Unable to make required/mandatory "custom" filter field

[jerinpetergeorge]

How can we make a "custom filter method/field" as a required/mandatory field?

Describe the Bug

Consider that I have this filter class

@strawberry_django.filter(PayoutExecution, lookups=True)
class AnyFilter:
    status: BaseFilterLookup[PayoutExecutionStatus]  # Mandatory

    @strawberry_django.filter_field()
    def custom_field(
        self,
        info: Info,
        value: BaseFilterLookup[PayoutExecutionStatus],
        prefix: str,
        queryset: QuerySet,
    ):
        return Q(field_1=True)

For this AnyFilter type, the generated schema looks like this

# schema.graphql
input AnyFilter {
  status: PayoutExecutionStatusBaseFilterLookup!
  AND: AnyFilter
  OR: AnyFilter
  NOT: AnyFilter
  DISTINCT: Boolean
  customField: PayoutExecutionStatusBaseFilterLookup
}

As we can see, the customField is not required/mandatory - how can we fix that?

System Information

• Operating system: Linux
• Python version: Python 3.11.5
• Strawberry version (if applicable): strawberry-graphql-django==0.52.1

[bellini666]

Hi @jerinpetergeorge ,

Hrm, I don't think it is currently possible... maybe we can add a required: bool = False in the filter_field to allow that situation?

[jerinpetergeorge]

Yeah, thats something I was also thinking. But, what about the filter(...) method? its supposed to be True always?, Maybe? @bellini666

[bellini666]

In theory, the filter method should respect what is defined. If someone is overriding it, they also need to take that in consideration