无法在 subproces.call 中运行 ffmpeg

Question

所以，我有一个简单的类，我试图将来自终端 ffmpeg 命令的字符串响应保存到对象属性中：

import os
import subprocess

class Movie(object):

    absolute_path = None
    movie_info = None

    def __init__(self, path):
        self.absolute_path = "%s/%s" % (os.getcwd(), path)
        if(os.path.exists(self.absolute_path) is False):
            raise IOError("File does not exist")

    def get_movie_info(self):
        ffmpeg_command = "ffmpeg -i %s" % self.absolute_path
        self.movie_info = subprocess.call(ffmpeg_command)
        print self.movie_info

当我然后在 cmd 中运行此命令时：

import os
import sys
sys.path.append(os.getcwd())

from Encode.Movie import Movie

try:
    movie = Movie("tests/test_1.mpg")
    movie.get_movie_info()
except IOError as e:
    print e

我得到这个例外：

richard@richard-desktop:~/projects/hello-python$ python main.py 
Traceback (most recent call last):
  File "main.py", line 9, in <module>
    movie.get_movie_info()
  File "/home/richard/projects/hello-python/Encode/Movie.py", line 16, in get_movie_info
    self.movie_info = subprocess.call(ffmpeg_command)
  File "/usr/lib/python2.7/subprocess.py", line 493, in call
    return Popen(*popenargs, **kwargs).wait()
  File "/usr/lib/python2.7/subprocess.py", line 679, in __init__
    errread, errwrite)
  File "/usr/lib/python2.7/subprocess.py", line 1249, in _execute_child
    raise child_exception
OSError: [Errno 2] No such file or directory

路径是正确的，因为当我在 subprocess.call() 之前打印 self.absolute_path 时，我得到：

/home/richard/projects/hello-python/tests/test_1.mpg

而且这个文件是存在的。

Answer 1

问题是

ffmpeg_command = "ffmpeg -i %s" % self.absolute_path
self.movie_info = subprocess.call(ffmpeg_command)

• 您将单个字符串作为命令行，但省略了参数shell=True。

然而，推荐的方法是做

ffmpeg_command = ["ffmpeg", "-i", self.absolute_path]
self.movie_info = subprocess.call(ffmpeg_command)

为了分别给出命令和参数。这样，您在引用等方面就没有问题，并且省略了不必要的 shell 调用。