从打开的文件中获取文件名，而不是文件路径

Question

假设我使用文件对话框在 PyQt5 应用程序中打开了一个名为 file1.mp3 的文件，并将其分配给如下变量：

song = QFileDialog.getOpenFileName(self, "Open Song", "~", "Sound Files (*.mp3 *.ogg *.wav *.m4a)")
print(song[0])
url = QUrl.fromLocalFile(song[0])
self.playlist.addMedia(QMediaContent(url))

如何获取文件名而不是文件路径，以便将其显示在状态栏中？或者更好的是，有没有我可以使用或创建的类似 "now playing" 的函数？

Answer 1

编程不是魔法，你有一个文件路径，即： c://myfolder/song.mp3 - 假设您的音乐文件以歌曲命名，您必须解析歌曲名称的 url 并将状态栏 title/label 设置为您当前正在播放的歌曲玩。我建议您在 python 上学习 lvl 入门课程，然后再将 qt 框架混入其中。

Answer 2

Self-explanatory。您只需要对字符串进行切片。因为你在学习，我会把它切错，让你找出原因。

filepath = QFileDialog.getOpenFileName(self, "Open Song", "~", "Sound Files (*.mp3 *.ogg *.wav *.m4a)")[0]
filename = filepath.split("/")[-1]

print(filename)

之后你可以简单地使用

self.<statusbarname>.showMessage("Now playing {0} song or whatever".format(filename))

但是，这仅适用于 "some" 系统。如果您想在另一台计算机上使用该应用程序，您应该首先规范化路径（一些系统使用 // 和其他 \ 作为文件夹），然后使用安全的 built-in 命令对其进行切片。

import os # Careful with this library, Read the documentation first
filepath = os.path.normpath(filepath) # Normalize it
filename = filepath.split(os.sep) # Slice it

整个代码应该是这样的：

import os
filepath = QFileDialog.getOpenFileName(self, "Open Song", "~", "Sound Files (*.mp3 *.ogg *.wav *.m4a)")[0]
print(filepath)
filepath = os.path.normpath(filepath)
song = filepath.split(os.sep)
url = QUrl.fromLocalFile(filepath)
self.playlist.addMedia(QMediaContent(url))
self.<statusbarname>.showMessage("Now playing {0} song or whatever and it was at {1} folder".format(song, filepath))

Answer 3

有几种获取文件名的简单方法：

使用QUrl:

song, _ = QFileDialog.getOpenFileName(self, "Open Song", "~", "Sound Files (*.mp3 *.ogg *.wav *.m4a)")
print(song)
url = QUrl.fromLocalFile(song)
self.playlist.addMedia(QMediaContent(url))
your_statusbar.showMessage("now playing {}".format(url.fileName()))

使用QFileInfo:

song, _ = QFileDialog.getOpenFileName(self, "Open Song", "~", "Sound Files (*.mp3 *.ogg *.wav *.m4a)")
print(song)
url = QUrl.fromLocalFile(song)
self.playlist.addMedia(QMediaContent(url))
filename = QFileInfo(song).fileName()
your_statusbar.showMessage("now playing {}".format(filename))

使用pathlib:

song, _ = QFileDialog.getOpenFileName(self, "Open Song", "~", "Sound Files (*.mp3 *.ogg *.wav *.m4a)")
print(song)
url = QUrl.fromLocalFile(song)
self.playlist.addMedia(QMediaContent(url))

from pathlib import Path    

filename = Path(song).name
your_statusbar.showMessage("now playing {}".format(filename))

使用os:

song, _ = QFileDialog.getOpenFileName(self, "Open Song", "~", "Sound Files (*.mp3 *.ogg *.wav *.m4a)")
print(song)
url = QUrl.fromLocalFile(song)
self.playlist.addMedia(QMediaContent(url))

import os   

filename = song.rstrip(os.sep)
your_statusbar.showMessage("now playing {}".format(filename))

或：

song, _ = QFileDialog.getOpenFileName(self, "Open Song", "~", "Sound Files (*.mp3 *.ogg *.wav *.m4a)")
print(song)
url = QUrl.fromLocalFile(song)
self.playlist.addMedia(QMediaContent(url))

import os   

_ , filename = os.path.split(os.sep)
your_statusbar.showMessage("now playing {}".format(filename))

从打开的文件中获取文件名，而不是文件路径

Get file name from an opened file, not a file path

python

pyqt

pyqt5

qtmultimedia