Python 从打开的文件中获取文件名，而不是文件路径

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))

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))

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如何获取文件名而不是文件路径，以便在状态栏中显示？或者更好的是，我是否可以使用或创建一个类似“正在播放”的函数？

编程不是魔术，您有一个文件路径，即： c://myfolder/song.mp3-假设音乐文件以歌曲命名，则必须解析歌曲名称的url，并将状态栏标题/标签设置为当前播放的歌曲。我建议您在混合使用qt框架之前先学习python入门级lvl课程。

。你只需要把绳子切成薄片。因为你在学习，所以我会用错误的方式来切分，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，您可以简单地使用

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

整个代码应如下所示：

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))

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))

导入操作系统
filepath=QFileDialog.getOpenFileName（self，“打开歌曲”，“~”，“声音文件（*.mp3*.ogg*.wav*.m4a）”）[0]
打印（文件路径）
filepath=os.path.normpath（filepath）
song=filepath.split（os.sep）
url=QUrl.fromLocalFile（文件路径）
self.playlist.addMedia（QMediaContent（url））
self..showMessage（“正在播放{0}首歌曲或其他内容，它位于{1}文件夹中”。格式（歌曲，文件路径））

获取文件名有几种简单方法：

• 使用

  QUrl

  ：

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• 使用

  QFileInfo

  ：

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• 使用

  pathlib

  ：

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• 使用

  os

  ：

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或：

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为什么不在字符串上使用split来分隔“/”和“.”字符之间的短语。我知道如何在普通python中分割路径，但我的PyQt程序将路径转换为QUrl。os.path.split在QUrl上是否也能以同样的方式工作？抱歉，我仍在学习PyQt。

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))

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))

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))

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))