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Extraiga la URL de audio (mp3) del reproductor web usando Python

Estoy tratando de extraer el enlace URL de la página web para poder descargar el audio como mp3, pero sigo sin éxito... Aquí hay un código de página web:

código de la página web

Me gustaría obtener un valor del atributo 'scr' : 'https://radio-arch-pp.stv.livebox.sk/a520/00/0024/002424/00242487-1.mp3'

Ninguno de estos intentos imprime ese valor:

1)

 import requests from bs4 import BeautifulSoup re = requests.get("https://www.rtvs.sk/embed/audio/242500") soup = BeautifulSoup(re.text, 'html.parser') result = soup.findAll("div", id="player_audio_242500") print(result) for n in soup.findAll("div", id="player_audio_242500"): if 'src' in n.attrs: result = n['src'] else: result = n.text print(result) for n in soup.find_all('script'): print(n.get('src')) for n in soup.find_all('script'): if 'src' in n.attrs: result = n['src'] else: result = n.text print(result)
 from selenium import webdriver from selenium.webdriver.common.by import By from selenium.webdriver.common.keys import Keys from selenium.webdriver.chrome.service import Service ser = Service("chromedriver") op = webdriver.ChromeOptions() chrome_browser = webdriver.Chrome(service=ser, options=op) chrome_browser.maximize_window() chrome_browser.get('https://www.rtvs.sk/embed/audio/242500') enter = chrome_browser.find_element(by=By.CSS_SELECTOR, value="[id^=player_audio]") print(enter) print(enter.text)

¿Alguien puede ayudar por favor?

about 4 years ago · Juan Pablo Isaza
1 answers
Answer question

0

Eche un vistazo a su css selector , debería seleccionar el video y debería tomar su atributo src así:

 chrome_browser.find_element(By.CSS_SELECTOR, 'video').get_attribute("src")
about 4 years ago · Juan Pablo Isaza Report
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