Python爬取大量数据时,如何防止IP被封

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爬取了猪八戒上的一些数据可能是由于爬取的数据量有点多吧,结果我的IP被封了,需要自己手动来验证解封ip,但这显然阻止了我爬取更多的数据了。
Python爬取大量数据时,如何防止IP被封_第1张图片
下面是我写的爬取猪八戒的被封IP的代码

# coding=utf-8
import requests
from lxml import etree

def getUrl():
    for i in range(33):
        url = 'http://task.zbj.com/t-ppsj/p{}s5.html'.format(i+1)
        spiderPage(url)

def spiderPage(url):
    if url is None:
        return None

    htmlText = requests.get(url).text
    selector = etree.HTML(htmlText)
    tds = selector.xpath('//*[@class="tab-switch tab-progress"]/table/tr')
    try:
        for td in tds:
            price = td.xpath('./td/p/em/text()')
            href = td.xpath('./td/p/a/@href')
            title = td.xpath('./td/p/a/text()')
            subTitle = td.xpath('./td/p/text()')
            deadline = td.xpath('./td/span/text()')
            price = price[0] if len(price)>0 else ''    # python的三目运算 :为真时的结果 if 判定条件 else 为假时的结果
            title = title[0] if len(title)>0 else ''
            href = href[0] if len(href)>0 else ''
            subTitle = subTitle[0] if len(subTitle)>0 else ''
            deadline = deadline[0] if len(deadline)>0 else ''
            print price,title,href,subTitle,deadline
            print '---------------------------------------------------------------------------------------'
            spiderDetail(href)
    except:
        print '出错'


def spiderDetail(url):
    if url is None:
        return None

    try:
        htmlText = requests.get(url).text
        selector = etree.HTML(htmlText)
        aboutHref = selector.xpath('//*[@id="utopia_widget_10"]/div[1]/div/div/div/p[1]/a/@href')
        price = selector.xpath('//*[@id="utopia_widget_10"]/div[1]/div/div/div/p[1]/text()')
        title = selector.xpath('//*[@id="utopia_widget_10"]/div[1]/div/div/h2/text()')
        contentDetail = selector.xpath('//*[@id="utopia_widget_10"]/div[2]/div/div[1]/div[1]/text()')
        publishDate = selector.xpath('//*[@id="utopia_widget_10"]/div[2]/div/div[1]/p/text()')
        aboutHref = aboutHref[0] if len(aboutHref) > 0 else ''  # python的三目运算 :为真时的结果 if 判定条件 else 为假时的结果
        price = price[0] if len(price) > 0 else ''
        title = title[0] if len(title) > 0 else ''
        contentDetail = contentDetail[0] if len(contentDetail) > 0 else ''
        publishDate = publishDate[0] if len(publishDate) > 0 else ''
        print aboutHref,price,title,contentDetail,publishDate
    except:
      print '出错'

if '_main_':
  getUrl()

如何防止爬取数据的时候被网站封IP这里有一些套路.查了一些套路

1.修改请求头

之前的爬虫代码没有添加头部,这里我添加了头部,模拟成浏览器去访问网站

        user_agent = 'Mozilla/5.0 (Windows NT 10.0; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/53.0.2785.104 Safari/537.36 Core/1.53.4295.400'

        headers = {'User-Agent': user_agent}
        htmlText = requests.get(url, headers=headers, proxies=proxies).text

2.采用代理IP

当自己的ip被网站封了之后,只能采用代理ip的方式进行爬取,所以每次爬取的时候尽量用代理ip来爬取,封了代理还有代理。

# IP地址取自国内髙匿代理IP网站:http://www.xicidaili.com/nn/
# 仅仅爬取首页IP地址就足够一般使用

from bs4 import BeautifulSoup
import requests
import random

def get_ip_list(url, headers):
    web_data = requests.get(url, headers=headers)
    soup = BeautifulSoup(web_data.text, 'lxml')
    ips = soup.find_all('tr')
    ip_list = []
    for i in range(1, len(ips)):
        ip_info = ips[i]
        tds = ip_info.find_all('td')
        ip_list.append(tds[1].text + ':' + tds[2].text)
    return ip_list

def get_random_ip(ip_list):
    proxy_list = []
    for ip in ip_list:
        proxy_list.append('http://' + ip)
    proxy_ip = random.choice(proxy_list)
    proxies = {'http': proxy_ip}
    return proxies

if __name__ == '__main__':
    url = 'http://www.xicidaili.com/nn/'
    headers = {
        'User-Agent': 'Mozilla/5.0 (Windows NT 6.1; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/53.0.2785.143 Safari/537.36'
    }
    ip_list = get_ip_list(url, headers=headers)
    proxies = get_random_ip(ip_list)
    print(proxies)

