python - 如何防止 Instagram 临时请求被 selenium 阻塞？

Question

我正在尝试编写我自己的 python 脚本来找到一个帐户最受关注的追随者，它似乎工作正常，但是在一段时间或运行脚本超过 1-2 次之后，instagram 给了我一个重试错误，我搜索了并发现它的 Instagram 暂时阻止了我的 ip，因为我一次给出了许多请求。

有谁知道解决这个问题的方法？

我的代码

"""
WHAT DOES THIS SCRIPT ACTUALLY DO?:
This script enables you to scrape all your followers and then find X top followed followers.
--------------------------------------------------------------------------------------------
NOTICE:
Unfortunately it is very hard now a days to scrape social media sites, due to 
timeout issues, to many pings in a set time and other request restrictions.
So this script can only be ran 1-3 times a day.
I've tried also using exciting API's but all these are either too slow, or simply 
show a '428' to many requests error.
"""

import instaloader
from selenium import webdriver
import time
from webdriver_manager.chrome import ChromeDriverManager
from selenium.webdriver.common.keys import Keys
from rich.console import Console
from rich.table import Column, Table


# Global vars
L = instaloader.Instaloader()
URL = "https://www.instagram.com/{}/"
usernameGlobal = None
passwordGlobal = None
console = Console()


def get_followers():

    # Login
    while True: # Keep running if password/username was wrong
        try:
            global usernameGlobal, passwordGlobal
            
            print("\n"+"*-=-*"*5)
            usernameGlobal = input("> Enter your username: ")
            passwordGlobal = input("> Enter your password: ")
            
            L.login(usernameGlobal, passwordGlobal)
            
            print("\n"+"-"*28+"\n> Successfully Logged In!")
            print("> Please leave this program running in the background")
            print("> Until you see the 'FINISHED' message'"+"\n"+"-"*28)
            break
        except:
            print("\n"+"-"*28+"\n> Wrong Username / Password"+"\n"+"-"*28)
    
    # Obtain profile metadata
    profile = instaloader.Profile.from_username(L.context, usernameGlobal)
    
    follow_list = []
    
    # Loop through each follower and add to list
    for followee in profile.get_followers():
        follow_list.append(followee.username)
    
    return follow_list


def scrape_data(username):
    
    driver.get(URL.format(username))
    FOLLOWERS = 0
    
    try:
        try:
            FOLLOWERS = driver.find_element_by_xpath('/html/body/div[1]/section/main/div/header/section/ul/li[2]/a/span').text
        except: # For people who you don't follow but follow you and have private accounts
            FOLLOWERS = driver.find_element_by_xpath('/html/body/div[1]/section/main/div/header/section/ul/li[2]/span/span').text
    except:
        print("\n"+"-"*28+"\n> Please try this script again later!"+"\n"+"-"*28)
    
    result = ''.join([i for i in FOLLOWERS if i.isdigit()])
    
    return int(float(result))


def driver_login():
    
    driver.get("https://www.instagram.com")
    time.sleep(3)

    element = driver.find_element_by_xpath("//input[@name='username']")
    element.send_keys(usernameGlobal)

    element = driver.find_element_by_xpath("//input[@name='password']")
    element.send_keys(passwordGlobal)

    element.send_keys(Keys.RETURN)
    
    time.sleep(3)
    
    # -- This is for if you have two factor authentication enabled --
    # element = driver.find_element_by_xpath("//input[@name='verificationCode']")
    # key = input("Enter Activation key: ")
    # element.send_keys(key)
    
    # element.send_keys(Keys.RETURN)
    
    # time.sleep(3)


def output_result(size, result):
    
    n_input = 0
    
    # Get user to select how many of the top followed followers they want
    while True:
        try:
            print("\n"+"*-=-*"*10)
            n_input = int(input("> How many of your top followed followers do you want to see?\n> E.g 5 for top 5.\n> "))
            
            if n_input > size:
                continue
            
            break
        except:
            print("\n"+"-"*28+"\n> Invalid input. (Must be a number & less then your follower count)"+"\n"+"-"*28)
    
    # Make the table for a clean user friendly output and print it out
    table = Table(show_header=True, header_style="bold magenta")
    table.add_column("Your Followers", style="dim", width=12)
    table.add_column("There Follower Count")

    for x in range(n_input):
        table.add_row(
            list(result.keys())[x-1],
            list(result.values())[x-1]
        )

    console.print(table)
    return

    

if __name__ == "__main__":
    
    list_of_followers = get_followers()

    # Initialize the selenium driver
    driver = webdriver.Chrome(ChromeDriverManager().install())
    
    driver_login()
    
    result = {}
    
    for follower in list_of_followers:
        followers = scrape_data(follower)
        
        result[follower] = followers
        
    # Sort the dictionary by descending order
    result = dict(sorted(result.items(), key=lambda x: x[1], reverse=True))
    print("\n> FINISHED")
    
    driver.quit()
    
    output_result(len(list_of_followers), result)
    exit(0)
    

标签： pythonseleniumweb-scrapingpython-requestsinstagram

Answer 1

如果您使用代理，您可能会发出无限的请求。您可以从各个站点购买数以千计的代理，并在字典中轮换它们。

只需将代理列表添加到您的 GET 请求并享受：

proxyDict = { 
              "http"  : http_proxy, 
              "https" : https_proxy, 
              "ftp"   : ftp_proxy
            }

r = requests.get(url, headers=headers, proxies=proxyDict)

同样对于 Selenium，来自这个答案：

PROXY = "1.111.111.1:8080" #your proxy

chrome_options = WebDriverWait.ChromeOptions()
chrome_options.add_argument('--proxy-server=%s' % PROXY)

chrome = webdriver.Chrome(chrome_options=chrome_options)
chrome.get("instagram.com")