Requests python get 403

How to fix problem http error 403 in python 3 web scraping?

HTTP Error 403 is a common error encountered while web scraping using Python 3. It indicates that the server is refusing to fulfill the request made by the client, as the request lacks sufficient authorization or the server considers the request to be invalid. This error can be encountered for a variety of reasons, including the presence of IP blocking, CAPTCHAs, or rate limiting restrictions. In order to resolve the issue, there are several methods that can be implemented, including changing the User Agent, using proxies, and implementing wait time between requests.

Method 1: Changing the User Agent

If you encounter HTTP error 403 while web scraping with Python 3, it means that the server is denying you access to the webpage. One common solution to this problem is to change the user agent of your web scraper. The user agent is a string that identifies the web scraper to the server. By changing the user agent, you can make your web scraper appear as a regular web browser to the server.

Here is an example code that shows how to change the user agent of your web scraper using the requests library:

import requests url = 'https://example.com' headers = 'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/58.0.3029.110 Safari/537.36'> response = requests.get(url, headers=headers) print(response.content)

In this example, we set the User-Agent header to a string that mimics the user agent of the Google Chrome web browser. You can find the user agent string of your favorite web browser by searching for «my user agent» on Google.

By setting the User-Agent header, we can make our web scraper appear as a regular web browser to the server. This can help us bypass HTTP error 403 and access the webpage we want to scrape.

That’s it! By changing the user agent of your web scraper, you should be able to fix the problem of HTTP error 403 in Python 3 web scraping.

Method 2: Using Proxies

If you are encountering HTTP error 403 while web scraping with Python 3, it is likely that the website is blocking your IP address due to frequent requests. One way to solve this problem is by using proxies. Proxies allow you to make requests to the website from different IP addresses, making it difficult for the website to block your requests. Here is how you can fix HTTP error 403 in Python 3 web scraping with proxies:

Step 1: Install Required Libraries

You need to install the requests and bs4 libraries to make HTTP requests and parse HTML respectively. You can install them using pip:

pip install requests pip install bs4

Step 2: Get a List of Proxies

You need to get a list of proxies that you can use to make requests to the website. There are many websites that provide free proxies, such as https://free-proxy-list.net/ . You can scrape the website to get a list of proxies:

import requests from bs4 import BeautifulSoup url = 'https://free-proxy-list.net/' response = requests.get(url) soup = BeautifulSoup(response.text, 'html.parser') table = soup.find('table', 'id': 'proxylisttable'>) rows = table.tbody.find_all('tr') proxies = [] for row in rows: cols = row.find_all('td') if cols[6].text == 'yes': proxy = cols[0].text + ':' + cols[1].text proxies.append(proxy)

This code scrapes the website and gets a list of HTTP proxies that support HTTPS. The proxies are stored in the proxies list.

Step 3: Make Requests with Proxies

You can use the requests library to make requests to the website with a proxy. Here is an example code that makes a request to https://www.example.com with a random proxy from the proxies list:

import random url = 'https://www.example.com' proxy = random.choice(proxies) response = requests.get(url, proxies='https': proxy>) if response.status_code == 200: print(response.text) else: print('Request failed with status code:', response.status_code)

This code selects a random proxy from the proxies list and makes a request to https://www.example.com with the proxy. If the request is successful, it prints the response text. Otherwise, it prints the status code of the failed request.

Step 4: Handle Exceptions

You need to handle exceptions that may occur while making requests with proxies. Here is an example code that handles exceptions and retries the request with a different proxy:

import requests import random from requests.exceptions import ProxyError, ConnectionError, Timeout url = 'https://www.example.com' while True: proxy = random.choice(proxies) try: response = requests.get(url, proxies='https': proxy>, timeout=5) if response.status_code == 200: print(response.text) break else: print('Request failed with status code:', response.status_code) except (ProxyError, ConnectionError, Timeout): print('Proxy error. Retrying with a different proxy. ')

This code uses a while loop to keep retrying the request with a different proxy until it succeeds. It handles ProxyError , ConnectionError , and Timeout exceptions that may occur while making requests with proxies.

Method 3: Implementing Wait Time between Requests

When you are scraping a website, you might encounter an HTTP error 403, which means that the server is denying your request. This can happen when the server detects that you are sending too many requests in a short period of time, and it wants to protect itself from being overloaded.

One way to fix this problem is to implement wait time between requests. This means that you will wait a certain amount of time before sending the next request, which will give the server time to process the previous request and prevent it from being overloaded.

