Creating database file one directory above current

问题

In PHP, To refer to a higher directory I would use '../../test'; How would I do it in Python?

In this case, I'm using SQLite to create a database file.

import sqlite3
conn = sqlite3.connect('../data/test.db')

However; it says 'unable to open database file' which I'm considering as a directory access error. How do I refer to higher directory in Python?

回答1:

You might try this:

conn = sqlite3.connect(os.path.realpath('../data/test.db'))

回答2:

You need to find absolute path to your database file, using os.path.abspath and os.path.dirname. Then pass the absolute path to sqlite3.connect.

First (internal) os.path.dirname will give you current file's directory, second os.path.dirname will give you parent directory of current file's directory.

from os.path import join, dirname, abspath
db_path = join(dirname(dirname(abspath(__file__))), 'data/test.db')
sqlite3.connect(db_path)

来源：https://stackoverflow.com/questions/36784897/creating-database-file-one-directory-above-current