Django 更改下拉菜单

Question

下面开始代码，专门针对models和views -

class userchoice(models.Model):
choice = []
choice2 = list(userdata.objects.all().values_list('key'))
for x in choice2:
    for t in x:
        choice.append((t,t.capitalize()))

Job_ID = models.CharField(primary_key = True, max_length=200, choices=choice, default='key')
Group_ID= models.CharField(max_length=200, choices=choice, default='key')
Customer_name = models.CharField(max_length=200, choices=choice, default='key')
Planned_Duration = models.CharField(max_length=200, choices=choice, default='key')
Worker_name = models.CharField(max_length=200, choices=choice, default='key')
Job_note = models.CharField(max_length=200, choices=choice, default='key')
Customer_tp_name = models.CharField(max_length=200, choices=choice, default='key')

从 django.db 导入迁移、模型

class 迁移(migrations.Migration):

dependencies = [
    ('csvfileapp', '0008_userdata'),
]

operations = [
    migrations.CreateModel(
        name='userchoice',
        fields=[
            ('id', models.AutoField(auto_created=True, primary_key=True, serialize=False, verbose_name='ID')),
            ('Job_ID', models.CharField(default='key', max_length=200)),
            ('Group_ID', models.CharField(default='key', max_length=200)),
            ('Customer_name', models.CharField(default='key', max_length=200)),
            ('Planned_Duration', models.CharField(default='key', max_length=200)),
            ('Worker_name', models.CharField(default='key', max_length=200)),
            ('Job_note', models.CharField(default='key', max_length=200)),
            ('Customer_tp_name', models.CharField(default='key', max_length=200)),
        ],
    ),
]

Answer 1

到目前为止，这根本不是一个好的解决方案。因为选择可以随时更改（因为它正在获取数据库结果）所以任何更改都会导致新的迁移文件不好。按如下操作：

去掉那个部分，让它变得简单charField
在您的 forms.py 中启动您想要选择的字段与您想要的数据库查询结果。

总而言之，在视图和表单中处理这种情况，而不是数据库 orm 和迁移！

更新：

这是要使用的 forms.py 的代码示例：

class waypointForm(forms.Form):
    def __init__(self, user, *args, **kwargs):
        super(waypointForm, self).__init__(*args, **kwargs)
        self.fields['waypoints'] = forms.ChoiceField(
            choices=[(o.id, str(o)) for o in     Waypoint.objects.filter(user=user)]

这是stack link for it

Django 更改下拉菜单

Django changes dropdown menu

python

django

django-models

django-migrations

更新：