Here is an example code that shows how to implement wait time between requests using the time module:

import requests import time url = 'https://example.com' headers =  'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/58.0.3029.110 Safari/537.36'> for i in range(5): response = requests.get(url, headers=headers) print(response.status_code) time.sleep(1)

In this example, we are sending a request to https://example.com with headers that mimic a browser request. We are then using a for loop to send 5 requests with a 1 second delay between requests using the time.sleep() function.

Another way to implement wait time between requests is to use a random delay. This will make your requests less predictable and less likely to be detected as automated. Here is an example code that shows how to implement a random delay using the random module:

import requests import random import time url = 'https://example.com' headers =  'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/58.0.3029.110 Safari/537.36'> for i in range(5): response = requests.get(url, headers=headers) print(response.status_code) time.sleep(random.randint(1, 5))

In this example, we are sending a request to https://example.com with headers that mimic a browser request. We are then using a for loop to send 5 requests with a random delay between 1 and 5 seconds using the random.randint() function.

By implementing wait time between requests, you can prevent HTTP error 403 and ensure that your web scraping code runs smoothly.

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GET запрос с ошибкой 403

Делаю запрос. Статус запроса 403. Через браузер запрос выполнить получается. Везде пишут, что в такой ситуации надо указать User-Agent. Я указал, но никакого результата. Пробовал добавлять различные параметры в headers, но та же ошибка 403. Подскажите, что я делаю не так?

import requests url = 'https://www.mos.ru/altmosmvc/api/v1/taxi/getInfo/' param = {'pagenumber':6} headers={'User-Agent':'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/71.0.3578.98 Safari/537.36 OPR/58.0.3135.79'} req = requests.get(url,params=param,headers=headers) if req.status_code==200: print(req.json()) else: print(req.status_code) print(req.url)

Авторизация через twitter api завершается с ошибкой 403
Не могу авторизоваться в твиттере Uri uri = new.

GET запрос 403 ошибка
Здравствуйте, может кто подсказать почему при моем запросе появляется ответ 403? string url =.

GET запрос возвращает ошибку 403
Здравствуйте. Пытаюсь отправить GET запрос на простой php скрипт. Этот запрос успешно выполняется.

Get запрос возвращает 403 без сниффера
Столкнулся с такой проблемой, делаю обычный гет запрос на сайт — получаю или, но стоит включить.

import requests url = 'https://www.mos.ru/altmosmvc/api/v1/taxi/getInfo/' param = {'pagenumber':10} session = requests.Session() session.headers['User-Agent']='Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/71.0.3578.98 Safari/537.36 OPR/58.0.3135.79' req = session.get(url,params=param) if req.status_code==200: print(req.json()) else: print(req.status_code) print(req.url)

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import requests url = "https://www.mos.ru/altmosmvc/api/v1/taxi/getInfo/?Region=Москва&RegNum=&FullName=&LicenseNum=&Condition=&pagenumber=" header = { 'User-Agent': 'Mozilla/5.0 (Windows NT 6.3; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/72.0.3626.121 Safari/537.36', 'upgrade-insecure-requests': '1', 'cookie': 'mos_id=CllGxlx+PS20pAxcIuDnAgA=; session-cookie=158b36ec3ea4f5484054ad1fd21407333c874ef0fa4f0c8e34387efd5464a1e9500e2277b0367d71a273e5b46fa0869a; NSC_WBS-QUBG-jo-nptsv-WT-443=ffffffff0951e23245525d5f4f58455e445a4a423660; rheftjdd=rheftjddVal; _ym_uid=1552395093355938562; _ym_d=1552395093; _ym_isad=2' } req = requests.get(url + str(1), headers = header) if req.status_code==200: print(req.json()) else: print(req.status_code) print(req.url)

Источник

Получаю ошибку 403 в requests

Вы используете устаревший браузер. Этот и другие сайты могут отображаться в нем неправильно.
Необходимо обновить браузер или попробовать использовать другой.

