只想从Href收到短信

Question

我试着重新迭代代码。如果我打印履带名我得到全名，那么下一行什么也不做吗？在这里真的很挣扎。

帮助欣赏我应该做的事

import requests

from bs4 import BeautifulSoup

base_url = "http://www.harness.org.au"

webpage_response = requests.get('http://www.harness.org.au/racing/tracks/', "html.parser")

soup = BeautifulSoup(webpage_response.content, "html.parser")

# only finding one track
# soup.table to find all links for days racing
#harness_table = soup.row
# scraps a href that is an incomplete URL that im trying to get to

tracks = soup.find(class_="col-lg-10 col-md-10 col-sm-10 col-xs-10 content")

lists = []

links = tracks.find_all('a')

#Gets each track
for a in links:
    lists.append(base_url+a["href"])

# track1 = []
#purpose - just to get track name before going over other data
for link in lists:
  webpage = requests.get(link)
  track = BeautifulSoup(webpage.content, "html.parser")
  trackname = track.find_all(class_="pageTitle")
  track1 = trackname.get_text()
  print(trackname)

Answer 1

当您使用find_all时，它会返回一个元素列表，其中包含您希望它找到的任何标记和属性。因此，您需要从第一个元素中获取文本。或者您可以只使用find，这将返回第一个标记/属性。

所以试着改变一下：

#purpose - just to get track name before going over other data
for link in lists:
  webpage = requests.get(link)
  track = BeautifulSoup(webpage.content, "html.parser")
  trackname = track.find(class_="pageTitle")
  try:
      track1 = trackname.get_text()
  except:
      print ('No class="pageTitle" found.')
      track1 = ''
  print(track1)
No class="pageTitle" found.

输出：

Racing
Tracks
Albany
Albion Park
Albury
Ararat
Armidale
Bacchus Marsh
Ballarat
Bankstown
Bathurst
Benalla
...