生成代理ip,大家可以直接把这个代码拿去用

好了我用上面的代码给我生成了一批ip地址(有些ip地址可能无效,但只要不封我自己的ip就可以了,哈哈),然后我就可以在我的请求头部添加ip地址

给我们的请求添加代理ip

        proxies = {
            'http': 'http://124.72.109.183:8118',
            'http': 'http://49.85.1.79:31666'

        }
        user_agent = 'Mozilla/5.0 (Windows NT 10.0; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/53.0.2785.104 Safari/537.36 Core/1.53.4295.400'

        headers = {'User-Agent': user_agent}
        htmlText = requests.get(url, headers=headers, timeout=3, proxies=proxies).text

最后完整代码如下:

# coding=utf-8

import requests
import time
from lxml import etree

def getUrl():
    for i in range(33):
        url = 'http://task.zbj.com/t-ppsj/p{}s5.html'.format(i+1)
        spiderPage(url)


def spiderPage(url):
    if url is None:
        return None

    try:
        proxies = {
            'http': 'http://221.202.248.52:80',

        }
        user_agent = 'Mozilla/5.0 (Windows NT 10.0; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/53.0.2785.104 Safari/537.36 Core/1.53.4295.400'

        headers = {'User-Agent': user_agent}
        htmlText = requests.get(url, headers=headers,proxies=proxies).text

        selector = etree.HTML(htmlText)
        tds = selector.xpath('//*[@class="tab-switch tab-progress"]/table/tr')
        for td in tds:
            price = td.xpath('./td/p/em/text()')
            href = td.xpath('./td/p/a/@href')
            title = td.xpath('./td/p/a/text()')
            subTitle = td.xpath('./td/p/text()')
            deadline = td.xpath('./td/span/text()')
            price = price[0] if len(price)>0 else ''    # python的三目运算 :为真时的结果 if 判定条件 else 为假时的结果
            title = title[0] if len(title)>0 else ''
            href = href[0] if len(href)>0 else ''
            subTitle = subTitle[0] if len(subTitle)>0 else ''
            deadline = deadline[0] if len(deadline)>0 else ''
            print price,title,href,subTitle,deadline
            print '---------------------------------------------------------------------------------------'
            spiderDetail(href)
    except Exception,e:
        print '出错',e.message


def spiderDetail(url):
    if url is None:
        return None

    try:
        htmlText = requests.get(url).text
        selector = etree.HTML(htmlText)
        aboutHref = selector.xpath('//*[@id="utopia_widget_10"]/div[1]/div/div/div/p[1]/a/@href')
        price = selector.xpath('//*[@id="utopia_widget_10"]/div[1]/div/div/div/p[1]/text()')
        title = selector.xpath('//*[@id="utopia_widget_10"]/div[1]/div/div/h2/text()')
        contentDetail = selector.xpath('//*[@id="utopia_widget_10"]/div[2]/div/div[1]/div[1]/text()')
        publishDate = selector.xpath('//*[@id="utopia_widget_10"]/div[2]/div/div[1]/p/text()')
        aboutHref = aboutHref[0] if len(aboutHref) > 0 else ''  # python的三目运算 :为真时的结果 if 判定条件 else 为假时的结果
        price = price[0] if len(price) > 0 else ''
        title = title[0] if len(title) > 0 else ''
        contentDetail = contentDetail[0] if len(contentDetail) > 0 else ''
        publishDate = publishDate[0] if len(publishDate) > 0 else ''
        print aboutHref,price,title,contentDetail,publishDate
    except:
      print '出错'

if '_main_':
    getUrl()

Python爬取大量数据时,如何防止IP被封_第2张图片
数据全部爬取出来了,且我的IP也没有被封。当然防止被封IP肯定不止这些了,这还需要进一步探索!

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