daradan

Новичок

Здравствуйте!
Подскажите как побороть?
Имеется ссылка https://webapi[.]computeruniverse[.]net/api/catalog/topmenu/?lang=1&cachecountry=KZ
которая без проблем открывается через браузер и получается ответ в виде json.
Однако не получается эти данные получить через requests и постаянно status.code = 403 и в виде текста получаю следующий html

import requests params = < 'lang': '1', 'cachecountry': 'KZ', >response = requests.get('https://webapi.computeruniverse.net/api/catalog/topmenu/', params=params) print(response.status_code)

import requests cookies = < '_ALGOLIA': 'xxx', 'wtstp_sid': 'xxx', 'wtstp_eid': 'xxx', '_dy_c_exps': '', '_dy_c_att_exps': '', '_dycnst': 'xxx', '_dyid': 'xxx', '.Nop.Customer': 'xxx', 'dy_fs_page': 'www.computeruniverse.net%2Fen%2Fc%2Flaptops-tablet-pcs-pcs%2Flaptops-notebooks', '_dy_geo': 'KZ', '_dy_df_geo': 'Kazakhstan', '_dy_toffset': '0', '_dyid_server': 'xxx', '_dycst': 'xxx.', '__cf_bm': 'xxx', 'cu-edge-hints': xxx', '_dy_soct': 'xxx', >headers = < 'authority': 'www.computeruniverse.net', '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-language': 'en-US,en;q=0.9', 'cache-control': 'max-age=0', 'if-modified-since': 'Sat, 17 Dec 2022 03:23:59 GMT', 'sec-ch-ua': '"Opera";v="93", "Not/A)Brand";v="8", "Chromium";v="107"', 'sec-ch-ua-mobile': '?0', 'sec-ch-ua-platform': '"Linux"', 'sec-fetch-dest': 'document', 'sec-fetch-mode': 'navigate', 'sec-fetch-site': 'same-origin', 'sec-fetch-user': '?1', 'upgrade-insecure-requests': '1', 'user-agent': 'Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/107.0.0.0 Safari/537.36 OPR/93.0.0.0', >params = < 'lang': '1', 'cachecountry': 'KZ', >response = requests.get('https://webapi.computeruniverse.net/api/catalog/topmenu/', params=params, cookies=cookies, headers=headers) print(response.status_code)

Источник

Получаю ошибку 403 в requests

Вы используете устаревший браузер. Этот и другие сайты могут отображаться в нем неправильно.
Необходимо обновить браузер или попробовать использовать другой.

daradan

Новичок

Здравствуйте!
Подскажите как побороть?
Имеется ссылка https://webapi[.]computeruniverse[.]net/api/catalog/topmenu/?lang=1&cachecountry=KZ
которая без проблем открывается через браузер и получается ответ в виде json.
Однако не получается эти данные получить через requests и постаянно status.code = 403 и в виде текста получаю следующий html

import requests params = < 'lang': '1', 'cachecountry': 'KZ', >response = requests.get('https://webapi.computeruniverse.net/api/catalog/topmenu/', params=params) print(response.status_code)

import requests cookies = < '_ALGOLIA': 'xxx', 'wtstp_sid': 'xxx', 'wtstp_eid': 'xxx', '_dy_c_exps': '', '_dy_c_att_exps': '', '_dycnst': 'xxx', '_dyid': 'xxx', '.Nop.Customer': 'xxx', 'dy_fs_page': 'www.computeruniverse.net%2Fen%2Fc%2Flaptops-tablet-pcs-pcs%2Flaptops-notebooks', '_dy_geo': 'KZ', '_dy_df_geo': 'Kazakhstan', '_dy_toffset': '0', '_dyid_server': 'xxx', '_dycst': 'xxx.', '__cf_bm': 'xxx', 'cu-edge-hints': xxx', '_dy_soct': 'xxx', >headers = < 'authority': 'www.computeruniverse.net', '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-language': 'en-US,en;q=0.9', 'cache-control': 'max-age=0', 'if-modified-since': 'Sat, 17 Dec 2022 03:23:59 GMT', 'sec-ch-ua': '"Opera";v="93", "Not/A)Brand";v="8", "Chromium";v="107"', 'sec-ch-ua-mobile': '?0', 'sec-ch-ua-platform': '"Linux"', 'sec-fetch-dest': 'document', 'sec-fetch-mode': 'navigate', 'sec-fetch-site': 'same-origin', 'sec-fetch-user': '?1', 'upgrade-insecure-requests': '1', 'user-agent': 'Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/107.0.0.0 Safari/537.36 OPR/93.0.0.0', >params = < 'lang': '1', 'cachecountry': 'KZ', >response = requests.get('https://webapi.computeruniverse.net/api/catalog/topmenu/', params=params, cookies=cookies, headers=headers) print(response.status_code)

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