Question

我需要改变这个

data = models.CharField(max_length=500, null=True)

到这个

data = JSONField(null=True, blank=True, default={})

据我所知，我必须编写自定义迁移。最近的信息我找到了here，但我完全不知道如何处理RunPython，如果这是正确的事情。

Answer 1

您可以在接下来的步骤中执行此操作： 1.添加类型为JSONField的新字段并运行'makemigrations'：

data = models.CharField(max_length=500, null=True)
data_json = JSONField(null=True, blank=True, default={})

创建数据迁移（使用RunPython）

import json

from django.db import migrations

def forwards_func(apps, schema_editor):
    MyModel = apps.get_model("myapp", "MyModel")

    for obj in MyModel.objects.all():
        try:
            obj.data_json = json.loads(obj.data)
            obj.save()
        except json.decoder.JSONDecodeError as e:
            print('Cannot convert {} object'.format(obj.pk))


class Migration(migrations.Migration):

    dependencies = []  # WRITE YOUR LAST MIGRATION HERE

    operations = [
        migrations.RunPython(forwards_func, lambda apps, schema_editor: pass),
    ]

删除旧数据字段并运行'makemigrations'：

data_json = JSONField（null = True，blank = True，默认= {}）
重命名json字段并运行'makemigrations'：

data = JSONField（null = True，blank = True，默认= {}）

P.S。您可以对所有这些步骤使用一次迁移。我描述了更好理解的所有步骤。

Answer 2

假设您有这样的模型

class SampleModel(models.Model):
    name = models.CharField(max_length=120)
    age = models.IntegerField()
    address = models.CharField(max_length=100)

您需要将address更改为JSONField
因此，您必须定义temporary field，如下所示

from django.db import models
from jsonfield import JSONField


class SampleModel(models.Model):
    name = models.CharField(max_length=120)
    age = models.IntegerField()
    address = models.CharField(max_length=100)
    temp_address = JSONField(default={})

然后运行命令，
python manage.py makemigrations app_name和python manage.py migrate app_name

然后是DATA MIGRATION的角色。你需要做的是，运行命令python manage.py makemigrations app_name --empty
这将在您的目录中创建一个empty migration file。将该文件更改为此类的内容 0003_auto_20180320_1525.py （迁移文件）

# -*- coding: utf-8 -*-
# Generated by Django 1.11 on 2018-03-20 15:25
from __future__ import unicode_literals
import json
from django.db import migrations
from sample.models import SampleModel


class Migration(migrations.Migration):
    def forward_data_migration(apps, schema):
        for sample in SampleModel.objects.all():
            try:
                sample.temp_address = json.loads(sample.address)
                sample.save()
            except json.decoder.JSONDecodeError as e:
                print('Cannot convert {} object'.format(sample.pk))

    def revert_migration(apps, schema):
        pass

    dependencies = [
        ('sample', '0002_samplemodel_temp_address'),
    ]

    operations = [
        migrations.RunPython(
            code=forward_data_migration,
            reverse_code=revert_migration
        )
    ]

迁移此更改，然后像以前一样创建另一个空迁移文件，并编写RenameField脚本，如下所示的 0004_auto_20180320_1530.py

# -*- coding: utf-8 -*- # Generated by Django 1.11 on 2018-03-20 15:30 from __future__ import unicode_literals from django.db import migrations, models class Migration(migrations.Migration): dependencies = [ ('sample', '0003_auto_20180320_1525'), ] operations = [ migrations.RenameField( model_name='samplemodel', old_name='address', new_name='temp_address' ), migrations.RenameField( model_name='samplemodel', old_name='temp_address', new_name='address' ) ]

也移植这个。就是这样。

Django postgres从CharField到JSONField进行自定义迁移

2 个答案: