To open https://github.com/ytdl-org/youtube-dl/tree/master/youtube_dl/extractor ,bilibili.py in it,i get bilibili.py source code,it help nothing.
It is no use for a newbie like me to look over and over at bilibili.py's source code.
It is difficult for me to understand how youtube-dl get video's real url on bilibili?
What is the principle for bilibili.py to extract video's real url on bilibili?
There are 421 lines in bilibili.py's source code,please simplify bilibili.py as a short single function such as getUrlFromBilibili.
url = "https://www.bilibili.com/video/av52135588?from=search&seid=5625791913889140593"
real_url = getUrlFromBilibili(url)
print(real_url)
Please fulfil the getUrlFromBilibili function and give some comments in it to show principle to get video's real url on bilibili.
def getUrlFromBilibili(url):
#make the function short as soon as possible
Looks like you want to get only the url part without parameters. This can be achieved simply like this.
url = "https://www.bilibili.com/video/av52135588?from=search&seid=5625791913889140593"
real_url = url.split('?')[0]
//real_url = https://www.bilibili.com/video/av52135588
You have said you need to use functions for that and need know principles.
Bilibili site is like a youtube. There are lot videos in there. There for You have to program to get real url from any video url that you enter. I will show how to do that.
def getUrlFromBilibili(url):
real_url = url[:url.find('?')]
return real_url;
URL = input("Copy and Paste your Bilibili url = ")
#https://www.bilibili.com/video/av52135588?from=search&seid=5625791913889140593
print("Real url is = ", getUrlFromBilibili(URL))
output -:
Copy and Paste your Bilibili url = https://www.bilibili.com/video/av52135588?from=search&seid=5625791913889140593
Real url is = https://www.bilibili.com/video/av52135588
Now I am going to explain this code
#Start of the function
def getUrlFromBilibili(url): #This the function which we use to store our url
real_url = url[:url.find('?')] #In here we remove all characters after '?' in your url including '?'
return real_url; #After that we return our shorten url
#End of the function
URL = input("Copy and Paste your Bilibili url") #In here You can Enter any video url from Bilibili.com
print("Real url is = ", getUrlFromBilibili(URL)) #After all actions shorten url is displaying in here.
Importent
If you want to get real url from this url only -> https://www.bilibili.com/video/av52135588?from=search&seid=5625791913889140593 You can use this code
def getUrlFromBilibili(url):
real_url = url[:url.find('?')]
return real_url;
print("Real url is = ", getUrlFromBilibili("https://www.bilibili.com/video/av52135588?from=search&seid=5625791913889140593"))
output -:
Copy and Paste your Bilibili url = https://www.bilibili.com/video/av52135588?from=search&seid=5625791913889140593
Real url is = https://www.bilibili.com/video/av52135588
Related
I want to create an url (not a link) for a text field for a share functionality.
Like for example the one stackoverflow uses
I already hava function that produces the path part for the url, like
toUrl : Route -> String
toUrl route = ...
toUrl (Home (Just "hallo")) --> "/?b=hallo"
and using this string for a link as a href attribute works, but I'm wondering how I could create a complete url from this string.
PS: I'm using a single page application so I get an Url at the beginning.
What you say is "the path part for the url" isn't actually just the path part, but the path and query parts of the URL. Ideally you'd separate them so you can create a well-formed URL representation:
{ initialUrl
| path = "/"
, query = Just "b=hallo"
}
But since it's just a record, with no validation, it'll work if you just use it as the path as well. At least if you later just use Url.toString on it. Other operations might cause unexpected results.
{ initialUrl | path = "/?b=hallo" }
This url:
url = rtmp://xxxxxxxxxxxxxx.cloudfront.net/cfx/st/mp3:audios/lesson/2/cancion?Expires=1386332537&Signature=mysignature__&Key-Pair-Id=my-key-par-id
I would like to add the extension mp3 to all file name.
In this case the file name is cancion
The id of lesson is a dynamic value.
I would like to get this url something like:
url = rtmp://xxxxxxxxxxxxxx.cloudfront.net/cfx/st/mp3:audios/lesson/2/cancion.mp3?Expires=1386332537&Signature=mysignature__&Key-Pair-Id=my-key-par-id
Thanks!
You can parse the URI, edit the path, then return the value
require 'uri/http'
u = URI.parse('rtmp://xxxxxxxxxxxxxx.cloudfront.net/cfx/st/mp3:audios/lesson/2/cancion?Expires=1386332537&Signature=mysignature__&Key-Pair-Id=my-key-par-id')
u.path += ".mp3"
puts u.to_s
or use a simple regexp replace
u = 'rtmp://xxxxxxxxxxxxxx.cloudfront.net/cfx/st/mp3:audios/lesson/2/cancion?Expires=1386332537&Signature=mysignature__&Key-Pair-Id=my-key-par-id'
u.gsub('?', '.mp3?')
The second approach can be used only if you can assume the format of the input is always the same.
You can do simple gsub since this is URL and you can expect one occurrence of ? so simple do.
url.gsub!('?', '.mp3?')
Usually I would go regex here but no need from previously stated reason.
I am using
URL res = this.getClass().getClassLoader().getResource(dictionaryPath);
String path = res.getPath();
String path2 = path.substring(1);
because the output of the method getPath() returns sth like this:
/C:/Users/......
and I need this
C:/Users....
