fix: show original filename in CRUD detail view for MetadataFileField

get_field_display() used value.name.split('/')[-1] which gives the raw
storage path basename (UUID for new uploads).  Use str(value) instead,
which calls MetadataFieldFile.__str__ and returns original_filename from
the metadata row.

Co-Authored-By: Claude Sonnet 4.6 <noreply@anthropic.com>

sapl/crispy_layout_mixin.py

@@ -143,7 +143,7 @@ def get_field_display(obj, fieldname):
         if value:
             display = '<a href="{}">{}</a>'.format(
                 value.url,
-                value.name.split('/')[-1:][0])
+                str(value) or value.name.split('/')[-1:][0])
         else:
             display = ''
     elif 'ManyRelatedManager' in str_type_from_value\