作者mater2004 (mater)
看板Python
标题[问题] 抓网页UserWarning
时间Sat Mar 11 05:40:54 2017
import requests
res = requests.get('
http://money.cnn.com/2017/03/10/news/economy/february-j
obs-report/index.html')
from bs4 import BeautifulSoup
soup = BeautifulSoup(res)
print(soup.text)
---------------------
我在jupyter执行会出现
http://i.imgur.com/UIWbbPP.jpg
---------------------
我是照着实例做,网址改掉而已。
我只知道应该要印出没有tag的 内容
不懂问题在哪 我试着改这行 soup= BeautifulSoup((res), "html.parser")
改完後出现
http://i.imgur.com/Fptoc3L.jpg
但我看不懂
-------
另外html.parser是什麽?
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1F:推 woogee: soup = BeautifulSoup(res) 改 03/11 06:32
2F:→ woogee: soup = BeautifulSoup(res.text) 03/11 06:32
3F:→ woogee: requests是回传status code,要取得内容要用.text 03/11 06:33
4F:→ woogee: html.parser是新版BeautifulSoup要告诉他是用哪一种parser 03/11 06:34
5F:→ woogee: 抱歉,requests不是回传status code,我说错了 03/11 06:34
6F:→ woogee: 他是回传Response,要取得内容要用.text 03/11 06:35
7F:→ mater2004: 请问requests 回传 response要如何得知 03/11 16:23
8F:→ mater2004: response是浏览器开发人员里面的response吗 03/11 16:24
9F:→ zerof: response 是指 requests 的 response object.. 03/11 22:23
10F:推 woogee: response就是requests回传的东西叫做response 03/12 22:54