head>n    <title>Example Domain</title>nn    <meta
'
'
'
<p><a href="https://www.iana.org/domains/example">More information...</a></p>n</div>n</body>n</html>n'

Passing a valid user agent as a header parameter will quickly fix the problem.

Use Session Object to Solve urllib HTTP Error 403 Forbidden Message

Sometimes, even using a user agent won’t stop this error from occurring. The Session object of the requests module can then be used.

from random import seed
import requests

url = "https://stackoverflow.com/search?q=html+error+403"
session_obj = requests.Session()
response = session_obj.get(url, headers={"User-Agent": "Mozilla/5.0"})

print(response.status_code)

Output:

The above article finds the cause of the urllib.error.HTTPError: HTTP Error 403: Forbidden and the solution to handle it. mod_security basically causes this error as different web pages use different security mechanisms to differentiate between human and automated computers (bots).

Python is a rather easy-to-learn but powerful language enough to be used in a variety of applications ranging from AI and Machine Learning all the way to something as simple as creating web scraping bots. 

That said, random bugs and glitches are still the order of the day in a Python programmer’s life. In this article, we’re talking about the “urllib.error.httperror: HTTP error 403: Forbidden” when trying to scrape sites with Python and what you can to do fix the problem. 

Also read: Is Python case sensitive when dealing with identifiers?

Why does this happen?

While the error can be triggered by anything from a simple runtime error in the script to server issues on the website, the most likely reason is the presence of some sort of server security feature to prevent bots or spiders from crawling the site. In this case, the security feature might be blocking urllib, a library used to send requests to websites. 

How to fix this?

Here are two fixes you can try out. 

Disable mod_security or equivalent security features

As mentioned before, server-side security features can cause problems with web scrapers. Try setting your browser agent as follows to see if you can avoid the issue. 

from urllib.request import Request, urlopen

req = Request(
    url='enter request URL here', 
    headers={'User-Agent': 'Mozilla/5.0'}
)
webpage = urlopen(req).read()

A correctly defined browser agent should be able to scrape data from any site. 

Set a timeout

If you aren’t getting a response, try setting a timeout to prevent the server from mistaking your bot for a DDoS attacking and hence blocking all requests altogether. 

from urllib.request import Request, 
urlopen req = Request('enter request URL here', headers={'User-Agent': 'Mozilla/5.0'}) 
webpage = urlopen(req,timeout=10).read()

The aforementioned example sets a 10-second timeout between requests to not overload the server while maintaining good request frequency. 

Also read: How to fix Javascript error: ipython is not defined?

HOWTO Fetch Internet Resources Using The urllib Package¶

There is a French translation of an earlier revision of this HOWTO, available at urllib2 — Le Manuel manquant.

Introduction¶

You may also find useful the following article on fetching web resources with Python:

A tutorial on Basic Authentication, with examples in Python.

urllib.request is a Python module for fetching URLs (Uniform Resource Locators). It offers a very simple interface, in the form of the urlopen function. This is capable of fetching URLs using a variety of different protocols. It also offers a slightly more complex interface for handling common situations — like basic authentication, cookies, proxies and so on. These are provided by objects called handlers and openers.

urllib.request supports fetching URLs for many “URL schemes” (identified by the string before the «:» in URL — for example «ftp» is the URL scheme of «ftp://python.org/» ) using their associated network protocols (e.g. FTP, HTTP). This tutorial focuses on the most common case, HTTP.

For straightforward situations urlopen is very easy to use. But as soon as you encounter errors or non-trivial cases when opening HTTP URLs, you will need some understanding of the HyperText Transfer Protocol. The most comprehensive and authoritative reference to HTTP is RFC 2616. This is a technical document and not intended to be easy to read. This HOWTO aims to illustrate using urllib, with enough detail about HTTP to help you through. It is not intended to replace the urllib.request docs, but is supplementary to them.

Fetching URLs¶

The simplest way to use urllib.request is as follows:

If you wish to retrieve a resource via URL and store it in a temporary location, you can do so via the shutil.copyfileobj() and tempfile.NamedTemporaryFile() functions:

Many uses of urllib will be that simple (note that instead of an ‘http:’ URL we could have used a URL starting with ‘ftp:’, ‘file:’, etc.). However, it’s the purpose of this tutorial to explain the more complicated cases, concentrating on HTTP.

HTTP is based on requests and responses — the client makes requests and servers send responses. urllib.request mirrors this with a Request object which represents the HTTP request you are making. In its simplest form you create a Request object that specifies the URL you want to fetch. Calling urlopen with this Request object returns a response object for the URL requested. This response is a file-like object, which means you can for example call .read() on the response:

Note that urllib.request makes use of the same Request interface to handle all URL schemes. For example, you can make an FTP request like so:

In the case of HTTP, there are two extra things that Request objects allow you to do: First, you can pass data to be sent to the server. Second, you can pass extra information (“metadata”) about the data or about the request itself, to the server — this information is sent as HTTP “headers”. Let’s look at each of these in turn.

