| msg144166 - (view) |
Author: Mark Bucciarelli (Mark.Bucciarelli) |
Date: 2011-09-16 22:54 |
| 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. |
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| msg144220 - (view) |
Author: (poq) |
Date: 2011-09-17 21:26 |
| Works for me? $ python2.7 t.py Traceback (most recent call last): File "t.py", line 13, in 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 con.execute("insert into track (artist_id) values (1)") sqlite3.IntegrityError: foreign key constraint failed |
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| msg144222 - (view) |
Author: Mark Bucciarelli (Mark.Bucciarelli) |
Date: 2011-09-17 23:17 |
| 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). |
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| msg144225 - (view) |
Author: (poq) |
Date: 2011-09-18 02:57 |
| 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,)] |
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| msg145604 - (view) |
Author: Mark Bucciarelli (Mark.Bucciarelli) |
Date: 2011-10-15 18:52 |
| 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) |
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| msg145624 - (view) |
Author: (poq) |
Date: 2011-10-16 15:57 |
| 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) |
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| msg145625 - (view) |
Author: Mark Bucciarelli (Mark.Bucciarelli) |
Date: 2011-10-16 16:30 |
| 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. |
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| msg145760 - (view) |
Author: Ned Deily (ned.deily) *  |
Date: 2011-10-17 21:31 |
| 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? |
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