I really need the below address because some external library refuses to work with the slash at the beginning or with file:/ at the beginning or anything else.
I tried pretty much all the methods in URL like toString() toExternalPath() etc. and done the same with URI and none of it returns it like I need it. (I totally don't understand, why it keeps the slash at the beginning).
It is okay to do it on my machine with just erasing the first char. But a friend tried to run it on linux and since the addresses are different there, it does not work...
What should with such problem?
Convert the URL to a URI and use that in the File constructor:
URL res = this.getClass().getClassLoader().getResource(dictionaryPath);
File file = new File(res.toURI());
String fileName = file.getPath();
As long as UNIX paths are not supposed to contain drive letters, you may try this:
URL res = this.getClass().getClassLoader().getResource(dictionaryPath);
String path = res.getPath();
char a_char = text.charAt(2);
if (a_char==':') path = path.substring(1);
Convert to a URI, then use Paths.get().
URL res = this.getClass().getClassLoader().getResource(dictionaryPath);
String path = Paths.get(res.toURI()).toString();
You could probably just format the string once you get it.
something like this:
path2= path2[1:];
I was searching for one-line solution, so the best what i came up with was deleting it manually like this:
String url = this.getClass().getClassLoader().getResource(dictionaryPath).getPath().replaceFirst("/","");
In case if someone also needs to have it on different OS, you can make IF statement with
System.getProperty("os.name");
Removing the query string from a URL in Ruby could be done like this:
url.split('?')[0]
Where url is the complete URL including the query string (e.g. url = http://www.domain.extension/folder?schnoo=schnok&foo=bar).
Is there a faster way to do this, i.e. without using split, but rather using Rails?
edit: The goal is to redirect from http://www.domain.extension/folder?schnoo=schnok&foo=bar to http://www.domain.extension/folder.
EDIT: I used:
url = 'http://www.domain.extension/folder?schnoo=schnok&foo=bar'
parsed_url = URI.parse(url)
new_url = parsed_url.scheme+"://"+parsed_url.host+parsed_url.path
Easier to read and harder to screw up if you parse and set fragment & query to nil instead of rebuilding the URL.
parsed = URI::parse("http://www.domain.extension/folder?schnoo=schnok&foo=bar#frag")
parsed.fragment = parsed.query = nil
parsed.to_s
# => "http://www.domain.extension/folder"
url = 'http://www.domain.extension/folder?schnoo=schnok&foo=bar'
u = URI.parse(url)
p = CGI.parse(u.query)
# p is now {"schnoo"=>["schnok"], "foo"=>["bar"]}
Take a look on the : how to get query string from passed url in ruby on rails
You can gain performance using Regex
'http://www.domain.extension/folder?schnoo=schnok&foo=bar'[/[^\?]+/]
#=> "http://www.domain.extension/folder"
Probably no need to split the url. When you visit this link, you are pass two parameters to back-end:
http://www.domain.extension/folder?schnoo=schnok&foo=bar
params[:schnoo]=schnok
params[:foo]=bar
Try to monitor your log and you will see them, then you can use them in controller directly.
In rails I want to wrote some code to change this url string
https://img.skitch.com/20101222-kg5chjx4jetgcdeaug46hi6jpk.jpg
to
https://img.skitch.com/20101222-kg5chjx4jetgcdeaug46hi6jpk.preview.png
Should I use regular Expression to change it?
I'm new to Regexp, anyone can show me how to do this, and how to learn this stuff
thanks
If the extension is of fixed length, you're better off using string slicing.
url = "https://img.skitch.com/20101222-kg5chjx4jetgcdeaug46hi6jpk.jpg"
print url[0..-5] + ".preview" + url[-4..-1]
outputs
https://img.skitch.com/20101222-kg5chjx4jetgcdeaug46hi6jpk.preview.jpg
Or if your extensions are of variable length you can use rindex() to find the start of the extension.
url = "https://img.skitch.com/20101222-kg5chjx4jetgcdeaug46hi6jpk.jpeg"
dot_index = url.rindex(".")-1
print url[0..dot_index] + ".preview" + url[dot_index+1..-1]
outputs
https://img.skitch.com/20101222-kg5chjx4jetgcdeaug46hi6jpk.preview.jpeg
If you must use a regex then do it like this:
url = "https://img.skitch.com/20101222-kg5chjx4jetgcdeaug46hi6jpk.jpeg"
print url.gsub(/\.(\w{2,4})$/, ".preview.\\1")
outputs
https://img.skitch.com/20101222-kg5chjx4jetgcdeaug46hi6jpk.preview.jpeg
If you're sure the file ends with .jpg, you can to
url = "https://img.skitch.com/20101222-kg5chjx4jetgcdeaug46hi6jpk.jpg"
url.gsub(".jpg", ".preview.jpg")
Otherwise, you can get the filename, then append the extension.
url = "https://img.skitch.com/20101222-kg5chjx4jetgcdeaug46hi6jpk.jpg"
ext = File.extname(url)
url.gsub(ext, ".preview{ext}")
A string replace seems to be enough.
".jpg" -> ".preview.png"
Unfortunately I do not know ruby.
In python it'll be
new_url = url.replace(".jpg",".preview.png",1)
I think that it'll be similar in ruby. It seems to be sub() instead.
new_url = url.sub(".jpg",".preview.png")