Question

我正在从Windows命令提示符运行python脚本。它调用下面的函数，它使用LAME将MP3文件转换为波形文件。

def convert_mp3_to_wav(input_filename, output_filename):
    """
    converts the incoming mp3 file to wave file
    """
    if not os.path.exists(input_filename):
        raise AudioProcessingException, "file %s does not exist" % input_filename

    command = ["lame", "--silent", "--decode", input_filename, output_filename]

    process = subprocess.Popen(command, stdout=subprocess.PIPE, stderr=subprocess.PIPE)
    (stdout, stderr) = process.communicate()

    if process.returncode != 0 or not os.path.exists(output_filename):
        raise AudioProcessingException, stdout

    return output_filename

不幸的是，LAME总是在某些MP3上崩溃（并且不辜负它的名字）。 Windows出现“你的程序已崩溃”对话框，冻结了我的脚本。关闭Windows对话框后，将引发AudioProcessingException。我不想告诉Windows关闭，我只是喜欢脚本来提高异常然后转移到下一个MP3。

这有什么办法吗？最好是通过改变脚本而不是用Unix运行它。

我正在使用Windows 7和Python 2.6

Answer 1

经过一番谷歌搜索后，我偶然发现了这一点 http://www.activestate.com/blog/2007/11/supressing-windows-error-report-messagebox-subprocess-and-ctypes

它需要一些修补，但现在下面的方法不会让烦人的Windows消息:) 请注意subprocess.Popen中的creationflags = subprocess_flags

def convert_mp3_to_wav(input_filename, output_filename):

    if sys.platform.startswith("win"):
        # Don't display the Windows GPF dialog if the invoked program dies.
        # See comp.os.ms-windows.programmer.win32
        # How to suppress crash notification dialog?, Jan 14,2004 -
        # Raymond Chen's response [1]

        import ctypes
        SEM_NOGPFAULTERRORBOX = 0x0002 # From MSDN
        ctypes.windll.kernel32.SetErrorMode(SEM_NOGPFAULTERRORBOX);
        subprocess_flags = 0x8000000 #win32con.CREATE_NO_WINDOW?
    else:
        subprocess_flags = 0



    """
    converts the incoming mp3 file to wave file
    """
    if not os.path.exists(input_filename):
        raise AudioProcessingException, "file %s does not exist" % input_filename

    #exec("lame {$tmpname}_o.mp3 -f {$tmpname}.mp3 && lame --decode {$tmpname}.mp3 {$tmpname}.wav");
    command = ["lame", "--silent", "--decode", input_filename, output_filename]

    process = subprocess.Popen(command, stdout=subprocess.PIPE, stderr=subprocess.PIPE, creationflags=subprocess_flags)
    (stdout, stderr) = process.communicate()

    if process.returncode != 0 or not os.path.exists(output_filename):
        raise AudioProcessingException, stdout

    return output_filename

Answer 2

通过creationflags=subprocess.CREATE_NO_WINDOW为我工作。即使应用程序崩溃，也将返回正确的应用程序输出。

处理Windows中的子进程崩溃

2 个答案: