CODEWITHSUNDEEP
pythonpython-3.xpathlib
Bondrak
Bondrak

Reputation: 1590

Create new folder with pathlib and write files into it

I'm doing something like this:

import pathlib

p = pathlib.Path("temp/").mkdir(parents=True, exist_ok=True)

with p.open("temp."+fn, "w", encoding ="utf-8") as f:
    f.write(result)

Error message: AttributeError: 'NoneType' object has no attribute 'open'

Obviously, based on the error message, mkdir returns None.

Jean-Francois Fabre suggested this correction: 

p = pathlib.Path("temp/")
p.mkdir(parents=True, exist_ok=True)

with p.open("temp."+fn, "w", encoding ="utf-8") as f:
    ...

This triggered a new error message:

File "/Users/user/anaconda/lib/python3.6/pathlib.py", line 1164, in open opener=self._opener)
TypeError: an integer is required (got type str)

Upvotes: 146

Views: 227832

Answers (6)

Sylvain
Sylvain

Reputation: 799

You are almost there. The full pathlib version would be something like:

folder = "path/to/folder"
file_base = "file_name_base"
file_extension = ".extension"

p_folder = pathlib.Path(folder)
p_file_base = p_folder / file_base
p_file = p_file_base.with_suffix(file_extension)

p_file.parent.mkdir(parents=True, exist_ok=True)  # or `p_folder`

p_file.write_text("Your text here", encoding ="utf-8")

I decomposed all steps to make it as transparent as possible, you can shorten this code. For example:

p_file = pathlib.Path("path/to/folder/file_base")
p_file.parent.mkdir(parents=True, exist_ok=True)
p_file.with_suffix(".extension").write_text("Your text here", encoding ="utf-8")

Upvotes: 3

anapaulagomes
anapaulagomes

Reputation: 334

You can also do something like:

text_path = Path("random/results").resolve()
text_path.mkdir(parents=True, exist_ok=True)
(text_path / f"{title}.txt").write_text(raw_text)

Upvotes: 8

Yukihiko Shinoda
Yukihiko Shinoda

Reputation: 975

Maybe, this is the shortest code you want to do.

import pathlib

p = pathlib.Path("temp/")
p.mkdir(parents=True, exist_ok=True)
(p / ("temp." + fn)).write_text(result, encoding="utf-8")

In most cases, it doesn't need even open() context by using write_text() instead.

Upvotes: 23

Anton Protopopov
Anton Protopopov

Reputation: 31662

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)

Upvotes: 54

Davos
Davos

Reputation: 5405

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.

Upvotes: 16

Till
Till

Reputation: 4523

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

Upvotes: 166

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