如何独立于系统访问子文件夹中的文件？

Question

我希望将所有.mp3文件路径收集到一个文件夹（mp3）中，并将它们分配给字典中的键，这样无论系统文件路径如何，我的程序都能工作（目前，文件路径是硬编码的，这意味着该程序只在我自己的PC上工作）

以下是原始设置：

NotePaths = {
"C" : "C:/.../MP3/C4.mp3",
"D" : "C:/.../MP3/D4.mp3",
"E" : "C:.../MP3/E4.mp3",
"F" : "C:/.../MP3/F4.mp3",
"G" : "C:/.../MP3/G4.mp3",
"A" : "C:/.../MP3/A4.mp3",
"H" : "C:/.../MP3/H4.mp3"
}

以下是我试图使程序独立于系统文件夹结构工作的方法，只要程序所在的文件夹中有一个名为MP3的文件夹（其中包括所需的.MP3文件）：

NotePaths = {
"C" : "",
"D" : "",
"E" : "",
"F" : "",
"G" : "",
"A" : "",
"H" : ""
}

for root, dirs, files in os.walk("/MP3"):
    for file in files:
        if file.endswith(".mp3") and file.startswith("C"):
            NotePaths["C"].append(Str("(os.path.join(root, file))"))

        if file.endswith(".mp3") and file.startswith("D"):
            NotePaths["D"].append(Str("(os.path.join(root, file))"))

        if file.endswith(".mp3") and file.startswith("E"):
            NotePaths["E"].append(Str("(os.path.join(root, file))"))

        if file.endswith(".mp3") and file.startswith("F"):
            NotePaths["F"].append(Str("(os.path.join(root, file))"))

        if file.endswith(".mp3") and file.startswith("G"):
            NotePaths["G"].append(Str("(os.path.join(root, file))"))

        if file.endswith(".mp3") and file.startswith("A"):
            NotePaths["A"].append(Str("(os.path.join(root, file))"))

        if file.endswith(".mp3") and file.startswith("H"):
            NotePaths["H"].append(Str("(os.path.join(root, file))"))

试图运行程序（包括播放上述.mp3文件的playsound命令）会打开playsound模块并抛出以下错误：

Traceback (most recent call last):
  File "C:\...\Musical Quiz.py", line 67, in <module>
    playsound(currentPath)
  File "C:\...\AppData\Local\Programs\Python\Python38\lib\site-packages\playsound.py", line 35, in _playsoundWin
    winCommand('open "' + sound + '" alias', alias)
  File "C:\...\AppData\Local\Programs\Python\Python38\lib\site-packages\playsound.py", line 30, in winCommand
    '\n    ' + errorBuffer.value.decode())
UnicodeDecodeError: 'utf-8' codec can't decode byte 0xe4 in position 60: invalid continuation byte

文件路径按如下方式传递给playsound函数：

currentKey, currentPath = random.choice(list(NotePaths.items()))
playsound(currentPath)

更新：

我根据this answer重构了代码（谢谢！）。 我已经尝试通过修改playsound模块（如here所述）来查找实际问题，结果如下：

Traceback (most recent call last):
  File "C:\Users\...\Musical Quiz.py", line 67, in <module>
    playsound(currentPath)
  File "C:\Users\...\AppData\Local\Programs\Python\Python38\lib\site-packages\playsound.py", line 35, in _playsoundWin
    winCommand('open "' + sound + '" alias', alias)
  File "C:\Users\...\AppData\Local\Programs\Python\Python38\lib\site-packages\playsound.py", line 31, in winCommand
    raise PlaysoundException(exceptionMessage)
playsound.PlaysoundException: 
    Error 292 for command:
        open "" alias playsound_0.8907395801041198
    Der Befehl erfordert einen Alias-, Datei-, Treiber- oder Gertenamen.
    
##Attempted translation: the command requires an alias, file name, driver name or (I don't know what the last thing is)          

因此，似乎a）我的代码从一开始就不起作用，无法获得正确的文件路径，或者b）playsound不喜欢我正在做的事情

Answer 1

你核对过这个答案了吗Why do I have "'utf-8' codec can't decode byte 0xc4 in position 0: invalid continuation byte " problem and how do I solve it?

它概述了如何对此类错误进行故障排除，因为这是一种高级错误，实际错误可能是找不到文件或python版本控制错误

#######

另外，如果我可以对您的代码进行重构：

NotePaths = {
    "C" : "","D" : "","E" : "","F" : "","G" : "","A" : "","H" : ""
}

def file_startswith_key(file):
    """ Checking if file ends with wanted keys """
    for key in NotePaths:
        if file.startswith(key)
        return key, True
    return None, False

for root, dirs, files in os.walk("/MP3"):
    for file in files:
        key, bool_value = file_startswith_key(file)
        if file.endswith(".mp3") and bool_value:
            NotePaths[key].append(Str("(os.path.join(root, file))"))

这将允许您添加任意数量的键，而无需添加新的“if语句”