Question

我希望使用相似的语言对象生成多个国家/地区对象，而无需通过管理界面。

models.py

from django.db import models

class World(models.Model):
    country = models.CharField(max_length=200)
    Language = models.CharField(max_length=200)

class add_countries(models.Model):
    sub_part1 = ['USA','ENGLAND','AUSTRALIA','CANADA']
    for sub in sub_part1:
        sub = World.country
        World.Language = "English"

add_countries()

Answer 1

addCard = async ({
    number, expMonth, expYear, cvc,
  }) => {
    this.setState({ addingCardInProcess: true });
    try {
      const tokenObject = await stripe.createTokenWithCard({
        number, expMonth, expYear, cvc
      });
      firebase
        .database()
        .ref(`/stripe_customers/${uid()}/sources`)
        .push({ token: tokenObject.tokenId })
        .then(() => {
          this.setState({ addingCardInProcess: false });
          this.cardAlert(true);
        })
        .catch((err) => {
          this.setState({ addingCardInProcess: false });
          this.cardAlert(false, err.message);
        });
    } catch(err) {
      this.cardAlert(false, err.message);
      this.setState({ addingCardInProcess: false })
    }
  };

您可以将上面的代码放在方法中并调用它。

Answer 2

如果您希望有一些始终存在的数据库条目，您应该考虑使用data migrations。这将导致在迁移数据库时创建这些数据库行。基本的想法是这样的：

from django.db import migrations

def add_languages(apps, schema_editor):
    # We can't import the Person model directly as it may be a newer
    # version than this migration expects. We use the historical version.
    World = apps.get_model('yourappname', 'World')
    languages = {'English': ['USA','ENGLAND','AUSTRALIA','CANADA']}
    for language, countries in languages.items():
        for country in countries:
            World.objects.get_or_create(country=country, Language=language)

class Migration(migrations.Migration):

    dependencies = [
        ('yourappname', '0001_initial'),
    ]

    operations = [
        migrations.RunPython(add_languages),
    ]

迁移文件将位于您应用的迁移目录中（在本例中类似于project_root/yourappname/migrations/0002_insert_languages.py）。有关详细信息，请参阅文档。

从另一个类添加多个类对象

2 个答案: