获取youtube视频网址表单网页源码 [英] Get youtube video url form webpage source

问题描述

是否有可能从YouTube网页的源代码获取当前播放视频的下载网址？在此先感谢。请帮助我。

其实我有一个下载器，但是它现在不工作。当我尝试了浏览器中下载器提供的URL时，它返回一个错误页面，显示访问网页被拒绝。我为视频获得的网址是
http://r14---sn-h557snel.c.youtube.com/v ideoplayback GCR =在&安培; IP = 121.241.181.70&安培; UPN = OOcjKEO_S6I&安培; CP = U0hUTFZQT19KU0NONF9OSlNGOmFRT2JSUUs2bDVi&安培; ipbits = 8&安培; FEXP = 900148,920704,912806,928001,922403,922405,929901,913605,929104,913546,913556,908496 ，920201,913302,919009,911116,901451,902556和安培; MS = AU&安培; MT = 1357625350&安培; MV =米和ID = 31a79fcdbfc6c2f6&安培; sparams =厘泊，GCR，ID，IP，ipbits，ITAG，ratebypass，来源，UPN，到期&安培; expire = 1357651559& itag = 18& key = yt1& newshard = yes& source = youtube& ratebypass = yes& sver = 3 网址格式前几个月是
http：// oo --- preferred --- sn-h557snes-- -v24---lscache3.c.youtube.com/videoplayback?upn=AmmxhSpq470&sparams=cp,gcr,id,ip,ipbits,itag,ratebypass,source,upn,expire&fexp=917013,906437,916626,922401 ，920704,912806,927201,925706,922403,913546,913556,916805,920201,901451&安培; MS = AU&安培;到期= 1353668177&安培; ITAG = 18&安培; ipbits = 8&安培; GCR =在&安培; sver = 3及ratebypass = YES&安培;山= 1353646087& ip = 121.241.181.70& mv = m& source = youtube& key = yt1& cp = U0hUSFZRVl9MUUNONF9OS1pCOmhtUTVMak5oTVNy& id = c429fa5466ee9447& newshard = yes

所以我不知道下载器是否提供了正确的URL。请帮助我

解决方案

我已经编写了一个简单的python代码，可以从html代码中获取youtube href。
希望它可以帮助你：

 ＃！/ usr / bin / env python 
 
 import re 
 
 pattern ='。*<（iframe | param）。*（src | value）=（？P< link> http：//www.youtube.com/（embed | v）/ [A-ZA-Z0-9 / \.\&安培;; = \ + _-？] +）;？。* * GT * LT; /（IFRAME | PARAM） >。*'
 
 action = re.compile（pattern）
 result = action.findall（'< div>< iframe .....>< / iframe> ;< param ......>< / param>< / div>'）
 
打印结果
  

https://gist.github.com/Mortezaipo/ 5707738

Is it possible to get the download URL of currently playing video from source of YouTube webpage?Thanks in advance.Please help me.

Actually I have a downloader but it is not working now.When I have tried with the URL provided by the downloader in browser it returns an error page says "Access to the webpage was denied".The Url that I got for a video is as " http://r14---sn-h557snel.c.youtube.com/videoplayback?gcr=in&ip=121.241.181.70&upn=OOcjKEO_S6I&cp=U0hUTFZQT19KU0NONF9OSlNGOmFRT2JSUUs2bDVi&ipbits=8&fexp=900148,920704,912806,928001,922403,922405,929901,913605,929104,913546,913556,908496,920201,913302,919009,911116,901451,902556&ms=au&mt=1357625350&mv=m&id=31a79fcdbfc6c2f6&sparams=cp,gcr,id,ip,ipbits,itag,ratebypass,source,upn,expire&expire=1357651559&itag=18&key=yt1&newshard=yes&source=youtube&ratebypass=yes&sver=3"

Few months before the format of URL was " http://o-o---preferred---sn-h557snes---v24---lscache3.c.youtube.com/videoplayback?upn=AmmxhSpq470&sparams=cp,gcr,id,ip,ipbits,itag,ratebypass,source,upn,expire&fexp=917013,906437,916626,922401,920704,912806,927201,925706,922403,913546,913556,916805,920201,901451&ms=au&expire=1353668177&itag=18&ipbits=8&gcr=in&sver=3&ratebypass=yes&mt=1353646087&ip=121.241.181.70&mv=m&source=youtube&key=yt1&cp=U0hUSFZRVl9MUUNONF9OS1pCOmhtUTVMak5oTVNy&id=c429fa5466ee9447&newshard=yes "

So I don't know whether the downloader gives the correct URL. Please help me

解决方案

I have written a simple python code that can fetch youtube href from html code. hope it can help you:

#!/usr/bin/env python

import re

pattern = '.*<(iframe|param).*(src|value)="(?P<link>http://www.youtube.com/(embed|v)/[a-zA-Z0-9/\.\?&;=\+_-]+);?.*".*>.*</(iframe|param)>.*'

action = re.compile(pattern)
result = action.findall('<div><iframe.....></iframe><param......></param></div>')

print result 

https://gist.github.com/Mortezaipo/5707738

这篇关于获取youtube视频网址表单网页源码的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！