Mario,
It's not the best to re-open a 2.5 year old thread :) Better to start a new
one, link to the old one, and state what you think is different.
I believe your use case can be fit with a custom ruserver management
command to override the original as Shai suggested:
> from (...)runserver import Command as BaseCommand
>
> class Command(BaseCommand):
> def handle(self, *args, **options):
> options.setdefault('addrport', '0.0.0.0:8000')
> super(Command, self).handle(*args, **options)
You can change the options.setdefault line to set the port in addrport from
a new setting you invent and put in your config file, such as
MYPROJECTNAME_RUNSERVER_PORT = 8001
On Thu, 18 Jul 2019 at 22:21, Mario Frasca <[email protected]> wrote:
> was this rejected?
> my use case is: I have a single django program, which I run in multiple
> instances, each on a different port and connecting to a different
> database. for each instance, I have a separate `config_INSTANCE.py` file.
> the database settings fit in the `config` file, but I have to put the port
> in a batch file, or something else, outside the config file. it feels as
> if I were doing something which isn't allowed...
> I don't need to run the program on a single different port, I need
> different config files each stating a different port.
>
> On Tuesday, 17 January 2017 02:21:46 UTC-5, Shai Berger wrote:
>>
>> I am -1 on adding a setting to handle a use-case that can be handled by
>> users
>> with a 6-line file:
>>
>> On Monday 28 November 2016 16:05:39 Shai Berger wrote:
>> >
>> > It seems all you need in the overridden runserver is:
>> >
>> > from (...)runserver import Command as BaseCommand
>> >
>> > class Command(BaseCommand):
>> > def handle(self, *args, **options):
>> > options.setdefault('addrport', '0.0.0.0:8000')
>> > super(Command, self).handle(*args, **options)
>> >
>> > What am I missing?
>>
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Adam
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