So I have this awesome Django app which gives me satisfaction and such.
But now the problem, I want to use dumpdata (or something that does the same) to export a model with a nested other model in yaml format.
Lets say I have two models, Project and Questions. Each Project can have it's owns set of Questions.
The code looks something like this:
Project Model:
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)
Questions Model:
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()
The Project model has his own app and so has the Questions.
The structure looks like this:
- Django
- Apps
- Project
- Questions
Whenever I want to export my database I'll do the following:
./manage.py dumpdata --indent 4 --format yaml > dbdump.yaml
Although this works, and I can later import it with LoadData, its not what I want, the output of the yaml file looks crappy. I want to have a nice nested model looking yaml file for review purposes, below the crappy looking file:
Projects section:
- 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"]}
Questions section:
- 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"}
What I really want is to export the yaml file like so:
- 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"
}
}
}
To achieve this I implemented a custom serialiser inside projects:
- Django
- Apps
- Project
- Management
- Commands
- test.py
- Questions
The code looks like:
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)
I can run the code by running:
./manage.py test
Only this exports my models like so:
- 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
As you can see the above is not usable for import or even readable export...
Can you guys point me in the right direction on how to achieve a nested models dumpdata yaml export in django?
Thanks!
So I figured out how to handle the export/import.
To export to a proper yaml file format I did the following:
...
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
)
This export to a proper yaml file.
Then to import the yaml file I did the following:
...
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
And that's it!