我正在嘗試從我的播放列表中抓取曲目的所有鏈接。這是我的代碼from selenium import webdriver from time import sleepfrom bs4 import BeautifulSoupfrom urllib.request import urlopenimport replaylist = 'minimal_house'url = 'https://www.youtube.com/channel/UCt2GxiTBN_RiE-cbP0cmk5Q/playlists'html = urlopen(url)soup = BeautifulSoup(html , 'html.parser')tracks = soup.find(title = playlist).get('href')print(tracks)url = url + tracksprint(url)html = urlopen(url)soup = BeautifulSoup(html, 'html.parser')links = soup.find_all('a',attrs={'class':'yt-simple-endpoint style-scope ytd-playlist-panel-video-renderer'})print(links)我不能刮'a'; 也不由id; 也不按類名。
1 回答
Helenr
TA貢獻1780條經驗 獲得超4個贊
這是對我有用的混亂代碼：
from selenium import webdriver
from time import sleep
from bs4 import BeautifulSoup
from urllib.request import urlopen
import re
playlist = 'minimal_house'
url = 'https://www.youtube.com/channel/UCt2GxiTBN_RiE-cbP0cmk5Q/playlists'
html = urlopen(url)
soup = BeautifulSoup(html, 'html.parser')
tracks = soup.find('a', attrs={'title': playlist}).get('href')
print(tracks)
url = 'https://www.youtube.com' + str(tracks)
print(url)
html = urlopen(url)
soup = BeautifulSoup(html, 'html.parser')
links = soup.find_all('a')
links = set([link.get('href') for link in links if link.get('href').count('watch')])
print(links)
由于類名會根據設備請求更改，因此在這種情況下最好獲取所有鏈接。并且您需要使用 selenium 向下滾動以獲取所有列表。
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