Thomas Steinmaurer wrote:
> Hello,
>
> I have the following model:
>
> class ModelVersion(models.Model):
> model_version_id = models.AutoField(primary_key=True)
> major_version = models.IntegerField(null=True, blank=True)
> minor_version = models.IntegerField(null=True, blank=True)
> class Meta:
> db_table = u'model_version'
> def __unicode__(self):
> return str(self.major_version) + '.' + str(self.minor_version)
>
> class DataPackage(models.Model):
> data_package_id = models.AutoField(primary_key=True)
> physical_name = models.CharField(max_length=255)
> model_version = models.ForeignKey(ModelVersion)
> class Meta:
> db_table = u'data_package'
> def __unicode__(self):
> return self.physical_name
>
> class RelationalDatabase(DataPackage):
> data_package = models.OneToOneField(DataPackage, parent_link=True)
> database_instance_name = models.CharField(max_length=255)
> class Meta:
> db_table = u'relational_database'
> def __unicode__(self):
> return self.physical_name + '(' + self.database_instance_name + ')'
>
>
> So, RelationalDatabase is derived from DataPackage implemented with a
> OneToOneField, thus also has a model_version property.
>
>
> Now, in admin.py, when I have the following:
>
> class RelationalDatabaseAdmin(admin.ModelAdmin):
> list_display =
> ('data_package_id','physical_name','database_instance_name','model_version')
> list_filter = ['model_version']
> search_fields = ['database_instance_name']
>
> admin.site.register(RelationalDatabase, RelationalDatabaseAdmin);
>
>
> then model_version gets displayed in the list, but although list_filter
> is present, a filter box isn't available in the admin area.
>
>
> Is this supposed to work?
Ah, sorry, my fault. The mechanism is clever enough to display the
filter box only when there are at least two distinct values.
So, everything is fine.
Thomas
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