I am trying to create a function in Python 2.7.3 to open a SQLite database.

This is my code at the moment:

import sqlite3 as lite
import sys

db = r'someDb.sqlite'

def opendb(db):
    try:
        conn = lite.connect(db)
    except sqlite3.Error:
        print "Error open db.\n"
        return False
    cur = conn.cursor()
    return [conn, cur]

I have tried the code above and I have observed that the sqlite3 library opens the database declared if exists, or creates a new database if this one doesn't exist.

Is there a way to check if the database exists with sqlite3 methods or I have to use file operation like os.path.isfile(path)?

In Python 2, you'll have to explicitly test for the existence using os.path.isfile:

if os.path.isfile(db):

There is no way to force the sqlite3.connect function to not create the file for you.

For those that are using Python 3.4 or newer, you can use the newer URI path feature to set a different mode when opening a database. The sqlite3.connect() function by default will open databases in rwc, that is Read, Write & Create mode, so connecting to a non-existing database will cause it to be created.

Using a URI, you can specify a different mode instead; if you set it to rw, so Read & Write mode, an exception is raised when trying to connect to a non-existing database. You can set different modes when you set the uri=True flag when connecting and pass in a file: URI, and add a mode=rw query parameter to the path:

from urllib.request import pathname2url

try:
    dburi = 'file:{}?mode=rw'.format(pathname2url(db))
    conn = lite.connect(dburi, uri=True)
except sqlite3.OperationalError:
    # handle missing database case

See the SQLite URI Recognized Query Parameters documentation for more details on what parameters are accepted.