Reputation: 90
I have a list of URLs to video files, and I want to get the time length (duration) of those videos. They are all in the mp4 format.
Any support material I seem to be able to find relates to local video files. Is this even possible in Python? I'm totally confused about how to go around solving this one.
Upvotes: 2
Views: 4096
Reputation: 11
Try this Class that I wrote:
# coding:utf-8
import struct
import requests
class Mp4info:
def __init__(self, file):
self.file = file
self.seek = 0
self.duration = 0
self.s = requests.session()
self.timeout = 6
self.s.headers = {
'Connection': 'keep-alive',
'Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,image/apng,*/*;q=0.8',
'Accept-Encoding': 'gzip, deflate',
'Accept-Language': 'zh-CN,zh;q=0.9,en-US;q=0.8,en;q=0.7',
'User-Agent': 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_13_4) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/66.0.3359.139 Safari/537.36'
}
# 设置请求头 set request header
# 传入的seek表示代表需要跳过的字节数量 use seek to skip initial data
# 在这里进行判断是为了后续获取视频的宽高信息预留的 the condition here is for reserving space for getting the media data
def _set_headers(self, seek, type):
if type in ['moov', 'duration']:
self.s.headers['Range'] = 'bytes={}-{}'.format(seek, seek + 7)
def _send_request(self):
try:
data = self.s.get(url=self.file, stream=True,
timeout=self.timeout).raw.read()
except requests.Timeout:
raise '连接超时:超过6秒(默认)服务器没有响应任何数据!' # timeout 6 seconds, the server fails to respond and assumes there is no data
return data
def _find_moov_request(self):
self._set_headers(self.seek, type='moov')
data = self._send_request()
size = int(struct.unpack('>I', data[:4])[0])
flag = data[-4:].decode('ascii')
return size, flag
def _find_duration_request(self):
# 4+4是moov的大小和标识,跳过20个字符,直接读到time_scale,duration # 4+4 is the first 8 characters denoting charset, skip the next 20 to time_scale and duration
self._set_headers(seek=self.seek+4+4+20, type='duration')
data = self._send_request()
time_scale = int(struct.unpack('>I', data[:4])[0])
duration = int(struct.unpack('>I', data[-4:])[0])
return time_scale, duration
def get_duration(self):
while True:
size, flag = self._find_moov_request()
if flag == 'moov':
time_scale, duration = self._find_duration_request()
self.duration = duration/time_scale
return self.duration
else:
self.seek += size
if __name__ == '__main__':
url = 'http://tekeye.uk/html/images/Joren_Falls_Izu_Japan.mp4'
file = Mp4info(url)
a = file.get_duration()
print(a)
Upvotes: 1
Reputation: 6682
I have find a solution, if you run the following command in terminal you will get the duration of the online video.
ffprobe -i https://cldup.com/po79gkocrO.mp4 -show_entries format=duration -v quiet -of csv="p=0"
And I believe it can be used in python script.
Upvotes: 1
Reputation: 1070
import subprocess
def getLength(filename):
result = subprocess.Popen(["ffprobe", filename],
stdout = subprocess.PIPE, stderr = subprocess.STDOUT)
return [x for x in result.stdout.readlines() if "Duration" in x]
This is not my code. I haven't ever worked in python. I got the following answer from How to get video duration in Python or Django?
Upvotes: 0