If I read http://bugs.python.org/issue10740#msg132470 correctly, the foreign_keys PRAGMA is a no-op b/c "The python sqlite module automatically commits open transactions when it encounters a DDL statement."

Entering as a separate issue as I'm not sure solving 10740 is required to solve this one.

Works for me?

$ python2.7 t.py
Traceback (most recent call last):
  File "t.py", line 13, in <module>
    con.execute("insert into track (artist_id) values (1)")
sqlite3.IntegrityError: foreign key constraint failed
$ python3.2 t.py
Traceback (most recent call last):
  File "t.py", line 13, in <module>
    con.execute("insert into track (artist_id) values (1)")
sqlite3.IntegrityError: foreign key constraint failed

huh.  is it already on in your sqlite  install?

$ sqlite3
SQLite version 3.7.7.1 2011-06-28 17:39:05
Enter ".help" for instructions
Enter SQL statements terminated with a ";"
sqlite> pragma foreign_keys;
0
sqlite> 

is what i get (it's off).

Nope.

$ sqlite3
SQLite version 3.7.4
Enter ".help" for instructions
Enter SQL statements terminated with a ";"
sqlite> pragma foreign_keys;
0
sqlite>

$ python
Python 2.7.1+ (r271:86832, Apr 11 2011, 18:13:53) 
[GCC 4.5.2] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import sqlite3
>>> c = sqlite3.connect(':memory:')
>>> list(c.execute('pragma foreign_keys'))
[(0,)]
>>> list(c.execute('pragma foreign_keys = on'))
[]
>>> list(c.execute('pragma foreign_keys'))
[(1,)]

Something strange is going on.

I just built pysql 2.6.3 from source, and now my unit test gives the expected "IntegrityError: foreign key constraint failed" message.

poq, what do you get when you run this script:

import sqlite3
print "sqlite3.version_info =", sqlite3.version_info
print "sqlite3.sqlite_version_info =", sqlite3.sqlite_version_info
from pysqlite2 import dbapi2 as sqlite3
print "pysqlite2.version_info =", sqlite3.version_info
print "pysqlite2.sqlite_version_info =", sqlite3.sqlite_version_info

I get the following:

sqlite3.version_info = (2, 6, 0)
sqlite3.sqlite_version_info = (3, 6, 12)
pysqlite2.version_info = (2, 6, 3)
pysqlite2.sqlite_version_info = (3, 7, 8)

sqlite3.version_info = (2, 6, 0)
sqlite3.sqlite_version_info = (3, 7, 4)
pysqlite2.version_info = (2, 6, 0)
pysqlite2.sqlite_version_info = (3, 7, 4)

BAD : sqlite3.sqlite_version_info = (3, 6, 12)
GOOD: sqlite3.sqlite_version_info = (3, 7, 4)
GOOD: sqlite3.sqlite_version_info = (3, 7, 8)

I guess this is the cause of the different behavior.

sqlite_version_info is the version of the underlying sqlite3 database library.

So, the sqlite3 is loading an older dynamic lib on my OSX.

Mark, you don't say what version of OS X or which Python you are using but, if you are using an Apple-supplied Python on 10.6 or 10.7 or if you are using python.org 64-bit/32-bit installers for 2.7 or 3.2, the Python standard library sqlite3 modules dynamically links to the system-supplied /usr/lib/libsqlite3.dylib.  On 10.6.8 Snow Leopard that is 3.6.12.  On Lion (10.7.2) that has been updated by Apple to 3.7.5.  (BTW, the most recent 32-bit-only python.org installers for 3.2.x and 2.7.x statically link a version of 3.7.4 sqlite3 in order to work on older versions of OS X which have much older or even no system versions of sqlite3.)

Can this issue be closed?