Question

我有一个正在迭代的链接列表，如下所示

https://www.loc.gov/item/2015669100/
https://www.loc.gov/item/2015669101/
https://www.loc.gov/item/2015669102/
https://www.loc.gov/item/2015669103/
https://www.loc.gov/item/2015669104/
https://www.loc.gov/item/2015669105/
https://www.loc.gov/item/2015669106/
https://www.loc.gov/item/2015669107/
https://www.loc.gov/item/2015669108/
https://www.loc.gov/item/2015669109/

如果您查看这些链接，则可以看到它具有视频和可下载的XML文件。我的任务是从视频下载音频并从一页一起下载XML文件。

我的问题是，如何从此类音频文件中获取音频？

下面是我当前的代码。

from bs4 import BeautifulSoup
from urllib.request import Request, urlopen

base_html = "https://www.loc.gov/collections/civil-rights-history-project/?sp={}"

for i in range(1,8):
    html = base_html.format(i)
    req = Request(html, headers={'User-Agent': 'Mozilla/5.0'})
    soup = BeautifulSoup(urlopen(req).read(), 'html.parser')
    
    pages = soup.findAll('div', attrs={'class' : 'item-description'})
    for div in pages:
            crawl_p = div.find('a')['href']
            #some logic here

Answer 1

在站点上，看起来像是通过m3u8 url通过传输段流过的视频和音频文件（.ts文件）。

<source>包含m3u8流的URL。标签是使用标签中的属性来标识的（这里是type属性）。

import subprocess
from bs4 import BeautifulSoup
import requests as r

sess = r.session()
site_url = "https://www.loc.gov/item/2015669100/"
request = sess.get(site_url)
#print(request.content)

soup = BeautifulSoup(request.content, 'html5lib')
m3u8_url = soup.find('source',attrs={'type' : 'application/x-mpegURL'})['src']
print(str(m3u8_url))

这会删除m3u8网址https://tile.loc.gov/streaming-services/iiif/service:afc:afc2010039:afc2010039_crhp0001:afc2010039_crhp0001_mv04/full/full/0/full/default.m3u8

ffmpeg可以从m3u8文件中下载流（视频或音频）。也可以从python运行。

subprocess.call(['ffmpeg','-i',m3u8_url,'-vn','-map','a','output.ts' ])

请不要忘记用[]包装命令。引号内的每个单词代表一个以空格分隔的命令。完整的ffmpeg命令为ffmpeg -i m3u8_url -vn -map a output.ts

这是完整的代码。只要确保包含ffmpeg路径变量，否则子进程将引发错误。文件很大，因此下载音频文件可能需要一段时间。

import subprocess
from bs4 import BeautifulSoup
import requests as r

sess = r.session()
site_url = "https://www.loc.gov/item/2015669100/"
request = sess.get(site_url)
#print(request.content)

soup = BeautifulSoup(request.content, 'html5lib')
m3u8_url = soup.find('source',attrs={'type' : 'application/x-mpegURL'})['src']
print(str(m3u8_url))

subprocess.call(['ffmpeg','-i',m3u8_url,'-vn','-map','a','output.ts' ])

如何从Python中的视频链接下载音频

1 个答案: