python - python 3.4.3 中 urllib.httperror 的语法错误

Question

我正在关注一本书《Web Scraping with Python》，我正在尝试这个:

• 我在 OSX 上使用 python 3.4.3 的虚拟环境中
• BeautifulSoup 库已安装

当我尝试这个时:

from urllib.request import urlopen
from bs4 import BeautifulSoup
from urllib.error import HTTPError

html = urlopen("http://www.pythonscraping.com/exercises/exercise1.html")
except urllib.error.HTTPError as e:
    print(e.code)
    if html is none:
        print("url is not found")

else:
    bsObj = BeautifulSoup(html.read());
    print(bsObj

当我运行它时，出现以下错误:

    (scrapingEnv)Macintosh:scrapingenv nicolas$ python3 scrapetest.py
  File "scrapetest.py", line 6
    except urllib.error.HTTPError as e:
         ^
SyntaxError: invalid syntax

我还在第 6 行尝试了“except urllib.HTTPError”，但没有成功。

我做错了什么？

Answer 1

您缺少 try 语句

from urllib.request import urlopen
from bs4 import BeautifulSoup
from urllib.error import HTTPError

try:
     html = urlopen("http://www.pythonscraping.com/exercises/exercise1.html")
except urllib.error.HTTPError as e:
    print(e.code)
    if html is None:
        print("url is not found")

编辑:您还应该将 none 更改为 None