这个playsound模块的代码中的错误是什么，即使语法与官方网站上相同?

Question

我已经检查了路径变量，一切都适用于其他函数，但它仅显示此模块的错误。代码:

from playsound import playsound
playsound('E://PYTHON//Music.mp3')

错误如下:

PS C:UsersHP> & C:/Users/HP/AppData/Local/Programs/Python/Python39/python.exe e:/PYTHON/Playsound.py
Error 263 for command:
open E://PYTHON//Music.mp3
The specified device is not open or is not recognized by MCI.
Error 263 for command:
close E://PYTHON//Music.mp3
Failed to close the file: E://PYTHON//Music.mp3
Traceback (most recent call last):
File "e:PYTHONPlaysound.py", line 2, in <module>
playsound('E://PYTHON//Music.mp3')
File "C:UsersHPAppDataLocalProgramsPythonPython39libsite-packagesplaysound.py", line 72, in _playsoundWin
winCommand(u'open {}'.format(sound))
File "C:UsersHPAppDataLocalProgramsPythonPython39libsite-packagesplaysound.py", line 64, in winCommand
raise PlaysoundException(exceptionMessage)
playsound.PlaysoundException:
Error 263 for command:
open E://PYTHON//Music.mp3
The specified device is not open or is not recognized by MCI.
PS C:UsersHP> & C:/Users/HP/AppData/Local/Programs/Python/Python39/python.exe e:/PYTHON/Playsound.py
Error 263 for command:
open E://PYTHON//Music.mp3
The specified device is not open or is not recognized by MCI.
Error 263 for command:
close E://PYTHON//Music.mp3
The specified device is not open or is not recognized by MCI.
Failed to close the file: E://PYTHON//Music.mp3
Traceback (most recent call last):
File "e:PYTHONPlaysound.py", line 2, in <module>
playsound('E://PYTHON//Music.mp3')
File "C:UsersHPAppDataLocalProgramsPythonPython39libsite-packagesplaysound.py", line 72, in _playsoundWin
winCommand(u'open {}'.format(sound))
File "C:UsersHPAppDataLocalProgramsPythonPython39libsite-packagesplaysound.py", line 64, in winCommand
raise PlaysoundException(exceptionMessage)
playsound.PlaysoundException:
Error 263 for command:
open E://PYTHON//Music.mp3
The specified device is not open or is not recognized by MCI.

Answer 1

pip install playsound==1.2.2

指定的设备未打开或MCI无法识别- playsound 1.3.0

Answer 2

我在Windows 10下使用Python 3.8.8。

我的音频文件是D:mypathtowavmysound.wav

playsound ('mysound.wav')导致'指定的设备未打开…

我试图提供完整的路径，但后来意识到os.path.dirname(__file__)返回一个空字符串。

os.getcwd()返回了正确的目录，但是遇到了一个不同的问题:当我将目录名与文件名拼接在一起时，我得到了D:mypathtowavmysound.wav，但这也不适用于playsound。我得到错误:'不能在引号括起来的字符串后指定额外的字符。'

最后，我使用正斜杠对整个路径进行硬编码:D:/my/path/to/wav/mysound.wav现在playsound()正常工作了

这只是一个快速和肮脏的解决方案，因为我的项目是实验性的。但我想说的是，playsound()对文件名很挑剔。

Answer 3

在我的例子中，我只是使用playsound('./' + audio_filename)，因为音频和.py文件位于相同的路径或目录中。您可以使用../后退。

Answer 4

我找到了解决方案，我们需要将文件添加到当前python程序所在的文件夹中所以我们确实需要添加文件的路径

from playsound import playsound
playsound('Music.mp3')