Data¶

Sometimes you want to send data to a URL (often the URL will refer to a CGI (Common Gateway Interface) script or other web application). With HTTP, this is often done using what’s known as a POST request. This is often what your browser does when you submit a HTML form that you filled in on the web. Not all POSTs have to come from forms: you can use a POST to transmit arbitrary data to your own application. In the common case of HTML forms, the data needs to be encoded in a standard way, and then passed to the Request object as the data argument. The encoding is done using a function from the urllib.parse library.

Note that other encodings are sometimes required (e.g. for file upload from HTML forms — see HTML Specification, Form Submission for more details).

If you do not pass the data argument, urllib uses a GET request. One way in which GET and POST requests differ is that POST requests often have “side-effects”: they change the state of the system in some way (for example by placing an order with the website for a hundredweight of tinned spam to be delivered to your door). Though the HTTP standard makes it clear that POSTs are intended to always cause side-effects, and GET requests never to cause side-effects, nothing prevents a GET request from having side-effects, nor a POST requests from having no side-effects. Data can also be passed in an HTTP GET request by encoding it in the URL itself.

This is done as follows:

Notice that the full URL is created by adding a ? to the URL, followed by the encoded values.

We’ll discuss here one particular HTTP header, to illustrate how to add headers to your HTTP request.

Some websites 1 dislike being browsed by programs, or send different versions to different browsers 2. By default urllib identifies itself as Python-urllib/x.y (where x and y are the major and minor version numbers of the Python release, e.g. Python-urllib/2.5 ), which may confuse the site, or just plain not work. The way a browser identifies itself is through the User-Agent header 3. When you create a Request object you can pass a dictionary of headers in. The following example makes the same request as above, but identifies itself as a version of Internet Explorer 4.

The response also has two useful methods. See the section on info and geturl which comes after we have a look at what happens when things go wrong.

Handling Exceptions¶

urlopen raises URLError when it cannot handle a response (though as usual with Python APIs, built-in exceptions such as ValueError , TypeError etc. may also be raised).

HTTPError is the subclass of URLError raised in the specific case of HTTP URLs.

The exception classes are exported from the urllib.error module.

URLError¶

Often, URLError is raised because there is no network connection (no route to the specified server), or the specified server doesn’t exist. In this case, the exception raised will have a ‘reason’ attribute, which is a tuple containing an error code and a text error message.

HTTPError¶

Every HTTP response from the server contains a numeric “status code”. Sometimes the status code indicates that the server is unable to fulfil the request. The default handlers will handle some of these responses for you (for example, if the response is a “redirection” that requests the client fetch the document from a different URL, urllib will handle that for you). For those it can’t handle, urlopen will raise an HTTPError . Typical errors include ‘404’ (page not found), ‘403’ (request forbidden), and ‘401’ (authentication required).

See section 10 of RFC 2616 for a reference on all the HTTP error codes.

The HTTPError instance raised will have an integer ‘code’ attribute, which corresponds to the error sent by the server.

Error Codes¶

Because the default handlers handle redirects (codes in the 300 range), and codes in the 100–299 range indicate success, you will usually only see error codes in the 400–599 range.

http.server.BaseHTTPRequestHandler.responses is a useful dictionary of response codes in that shows all the response codes used by RFC 2616. The dictionary is reproduced here for convenience

When an error is raised the server responds by returning an HTTP error code and an error page. You can use the HTTPError instance as a response on the page returned. This means that as well as the code attribute, it also has read, geturl, and info, methods as returned by the urllib.response module:

Wrapping it Up¶

So if you want to be prepared for HTTPError or URLError there are two basic approaches. I prefer the second approach.

Number 1В¶

The except HTTPError must come first, otherwise except URLError will also catch an HTTPError .

Number 2В¶

info and geturl¶

The response returned by urlopen (or the HTTPError instance) has two useful methods info() and geturl() and is defined in the module urllib.response ..

geturl — this returns the real URL of the page fetched. This is useful because urlopen (or the opener object used) may have followed a redirect. The URL of the page fetched may not be the same as the URL requested.

info — this returns a dictionary-like object that describes the page fetched, particularly the headers sent by the server. It is currently an http.client.HTTPMessage instance.

Typical headers include ‘Content-length’, ‘Content-type’, and so on. See the Quick Reference to HTTP Headers for a useful listing of HTTP headers with brief explanations of their meaning and use.

