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SilviaAmAmsvenvandescheur
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🐛 [#303] Handle case where zaaktype identificatie is an empty string
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  • backend/src/openarchiefbeheer/zaken

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backend/src/openarchiefbeheer/zaken/utils.py

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@@ -4,6 +4,7 @@
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from typing import TYPE_CHECKING, Callable, Generator, Literal
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from django.conf import settings
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from django.utils.translation import gettext as _
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from ape_pie import APIClient
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from djangorestframework_camel_case.parser import CamelCaseJSONParser
@@ -108,7 +109,7 @@ def retrieve_zaaktypen_choices() -> list[DropDownChoice]:
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.distinct()
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)
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zaaktypen_choices = [
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{"label": key, "value": ",".join(value)}
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{"label": key or _("(no identificatie)"), "value": ",".join(value)}
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for key, value in zaaktypen.items()
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if key in zaaktypes_to_include
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]

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