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.
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)