using pathlib to open shelved files in python

Question

I use pathlib to open textfiles that exists in a different directory, but get this error

TypeError:`"Unsupported operand type for +:'WindowsPath' and 'str'"

when I try opening a shelved file that exist in a score folder in the current directory like this.

from pathlib import *
import shelve

def read_shelve():
    data_folder = Path("score")
    file = data_folder / "myDB"
    myDB = shelve.open(file)
    return myDB

What am I doing wrong or is there another way to achieve this?

Apparently `shelve.open` hasn't been updated to support input paths based on the `__fspath__` protocol. So you'll have to manually pass `os.fspath(file)` instead of `file`. — Eryk Sun, Apr 23 '20 at 17:38

score 1 · Accepted Answer · answered Apr 25 '20 at 17:44

shelve.open() requires filename as string but you provide WindowsPath object created by Path.

Solution would be to simply convert Path to string following pathlib documentation guidelines:

from pathlib import *
import shelve

def read_shelve():
    data_folder = Path("score")
    file = data_folder / "myDB"
    myDB = shelve.open(str(file))
    return myDB

using pathlib to open shelved files in python

1 Answers1