How do I pass the None value explicitly? about typer

[jonaslb]

Semantically None is used to communicate the absence of a specified value in typer. So you can't as far as I can tell set it explicitly or enable functionality to allow it.

My proposal is that you either use a special value (for example -1 or 0) to indicate the limit, or to create a special --disable-limit flag, like so:

from typing import Optional

import typer

def main1(
    limit: Optional[int] = typer.Option(11, help="A limit value of 0 disables the limit.")
):
    if limit > 0:
        print(limit)
    else:
        print("unlimited")

def main2(
    limit: Optional[int] = 11,
    disable_limit: bool = typer.Option(False, "--disable-limit")
):
    if disable_limit:
        limit = None
    print(limit)

if __name__ == "__main__":
    typer.run(main1)

[kuykendall-ben]

Update here: with custom parsers, you should be able to implement this as of typer 0.8.0!

from typing import Optional
import typer

def int_or_none(raw: str) -> Optional[int]:
    return None if raw == 'None' else int(raw)

def main(
    limit: Optional[int] = typer.Option(11, parser=int_or_none),
):
    print(limit)

if __name__ == "__main__":
    typer.run(main)

Usage:

$ /usr/bin/python3 main.py       
11
$ /usr/bin/python3 main.py --limit None
None
$ /usr/bin/python3 main.py --limit 3   
3

The name of the parser is even displayed in the help string:

$ /usr/bin/python3 main.py --help      
Usage: main.py [OPTIONS]

Options:
  --limit INT_OR_NONE  [default: 11]
  --help               Show this message and exit.

You could modify the metavar or help field of your Option for additional clarify.