Openers and Handlers¶

When you fetch a URL you use an opener (an instance of the perhaps confusingly named urllib.request.OpenerDirector ). Normally we have been using the default opener — via urlopen — but you can create custom openers. Openers use handlers. All the “heavy lifting” is done by the handlers. Each handler knows how to open URLs for a particular URL scheme (http, ftp, etc.), or how to handle an aspect of URL opening, for example HTTP redirections or HTTP cookies.

You will want to create openers if you want to fetch URLs with specific handlers installed, for example to get an opener that handles cookies, or to get an opener that does not handle redirections.

To create an opener, instantiate an OpenerDirector , and then call .add_handler(some_handler_instance) repeatedly.

Alternatively, you can use build_opener , which is a convenience function for creating opener objects with a single function call. build_opener adds several handlers by default, but provides a quick way to add more and/or override the default handlers.

Other sorts of handlers you might want to can handle proxies, authentication, and other common but slightly specialised situations.

install_opener can be used to make an opener object the (global) default opener. This means that calls to urlopen will use the opener you have installed.

Opener objects have an open method, which can be called directly to fetch urls in the same way as the urlopen function: there’s no need to call install_opener , except as a convenience.

Basic Authentication¶

To illustrate creating and installing a handler we will use the HTTPBasicAuthHandler . For a more detailed discussion of this subject – including an explanation of how Basic Authentication works — see the Basic Authentication Tutorial.

When authentication is required, the server sends a header (as well as the 401 error code) requesting authentication. This specifies the authentication scheme and a ‘realm’. The header looks like: WWW-Authenticate: SCHEME realm=»REALM» .

The client should then retry the request with the appropriate name and password for the realm included as a header in the request. This is ‘basic authentication’. In order to simplify this process we can create an instance of HTTPBasicAuthHandler and an opener to use this handler.

The HTTPBasicAuthHandler uses an object called a password manager to handle the mapping of URLs and realms to passwords and usernames. If you know what the realm is (from the authentication header sent by the server), then you can use a HTTPPasswordMgr . Frequently one doesn’t care what the realm is. In that case, it is convenient to use HTTPPasswordMgrWithDefaultRealm . This allows you to specify a default username and password for a URL. This will be supplied in the absence of you providing an alternative combination for a specific realm. We indicate this by providing None as the realm argument to the add_password method.

The top-level URL is the first URL that requires authentication. URLs “deeper” than the URL you pass to .add_password() will also match.

In the above example we only supplied our HTTPBasicAuthHandler to build_opener . By default openers have the handlers for normal situations – ProxyHandler (if a proxy setting such as an http_proxy environment variable is set), UnknownHandler , HTTPHandler , HTTPDefaultErrorHandler , HTTPRedirectHandler , FTPHandler , FileHandler , DataHandler , HTTPErrorProcessor .

top_level_url is in fact either a full URL (including the ‘http:’ scheme component and the hostname and optionally the port number) e.g. «http://example.com/» or an “authority” (i.e. the hostname, optionally including the port number) e.g. «example.com» or «example.com:8080» (the latter example includes a port number). The authority, if present, must NOT contain the “userinfo” component — for example «joe:password@example.com» is not correct.

Proxies¶

urllib will auto-detect your proxy settings and use those. This is through the ProxyHandler , which is part of the normal handler chain when a proxy setting is detected. Normally that’s a good thing, but there are occasions when it may not be helpful 5. One way to do this is to setup our own ProxyHandler , with no proxies defined. This is done using similar steps to setting up a Basic Authentication handler:

Currently urllib.request does not support fetching of https locations through a proxy. However, this can be enabled by extending urllib.request as shown in the recipe 6.

HTTP_PROXY will be ignored if a variable REQUEST_METHOD is set; see the documentation on getproxies() .

Sockets and Layers¶

The Python support for fetching resources from the web is layered. urllib uses the http.client library, which in turn uses the socket library.

As of Python 2.3 you can specify how long a socket should wait for a response before timing out. This can be useful in applications which have to fetch web pages. By default the socket module has no timeout and can hang. Currently, the socket timeout is not exposed at the http.client or urllib.request levels. However, you can set the default timeout globally for all sockets using

This document was reviewed and revised by John Lee.

Google for example.

Browser sniffing is a very bad practice for website design — building sites using web standards is much more sensible. Unfortunately a lot of sites still send different versions to different browsers.

The user agent for MSIE 6 is ‘Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1; .NET CLR 1.1.4322)’

For details of more HTTP request headers, see Quick Reference to HTTP Headers.

In my case I have to use a proxy to access the internet at work. If you attempt to fetch localhost URLs through this proxy it blocks them. IE is set to use the proxy, which urllib picks up on. In order to test scripts with a localhost server, I have to prevent urllib from using the proxy.

urllib opener for SSL proxy (CONNECT method): ASPN Cookbook Recipe.

