Question

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)?

Answer 1

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.