获取图片比获取文字更加复杂，获取文字在网页当中可以直接一次性的读取到文字；获取图片是获取图片的链接地址，然后通过链接地址下载到本地。

第一步：如何获取图片的链接地址

打开图片新闻的地址：https://www.infoq.com/presentations

1. 先获取图片地址：右键-查看源代码，获取新闻可以用文字的关键字搜索，作为图片在网页嵌入肯定会有img这样的标签，img后面跟着的链接地址就是我们要找的图片链接地址。

2.代码实现

import osimport shutilfrom bs4 import BeautifulSoup # 编码解码库import requests # 请求访问库# 用 dict 定义http头，伪装浏览器访问，避免被拒之门外headers = {    "Accept": "text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,*/*;q=0.8",    "Accept-Language": "zh-CN,zh;q=0.8",    "Connection": "close",    "Cookie": "_gauges_unique_hour=1; _gauges_unique_day=1; _gauges_unique_month=1; _gauges_unique_year=1; _gauges_unique=1",    "Referer": "https://www.infoq.com/presentations",    "Upgrade-Insecure-Requests": "1",    "User-Agent": "Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/57.0.2987.98 Safari/537.36 LBBROWSER"}url='https://www.infoq.com/presentations'def download_pic(img_url,image_localpath):    response=requests.get(img_url,stream=True)    if response.status_code==200:        with open (image_localpath,'wb') as f:            response.raw.deconde_content=True            shutil.copyfileobj(response.raw,f)#shutil库用来配合requests模块，他把response返回的内容写入到文件中#获取演讲图片,requests库封装的接口只提供了浏览的功能，如果想要下载的话需要提供stream这样的参数，stream设置成true才能和网页建立一个链接，再用response.content这样的方法进行下载def craw3(url):    response=requests.get(url,headers=headers)    soup=BeautifulSoup(response.text,'lxml')    for pic_href in soup.find_all('div',class_='card__content'):        for pic in pic_href.find_all('img'):            imgurl=pic.get('src')            dir=os.path.join('.')            filename=os.path.basename(imgurl)#该功能将链接地址只获取图片名称            imgpath=os.path.join(dir,filename)            #http://a.com/b/c.png只获取c.png文件名称            #c.png在和路径结合            print('开始下载 %s'%imgurl)            download_pic(imgurl,imgpath)craw3(url)

注意：

requests库封装的接口只提供了浏览的功能，如果想要下载的话需要提供stream这样的参数，stream设置成true才能和网页建立一个链接，再用response.content这样的方法进行下载

3.翻页功能

import osimport shutilfrom bs4 import BeautifulSoup # 编码解码库import requests # 请求访问库# 用 dict 定义http头，伪装浏览器访问，避免被拒之门外headers = {    "Accept": "text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,*/*;q=0.8",    "Accept-Language": "zh-CN,zh;q=0.8",    "Connection": "close",    "Cookie": "_gauges_unique_hour=1; _gauges_unique_day=1; _gauges_unique_month=1; _gauges_unique_year=1; _gauges_unique=1",    "Referer": "https://www.infoq.com/presentations",    "Upgrade-Insecure-Requests": "1",    "User-Agent": "Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/57.0.2987.98 Safari/537.36 LBBROWSER"}url='https://www.infoq.com/presentations'def download_pic(img_url,image_localpath):    response=requests.get(img_url,stream=True)    if response.status_code==200:        with open (image_localpath,'wb') as f:            response.raw.deconde_content=True            shutil.copyfileobj(response.raw,f)#shutil库用来配合requests模块，他把response返回的内容写入到文件中#获取演讲图片,requests库封装的接口只提供了浏览的功能，如果想要下载的话需要提供stream这样的参数，stream设置成true才能和网页建立一个链接，再用response.content这样的方法进行下载def craw3(url):    response=requests.get(url,headers=headers)    soup=BeautifulSoup(response.text,'lxml')    for pic_href in soup.find_all('div',class_='card__content'):        for pic in pic_href.find_all('img'):            imgurl=pic.get('src')            dir=os.path.join('.')            filename=os.path.basename(imgurl)#该功能将链接地址只获取图片名称            imgpath=os.path.join(dir,filename)            #http://a.com/b/c.png只获取c.png文件名称            #c.png在和路径结合            print('开始下载 %s'%imgurl)            download_pic(imgurl,imgpath)#craw3(url)j=0for i in range(12,46,12):    url='https://www.infoq.com/presentations'+str(i)    j+=1    print('第%d页'%j)craw3(url)

问题

考虑多线程实现

做梦梦到了海豚在跳跃