Источник

problem:
The urllib.request-urlopen () method is often used to open the source code of a web page and then analyze the source code of the page, but it will throw an “HTTP Error 403: Forbidden” exception for some websites
For example, when the following statement is executed,

 urllib.request.urlopen("http://blog.csdn.net/eric_sunah/article/details/11099295")

will appear the following exception:

  File "D:Python32liburllibrequest.py", line 475, in open
    response = meth(req, response)
  File "D:Python32liburllibrequest.py", line 587, in http_response
    'http', request, response, code, msg, hdrs)
  File "D:Python32liburllibrequest.py", line 513, in error
    return self._call_chain(*args)
  File "D:Python32liburllibrequest.py", line 447, in _call_chain
    result = func(*args)
  File "D:Python32liburllibrequest.py", line 595, in http_error_default
    raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 403: Forbidden

Analysis:
Appear abnormal, because if use urllib request. Open a URL urlopen way, the server will only receive a simple access request for the page, but the server does not know to send the request to use the browser, operating system, hardware platform, such as information, and often lack the information request are normal access, such as the crawler.
In order to prevent this abnormal access, some websites will verify the UserAgent in the request information (its information includes hardware platform, system software, application software, and user preferences). If the UserAgent is abnormal or does not exist, then the request will be rejected (as shown in the error message above).
So you can try to add the UserAgent’s information
to the request
Solution:
For Python 3.x, adding the UserAgent information to the request is simple as follows

#HTTPError: HTTP Error 403: Forbidden error appears if the following line is not added
    #The main reason is that the site is forbidden to crawl, you can add header information to the request, pretending to be a browser to access the User-Agent, specific information can be found through the Firefox FireBug plugin.
    headers = {'User-Agent':'Mozilla/5.0 (Windows NT 6.1; WOW64; rv:23.0) Gecko/20100101 Firefox/23.0'}
    req = urllib.request.Request(url=chaper_url, headers=headers)
    urllib.request.urlopen(req).read()

The urllib. Request. Urlopen. Read () to replace the above code, for problems page can normal visit

Вопрос:

Метод urllib.request.urlopen () часто используется для открытия исходного кода веб-страницы, а затем для анализа исходного кода этой страницы, но для некоторых веб-сайтов при использовании этого метода возникает исключение «Ошибка HTTP 403: запрещено».

Например, при выполнении следующего оператора

 urllib.request.urlopen("http://blog.csdn.net/eric_sunah/article/details/11099295")

Появится следующее исключение:

  File "D:Python32liburllibrequest.py", line 475, in open
    response = meth(req, response)
  File "D:Python32liburllibrequest.py", line 587, in http_response
    'http', request, response, code, msg, hdrs)
  File "D:Python32liburllibrequest.py", line 513, in error
    return self._call_chain(*args)
  File "D:Python32liburllibrequest.py", line 447, in _call_chain
    result = func(*args)
  File "D:Python32liburllibrequest.py", line 595, in http_error_default
    raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 403: Forbidden

анализ:

Вышеуказанное исключение происходит потому, что если вы откроете URL-адрес с помощью urllib.request.urlopen, сервер получит только простой запрос на доступ к странице, но сервер не знает браузер, использованный для отправки этого запроса. Операционная система, аппаратная платформа и другая информация, а также запросы, в которых отсутствует эта информация, часто имеют ненормальный доступ, например, сканеры.

Чтобы предотвратить такой ненормальный доступ, некоторые веб-сайты будут проверять UserAgent (егоИнформационный пакетВключая аппаратную платформу, системупрограммное обеспечение、применениеИ личные предпочтения пользователя), если UserAgent является ненормальным или не существует, то этот запрос будет отклонен (как показано в сообщении об ошибке выше)

Таким образом, вы можете попытаться добавить информацию UserAgent к запросу.

Программа:

Для Python 3.x добавить информацию UserAgent к запросу очень просто, код выглядит следующим образом

# Если вы не добавите следующую строку, urllib2.HTTPError: Ошибка HTTP 403: появится запрещенная ошибка
         # В основном потому, что сайт запрещает сканирование, вы можете добавить информацию заголовка к запросу и притвориться браузером для доступа к User-Agent. Конкретную информацию можно запросить через плагин Firefox FireBug
    headers = {'User-Agent':'Mozilla/5.0 (Windows NT 6.1; WOW64; rv:23.0) Gecko/20100101 Firefox/23.0'}
    req = urllib.request.Request(url=chaper_url, headers=headers)
    urllib.request.urlopen(req).read()

После замены urllib.request.urlopen.read () на приведенный выше код, страница с проблемой может быть открыта как обычно