如何在 python 中编写一个文件，并以变量作为路径？

发布于 2025-01-19 00:08:36 字数 592 浏览 5 评论 0原文

我正在尝试打开一个文件，该文件在程序中保存之前。然后，我想将一些文本写入文件中。但这给了我以下错误，我已经在Google和Stackoverflow上寻找了解决方案，但是解决方案没有用。

OSError: [Errno 22] Invalid argument: "<_io.TextIOWrapper name='C:/Users/kevin/Music/playlist.txt' mode='w' encoding='cp1252'>"

和我的代码：

def create_playlist():
playlist_songs = filedialog.askopenfilenames(initialdir=r'C:\Users\kevin\Music')
str(playlist_songs)
playlist_file = str(filedialog.asksaveasfile(mode="w",defaultextension=".txt"))
with open (playlist_file, 'w') as f:
    f.write(playlist_songs)

希望您能帮助我。感谢您提前的帮助。

原文

I am trying to open a file, which I save before, in the program. Then I want to write some text, into the file. But it gives me the following Error, I have already looked for solutions in google and also here on stackoverflow, but the solutions didn't work.

OSError: [Errno 22] Invalid argument: "<_io.TextIOWrapper name='C:/Users/kevin/Music/playlist.txt' mode='w' encoding='cp1252'>"

and my Code:

def create_playlist():
playlist_songs = filedialog.askopenfilenames(initialdir=r'C:\Users\kevin\Music')
str(playlist_songs)
playlist_file = str(filedialog.asksaveasfile(mode="w",defaultextension=".txt"))
with open (playlist_file, 'w') as f:
    f.write(playlist_songs)

I hope you can help me. I thank you for your help in advance.

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深海夜未眠 2025-01-26 00:08:36

playlist_file变量包含字符串“＆lt; _io.textiowrapper name ='c：/users/kevin/music/music/playlist.txt.txt'mode ='w'sopoding ='w'encoding ='cp1252'＆gt; gt; “;不仅“ c：/users/kevin/music/playlist.txt”，导致问题。

只需添加：

playlist_file = playlist_file[25: playlist_file.index("' ")]

以便您的代码成为

def create_playlist():
    playlist_songs = filedialog.askopenfilenames(initialdir=r'C:\Users\kevin\Music')
    playlist_file = str(filedialog.asksaveasfile(mode="w",defaultextension=".txt"))
    playlist_file = playlist_file[25: playlist_file.index("' ")]
    with open (playlist_file, 'w') as f:
        f.write(playlist_songs)

可运行的示例：

from tkinter import filedialog

playlist_songs = filedialog.askopenfilenames(initialdir=r'C:\Users\kevin\Music')
playlist_file = str(filedialog.asksaveasfile(mode="w",defaultextension=".txt"))
playlist_file = playlist_file[25: playlist_file.index("' ")]
with open (playlist_file, 'w') as f:
    f.write(playlist_songs)

The playlist_file variable contains the string "<_io.TextIOWrapper name='C:/Users/kevin/Music/playlist.txt' mode='w' encoding='cp1252'>"; not just "C:/Users/kevin/Music/playlist.txt", causing the issue.

Simply add:

playlist_file = playlist_file[25: playlist_file.index("' ")]

so that your code becomes

def create_playlist():
    playlist_songs = filedialog.askopenfilenames(initialdir=r'C:\Users\kevin\Music')
    playlist_file = str(filedialog.asksaveasfile(mode="w",defaultextension=".txt"))
    playlist_file = playlist_file[25: playlist_file.index("' ")]
    with open (playlist_file, 'w') as f:
        f.write(playlist_songs)

Runnable example:

from tkinter import filedialog

playlist_songs = filedialog.askopenfilenames(initialdir=r'C:\Users\kevin\Music')
playlist_file = str(filedialog.asksaveasfile(mode="w",defaultextension=".txt"))
playlist_file = playlist_file[25: playlist_file.index("' ")]
with open (playlist_file, 'w') as f:
    f.write(playlist_songs)

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