create new folder in python with pathlib and write files into it

You could try:

p = pathlib.Path("temp/")
p.mkdir(parents=True, exist_ok=True)
fn = "test.txt" # I don't know what is your fn
filepath = p / fn
with filepath.open("w", encoding ="utf-8") as f:
    f.write(result)

You shouldn't give a string as path. It is your object filepath which has the method open.

source

You can directly initialize filepath and create parent directories for parent attribute:

from pathlib import Path

filepath = Path("temp/test.txt")
filepath.parent.mkdir(parents=True, exist_ok=True)

with filepath.open("w", encoding ="utf-8") as f:
    f.write(result)

The pathlib module offers an open method that has a slightly different signature to the built-in open function.

pathlib:

Path.open(mode='r', buffering=-1, encoding=None, errors=None, newline=None)

The built-in:

open(file, mode='r', buffering=-1, encoding=None, errors=None, newline=None, closefd=True, opener=None)

In the case of this p = pathlib.Path("temp/") it has created a path p so calling p.open("temp."+fn, "w", encoding ="utf-8") with positional arguments (not using keywords) expects the first to be mode, then buffering, and buffering expects an integer, and that is the essence of the error; an integer is expected but it received the string 'w'.

This call p.open("temp."+fn, "w", encoding ="utf-8") is trying to open the path p (which is a directory) and also providing a filename which isn't supported. You have to construct the full path, and then either call the path's open method or pass the full path into the open built-in function.