As we all know, displaying a method return value as boolean in the Django admin is easily done by setting the boolean
attribute:
众所周知,通过设置boolean属性,可以轻松地在Django管理员中显示方法返回值为boolean:
class MyModel(models.Model):
def is_something(self):
if self.something == 'something':
return True
return False
is_something.boolean = True
How can you achieve the same effect for a property, like in the following case?
如何在财产上实现同样的效果,例如以下情况?
class MyModel(models.Model):
@property
def is_something(self):
if self.something == 'something':
return True
return False
4 个解决方案
#1
18
Waiting for better solutions to come up, I've solved it in the following way:
等待更好的解决方案出现,我已经通过以下方式解决了这个问题:
class MyModel(models.Model):
def _is_something(self):
if self.something == 'something':
return True
return False
_is_something.boolean = True
is_something = property(_is_something)
I'll then reference the _is_something
method in the ModelAdmin
subclass:
然后,我将引用ModelAdmin子类中的_is_something方法:
class MyModelAdmin(admin.ModelAdmin):
list_display = ['_is_something']
And the is_something
property otherwise:
而is_something属性否则:
if my_model_instance.is_something:
print("I'm something")
#2
15
this is the simplest way I found, directly in the ModelAdmin:
这是我发现的最简单的方法,直接在ModelAdmin中:
class MyModelAdmin(admin.ModelAdmin):
def is_something(self, instance):
return instance.something == "something"
is_something.boolean = True
is_something.short_description = u"Is something"
list_display = ['is_something']
#3
2
You need to create a shadowing
function to the property in the model. What I mean is that you will need to recreate a function in the ModelAdmin class with the same name as the property defined in the main Model.
您需要为模型中的属性创建阴影函数。我的意思是你需要在ModelAdmin类中重新创建一个与主模型中定义的属性相同的函数。
Example:
例:
# Model
class Product(models.Model):
@property
def in_stock(self):
# boolean check return
return self.quantity > 0
...
...
# Django-modeladmin
class ProductAdmin(admin.ModelAdmin):
list_display = ('in_stock', ...)
def in_stock(self, instance):
return instance.in_stock
in_stock.boolean = True
#4
-1
If you define is_something
as a property, it will be an immutable object, instead of a function, but that object contains a reference to the decorated getter in the fget
attribute. I think that the Django admin interface use the getter of that property, thus this may works
如果将is_something定义为属性,它将是一个不可变对象,而不是一个函数,但该对象包含对fget属性中的装饰getter的引用。我认为Django管理界面使用该属性的getter,因此这可能有效
class MyModel(models.Model):
@property
def is_something(self):
if self.something == 'something':
return True
return False
is_something.fget.boolean = True
#1
18
Waiting for better solutions to come up, I've solved it in the following way:
等待更好的解决方案出现,我已经通过以下方式解决了这个问题:
class MyModel(models.Model):
def _is_something(self):
if self.something == 'something':
return True
return False
_is_something.boolean = True
is_something = property(_is_something)
I'll then reference the _is_something
method in the ModelAdmin
subclass:
然后,我将引用ModelAdmin子类中的_is_something方法:
class MyModelAdmin(admin.ModelAdmin):
list_display = ['_is_something']
And the is_something
property otherwise:
而is_something属性否则:
if my_model_instance.is_something:
print("I'm something")
#2
15
this is the simplest way I found, directly in the ModelAdmin:
这是我发现的最简单的方法,直接在ModelAdmin中:
class MyModelAdmin(admin.ModelAdmin):
def is_something(self, instance):
return instance.something == "something"
is_something.boolean = True
is_something.short_description = u"Is something"
list_display = ['is_something']
#3
2
You need to create a shadowing
function to the property in the model. What I mean is that you will need to recreate a function in the ModelAdmin class with the same name as the property defined in the main Model.
您需要为模型中的属性创建阴影函数。我的意思是你需要在ModelAdmin类中重新创建一个与主模型中定义的属性相同的函数。
Example:
例:
# Model
class Product(models.Model):
@property
def in_stock(self):
# boolean check return
return self.quantity > 0
...
...
# Django-modeladmin
class ProductAdmin(admin.ModelAdmin):
list_display = ('in_stock', ...)
def in_stock(self, instance):
return instance.in_stock
in_stock.boolean = True
#4
-1
If you define is_something
as a property, it will be an immutable object, instead of a function, but that object contains a reference to the decorated getter in the fget
attribute. I think that the Django admin interface use the getter of that property, thus this may works
如果将is_something定义为属性,它将是一个不可变对象,而不是一个函数,但该对象包含对fget属性中的装饰getter的引用。我认为Django管理界面使用该属性的getter,因此这可能有效
class MyModel(models.Model):
@property
def is_something(self):
if self.something == 'something':
return True
return False
is_something.fget.boolean = True