在python中循环以从URL获取标题标记

Question

import urllib.request as urllib2
from bs4 import BeautifulSoup

a = "https://player.vimeo.com/video/1234"

soup = BeautifulSoup(urllib2.urlopen(a))
print (a + soup.title.string)

我想通过一个循环获取URL的标题，每次获取URL标题时，循环都会添加一个数字

我得到了https://player.vimeo.com/video/1234的标题，然后是https://player.vimeo.com/video/1235等等

Answer 1

如果你有更多的url，添加到lst。你得到了所有的标题。您可以尝试以下脚本：

import urllib.request as urllib2
from bs4 import BeautifulSoup

lst = ["https://player.vimeo.com/video/1234","https://player.vimeo.com/video/1235"]
title = []
for a in lst:    
    soup = BeautifulSoup(urllib2.urlopen(a), 'lxml')
    title.append(soup.title.string)

print(title)

输出将是：

['Diving catch from Chris Bodenner on Vimeo', 'Hit with box from Chris Bodenner on Vimeo']

或

import urllib.request as urllib2
from bs4 import BeautifulSoup

lst = ["https://player.vimeo.com/video/1234","https://player.vimeo.com/video/1235"]
title = []
for a in lst:    
    soup = BeautifulSoup(urllib2.urlopen(a), 'lxml')
    title.append(soup.title.string)
    print (a + " : " + soup.title.string)

输出将是：

https://player.vimeo.com/video/1234 : Diving catch from Chris Bodenner on Vimeo
https://player.vimeo.com/video/1235 : Hit with box from Chris Bodenner on Vimeo

Answer 2

您可以这样做：

import urllib.request as urllib2
from bs4 import BeautifulSoup

start_idx, end_idx = 1234, 1245

for idx in range(start_idx, end_idx):
  a = f"https://player.vimeo.com/video/{idx}"
  soup = BeautifulSoup(urllib2.urlopen(a))
  print (f"for url:{a}, title: {soup.title.string}")

根据需要正确设置start_idx和end_idx

另外，您可能需要处理由于禁止访问某些URL而可能出现的HTTPError