如何将包含 ForeignKey 模型关系的 Django 模型导出到组合文本文件中

Question

我有一个 Django 模型，它有几个 ForeignKey 模型关系，我需要将它们组合成单个记录。

如果您查看下面的 plaque 模型，您会注意到我与 veteran 模型有 ForeignKey 关系。我创建了一个老兵模型并没有真正考虑清楚，现在需要将老兵字段添加到我的斑块模型中，但需要先导出所有数据，重建我的斑块模型，然后重新导入数据。

我已经安装了django-import-export，它适用于单个模型。我导出了我的牌匾数据，并在退伍军人字段中添加了相关退伍军人的身份证号。如果可能的话，我想做的实际上是导出我的斑块模型，而不是相关的退伍军人 ID 号，而是来自退伍军人模型的实际数据（名字、姓氏等）。如果有帮助，我也在使用 PostgreSQL。有任何想法吗？谢谢你。

class Plaque(models.Model):
     created_by = models.ForeignKey(settings.AUTH_USER_MODEL, on_delete=models.CASCADE, related_name='plaques', default=1)
     group = models.ForeignKey(Group, blank=True, null=True, verbose_name='Group Name')
     veteran = models.ForeignKey(Veteran, blank=True, null=True)
     ...

更新 - 这是我的整个资源模型解决方案。这将导出一个包含我所有 ForeignKey 模型字段的字符串值的文件。

from import_export import resources, fields
from import_export.widgets import ForeignKeyWidget
from import_export.admin import ImportExportModelAdmin
from .models import Plaque, Group, Veteran, Service, Branch, War, Rank
from django.contrib import admin


class PlaqueResource(resources.ModelResource):
    vet_first = fields.Field(column_name='Veteran First', attribute='veteran', widget=ForeignKeyWidget(Veteran, field='first_name'))
    vet_middle = fields.Field(column_name='Veteran Middle', attribute='veteran', widget=ForeignKeyWidget(Veteran, field='middle_name'))
    vet_last = fields.Field(column_name='Veteran Last', attribute='veteran', widget=ForeignKeyWidget(Veteran, field='last_name'))
    vet_display = fields.Field(column_name='Veteran Display Name', attribute='veteran', widget=ForeignKeyWidget(Veteran, field='display_name'))
    vet_nick = fields.Field(column_name='Veteran Nickname', attribute='veteran', widget=ForeignKeyWidget(Veteran, field='nickname'))
    vet_bio = fields.Field(column_name='Veteran Biography', attribute='veteran', widget=ForeignKeyWidget(Veteran, field='biography'))
    vet_branch = fields.Field(column_name='Veteran Branch - service model', attribute='service', widget=ForeignKeyWidget(Service, field='branch'))
    vet_rank = fields.Field(column_name='Veteran Rank - service model', attribute='service', widget=ForeignKeyWidget(Service, field='rank'))
    vet_war = fields.Field(column_name='Veteran War - service model', attribute='service', widget=ForeignKeyWidget(Service, field='war'))
    vet_branch2 = fields.Field(column_name='Veteran Branch - branch model', attribute='service', widget=ForeignKeyWidget(Branch, field='branch'))
    vet_rank2 = fields.Field(column_name='Veteran Rank - rank model', attribute='service', widget=ForeignKeyWidget(Rank, field='rank'))
    vet_war2 = fields.Field(column_name='Veteran War - war model', attribute='service', widget=ForeignKeyWidget(War, field='war'))
    group_name = fields.Field(column_name='Group Name', attribute='group', widget=ForeignKeyWidget(Group, field='group_name'))
    group_type = fields.Field(column_name='Group Type', attribute='group', widget=ForeignKeyWidget(Group, field='group_type'))

    class Meta:
        model = Plaque

Answer 1

这应该可行，(from the docs)

# Or however your resources look
class PlaqueResource(resources.ModelResource):

    class Meta:
        model = Plaque
        fields = (<other_fields>, 'veteran__first_name', 'veteran__last_name',)