What is causing this OverflowError in Django?

Posted by orokusaki on Stack Overflow See other posts from Stack Overflow or by orokusaki
Published on 2010-03-20T05:39:38Z Indexed on 2010/03/20 5:41 UTC
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I'm using a normal ModelForm.save() to create an object, and this exception comes up. It worked fine before until I added commit_manually, transaction.rollback() and transaction.commit() to my view. Has anyone else ran into this? Is this because of sqlite3?

OverflowError: long too big to convert

C:\Python26\Lib\site-packages\django-trunk\django\db\backends\sqlite3\base.py in execute, line 197

params: (203866156270872165269663274649746494334L,)
query: u'SELECT (1) AS "a", "auth_user"."id", "auth_user"."username", "auth_user"."first_name", "auth_user"."last_name", "auth_user"."email", "auth_user"."password", "auth_user"."is_staff", "auth_user"."is_active", "auth_user"."is_superuser", "auth_user"."last_login", "auth_user"."date_joined" FROM "auth_user" WHERE "auth_user"."id" = ? LIMIT 1'
self
<django.db.backends.sqlite3.base.SQLiteCursorWrapper object at 0x015D5A98>

Why would that L param be passed in, and

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    I'm using a normal ModelForm.save() to create an object, and this exception comes up. It worked fine before until I added commit_manually, transaction.rollback() and transaction.commit() to my view. Has anyone else ran into this? Is this because of sqlite3? OverflowError: long too big to convert C:\Python26\Lib\site-packages\django-trunk\django\db\backends\sqlite3\base… >>> More