[英]how to Django dumpdata nested models in yaml
所以我有這個很棒的Django應用,它給我滿意的感覺。 但現在出現問題了,我想使用dumpdata
(或執行相同操作的東西)以yaml格式導出具有嵌套的其他模型的模型。
可以說我有兩個模型, Project
和Questions
。 每個Project
都可以擁有自己的一組Questions
。
代碼看起來像這樣:
項目模型:
class Projects(SortableMixin):
"""
Some docstring
"""
slug = models.SlugField(
_('slug'),
help_text=_('Short name to address this projects from templates.'))
# Basic fields
object_id = models.PositiveIntegerField(blank=True, null=True)
name = models.TextField(_('name'), help_text=_('The question.'))
# Some other fields...
question = models.ForeignKey(Question, null=True, blank=True)
問題模型:
class Question(SortableMixin):
"""
Some docstring
"""
slug = models.SlugField(
_('slug'),
help_text=_('Short name to address this question from templates.'))
# Basic fields
object_id = models.PositiveIntegerField(blank=True, null=True)
name = models.TextField(_('name'), help_text=_('The question.'))
input = models.TextField()
Project
模型有自己的應用程序,而Questions
。 結構如下:
- Django
- Apps
- Project
- Questions
每當我要導出數據庫時,我都會執行以下操作:
./manage.py dumpdata --indent 4 --format yaml > dbdump.yaml
盡管這可行,並且以后可以使用LoadData
導入它,但它不是我想要的,但是yaml文件的輸出看起來很糟糕。 我想要一個漂亮的嵌套模型外觀yaml文件,以供查看,位於the腳的外觀文件下方:
項目部分:
- model: project.projects
pk: 1
fields: {slug: "slugproject1", object_id: 10, name: "some project 1", question: ["slugquestion1"]}
- model: project.projects
pk: 2
fields: {slug: "slugproject2", object_id: 11, name: "some project 2", question: ["slugquestion2"]}
- model: project.projects
pk: 3
fields: {slug: "slugproject3", object_id: 12, name: "some project 3", question: ["slugquestion3"]}
問題部分:
- model: question.question
pk: 1
fields: {slug: "slugquestion1", object_id: 100, name: "some question 1", input: "q1"}
- model: question.question
pk: 1
fields: {slug: "slugquestion2", object_id: 200, name: "some question 2", input: "q2"}
- model: question.question
pk: 1
fields: {slug: "slugquestion3", object_id: 300, name: "some question 3", input: "q3"}
我真正想要的是像這樣導出yaml文件:
- model: project.projects
pk: 1
fields: {
slug: "slugproject1",
object_id: 10,
name: "some project 1",
questions: {
model: question.question
pk: 1
fields: {
slug: "slugquestion1"
object_id: 100
name: "some question 1"
input: "q1"
}
}
}
- model: project.projects
pk: 2
fields: {
slug: "slugproject2",
object_id: 11,
name: "some project 2",
questions: {
model: question.question
pk: 2
fields: {
slug: "slugquestion2"
object_id: 200
name: "some question 2"
input: "q2"
}
}
}
- model: project.projects
pk: 3
fields: {
slug: "slugproject3",
object_id: 13,
name: "some project 3",
questions: {
model: question.question
pk: 3
fields: {
slug: "slugquestion3"
object_id: 300
name: "some question 3"
input: "q3"
}
}
}
為此,我在項目內部實現了自定義序列化程序:
- Django
- Apps
- Project
- Management
- Commands
- test.py
- Questions
代碼如下:
from django.core.management.base import BaseCommand, CommandError
from apps.project import Projects
from apps.questions import Question
from rest_framework import serializers
import yaml
class QuestionSerialier(serializers.ModelSerializer):
class Meta:
model = Question
fields = ('pk', 'slug', 'object_id', 'name', 'input')
class ProjectsSerializer(serializers.ModelSerializer):
questions = QuestionSerialier(many=True, read_only=True)
class Meta:
model = Projects
fields = ('pk', 'slug','object_id', 'name', 'questions')
class Command(BaseCommand):
help = ''
def add_arguments(self, parser):
pass
def handle(self, *args, **options):
with open('result.yaml', 'w') as yaml_file:
for i in Projects.objects.filter():
yaml.dump(ProjectsSerializer(i).data,
yaml_file,
default_flow_style=False,
allow_unicode=False,
encoding=None)
我可以通過運行以下代碼來運行代碼:
./manage.py test
只有這樣才能導出我的模型,如下所示:
- project: 1
pk: 1
slug: "slugproject1"
object_id: 10
name: "some project 1"
questions:
- !!python/object/apply:collections.OrderedDict
- - - pk
- - slug
- - object_id
- - name
- - input
- project: 2
pk: 2
slug: "slugproject2"
object_id: 11
name: "some project 2"
questions:
- !!python/object/apply:collections.OrderedDict
- - - pk
- - slug
- - object_id
- - name
- - input
- project: 3
pk: 3
slug: "slugproject3"
object_id: 12
name: "some project 3"
questions:
- !!python/object/apply:collections.OrderedDict
- - - pk
- - slug
- - object_id
- - name
- - input
如您所見,以上內容不適用於導入甚至可讀的導出...
你們能為我指出如何在Django中實現嵌套模型dumpdata yaml導出的正確方向嗎?
謝謝!
所以我想出了如何處理出口/進口。
要導出為適當的yaml文件格式,我執行了以下操作:
...
class Command(BaseCommand):
def handle(self, *args, **options):
with open('result.yaml', 'w') as yaml_file:
model = serializer.Meta.model
json_object = []
json_data = json.dumps(serializer(model_data).data)
json_object.append(json.loads(json_data))
yaml.dump(
json_object,
yaml_file,
default_flow_style=False,
allow_unicode=False,
encoding=None
)
這將導出到正確的yaml文件。
然后,要導入yaml文件,請執行以下操作:
...
class Command(BaseCommand):
def handle(self, *args, **options):
with open('result.yaml', 'r') as yaml_file:
yaml_list = yaml.load(yaml_file.read())
for data in yaml_list:
...process file
就是這樣!
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