manage.py syncdb: Error with the PostgreSQL version.

[Manuel Lanctot <sensoz@…>]

When I tried to initialize the database for the first time, I got the following error.

C:\django\blog>python manage.py syncdb
Traceback (most recent call last):
  File "manage.py", line 11, in <module>
    execute_manager(settings)
  File "C:\Python25\Lib\site-packages\django\core\management.py", line 1672, in execute_manager
    execute_from_command_line(action_mapping, argv)
  File "C:\Python25\Lib\site-packages\django\core\management.py", line 1571, in execute_from_command_line
    action_mapping[action](int(options.verbosity), options.interactive)
  File "C:\Python25\Lib\site-packages\django\core\management.py", line 504, in syncdb
    cursor = connection.cursor()
  File "C:\Python25\Lib\site-packages\django\db\backends\postgresql_psycopg2\base.py", line 57, in cursor
    postgres_version = [int(val) for val in cursor.fetchone()[0].split()[1].split('.')]
ValueError: invalid literal for int() with base 10: '1,'

Doing a SELECT version() returns: ('PostgreSQL 8.3.1, compiled by Visual C++ build 1400',)
So there is a trailing comma after the version number.

[Manuel Lanctot <sensoz@…>]

My solution was as followed.

In django\db\backends\postgresql_psycopg2\base.py, line 57:

Replace

postgres_version = [int(val) for val in cursor.fetchone()[0].split()[1].split('.')]

With

postgres_version = [int(val.strip(',')) for val in cursor.fetchone()[0].split()[1].split('.')]

[Russell Keith-Magee]

Essentially a dupe of #6433.