#python #flask #youtube-dl
#python #flask #youtube-dl
Вопрос:
Как загрузить видео с YouTube и других веб-сайтов и аудиофайлы через youtube-dl (загрузка интерфейса — веб-интерфейс)
Найдите и загрузите видео через интерфейс.
я должен написать скрипт на python и flask —
from flask import (
Flask, Response,
render_template,
request,
redirect,
flash,
url_for,
send_file,
session,
)
import subprocess
from ydl import get_media, verify, fetch_name
from zipper import zipping
import os
app = Flask(__name__)
app.secret_key = "supposed to be a secret"
@app.route("/return-file/")
def return_file():
import pdb
#pdb.set_trace()
num_choice = session.get("choice")
filename = session.get("filename")
url = session.get("url")
if num_choice == 1:
filename_formatted = filename ".mp3"
location = "media/Audio downloads/{}.mp3".format(session.get("id"))
if num_choice == 2:
#filename_formatted = filename ".mp4"
#cc = get_media(url, num_choice)
print(url)
print('==============================================================================')
#"youtube-dl", "--get-url", url
#subprocess.run(["youtube-dl", "--no-check-certificate", "--get-url", url])
#subprocess.run(["youtube-dl", "--no-check-certificate", url])
test = subprocess.run(["youtube-dl", "--no-check-certificate", "--get-filename", url])
print(test)
csv = '1,2,3n4,5,6n'
return Response(
csv,
mimetype="text/csv",
headers={"Content-disposition":
"attachment; filename=test"})
#return send_file('', attachment_filename="myplot.csv")
print('==============================================================================')
#subprocess.run(["youtube-dl", "--no-check-certificate", url])
#location = "media/{}.mp4".format(session.get("id"))
#if os.path.isdir(location):
#print('True')
if num_choice == 3 or num_choice == 4:
filename_formatted = filename ".zip"
location = "media/{}.zip".format(session.get("id"))
#return send_file(
#location, attachment_filename=filename_formatted, as_attachment=True
#)
@app.route("/", methods=["GET", "POST"])
def home_page():
"""
Displaying homepage
"""
title = "YDL | YouTube Downloader"
if request.method == "POST":
attempted_url = request.form["url"]
attempted_choice = int(request.form["submit"])
title = [attempted_url, attempted_choice]
if attempted_url != "":
if verify(attempted_url):
result_id = get_media(attempted_url, attempted_choice)
session["url"] = attempted_url
session["id"] = result_id
session["choice"] = attempted_choice
filename = fetch_name(attempted_url)
session["filename"] = filename
# return render_template('material-life.html', title = "Success {}".format(title))
# return render_template('material-life.html', title = result_id)
return redirect(url_for("return_file"))
else:
return render_template(
"material-life.html", title="YDL | Doesn't belong to YouTube"
)
else:
return render_template(
"material-life.html", title="YDL | URL shouldn't be empty"
)
return render_template("material-life.html", title=title)
@app.errorhandler(404)
def page_not_found(error):
"""
for anyone trying different links or searching for images within the server
"""
return (
render_template(
"error_template.html",
title="404 bud",
message="Time to make the chimi-fuckin'-changas. ",
subline="404, not there",
image_location=url_for("static", filename="images/deadpool-funny.jpg"),
),
404,
)
@app.errorhandler(400)
def bad_request(error):
"""
For handling situations where the server doesn't know what to do with the browser's request
"""
return (
render_template(
"error_template.html",
title="Aaaah ...",
message="나는 이해하지 못한다.",
subline="Yeah, the server couldn't understand what you asked for, Sorry",
image_location=url_for("static", filename="images/simpson-gangam.jpg"),
),
400,
)
if __name__ == "__main__":
app.run(debug=True)
- он работает на базовой загрузке консоли … но я хочу загрузить через интерфейс … без сохранения — прямая загрузка
Ответ №1:
Вы можете использовать pytube
модуль, который зависит от youtube-dl
. Кажется, вы создаете приложение, но с URL-адресом видео вы можете использовать приведенный ниже однострочник для загрузки видео:
from pytube import YouTube
YouTube(video_url).streams.first().download(filename='file_name')