Question

在我的models.py中，我有以下课程：

class AvailabilityTypes():
    STUDYING = 'STUDYING'
    WORKING = 'WORKING'
    SEARCHING = 'SEARCHING'
    FREELANCER = 'FREELANCER'

    types = (
        (STUDYING, 'Estudando'),
        (WORKING, 'Trabalhando'),
        (SEARCHING, 'Procurando por emprego'),
        (FREELANCER, 'Freelancer')
    )

    def get_types(self):
        return self.types.all()

我想以Django表单显示该选项。在我的forms.py文件中，我有以下代码段：

from django import forms
from .models import AvailabilityTypes

[...]

availability = forms.CharField(
        widget=forms.ChoiceField(
            choices=(AvailabilityTypes.types)
        )
    )

但是出现错误TypeError: 'DeferredAttribute' object is not iterable。我究竟做错了什么？另外，如果我尝试使用：

availability = forms.CharField(
        widget=forms.ChoiceField(
            choices=(AvailabilityTypes.get_types())
        )
    )

我收到错误TypeError: get_types() missing 1 required positional argument: 'self'.

我是Django和Python的新手，我可以使用一些光源。谢谢。

Answer 1

Sol 1.修正当前代码

首先修复您的方法get_types()：

class AvailabilityTypes():
...
# Above this Same as your code

def get_types(self):
    return self.types  # You need to return just types, it is a tuple, it doesn't has an attribute all(). In Django we usually use all() in querysets.

现在修复表单：

from django import forms
from .models import AvailabilityTypes

at_obj = AvailabilityTypes()  # Create an object of class AvailabilityTypes

[...]  # Guessing form class starts here

availability = forms.CharField(
    widget=forms.ChoiceField(
        # choices=AvailabilityTypes.get_types()  # You can't call a class's method on that class, you call it on that class's object
        choices=(at_obj.get_types())  # Call the method on object of the class not class itself
    )
)

Sol2。不要创建不必要的课程

在models.py中，无需创建用于容纳类型的类。您可以这样做：

...
# All your imports go above this
# These four variables are pointless in my opinion, but I will leave them be
STUDYING = 'STUDYING'
WORKING = 'WORKING'
SEARCHING = 'SEARCHING'
FREELANCER = 'FREELANCER'

# types is an constant so it should follow uppercase naming style
TYPES = (  
    (STUDYING, 'Estudando'),
    (WORKING, 'Trabalhando'),
    (SEARCHING, 'Procurando por emprego'),
    (FREELANCER, 'Freelancer')
)

# Create your models Here
...

现在在您的forms.py中：

...
from .models import TYPES  # Import the tuple from models

[...]

availability = forms.CharField(
    widget=forms.ChoiceField(choices=TYPES)  # Use the imported tuple directly here
)

Sol 3.使用模型形式（最简单）

在您的模型中：

from django.db import models

# types is an constant so it should follow uppercase naming style
TYPES = (  
    ('STUDYING', 'Estudando', ),
    ('WORKING', 'Trabalhando', ),
    ('SEARCHING', 'Procurando por emprego', ),
    ('FREELANCER', 'Freelancer', ),  # Unlike other languages your last element can have a trailing comma too, its optional but still do that. I has some advantages which I am not gonna explain here
)

# Create your models Here
class MyModel(models.Model):
    availability = models.CharField(max_length=63, choices=TYPES)
    # Other fields of your model
    ...

现在在您的forms.py中：

from django import forms
from .models import MyModel

class MyModelForm(forms.ModelForm):
    class Meta:
        model = MyModel
        fields = ('availability', )  # add other fields of your models too in the tuple

就是这样。完成后，在视图中使用表单。 Django将负责显示正确的选择，进行验证，显示相关的错误消息并将有效数据保存在数据库中。

TypeError：“ DeferredAttribute”对象不可迭代

1 个答案: