How to make sqlite3.connect() fail if the db file does not exist?

Question

sqlite3.connect() will create the db file if it does not exist. I'd like it to fail. Is there a way to do so?

Answer 1

The first way is to check the file path with os.path.isfile :

import sqlite3
import os

my_path = 'database.db' # or 'absolute_path/to/the/file'
if os.path.isfile(my_path):
    sqlite3.connect(my_path)

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Otherwise you can use the uri=True parameter to specify an opening mode and raise an error in case the file is missing. Without the mode, the file will be created if not exists, so you can use for example rw or ro to avoid the new file :

• to open with read & write mode :

  sqlite3.connect('file:database.db?mode=rw', uri=True)
  

• to open in read only mode :

  sqlite3.connect('file:database.db?mode=ro', uri=True)
  

This will raise the following error if the file doesn't exists :

sqlite3.OperationalError: unable to open database file

You can find more about these modes in these chapters of the doc :

https://www.sqlite.org/uri.html

https://www.sqlite.org/c3ref/open.html#urifilenamesinsqlite3open

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To be able to open files with special characters (special for an URI), another way is to use the following method :

import pathlib

my_db = pathlib.Path('path/to/data?ba=se.db').as_uri()
sqlite3.connect('{}?mode=rw'.format(my_db), uri=True)

# or sqlite3.connect(f'{my_db}?mode=rw', uri=True)  with f-string