How to check if a SQLite3 database exists in Python?

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.
" 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)`?

Accepted Answer

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.

How to check if a SQLite3 database exists in Python?

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