我有一个简单的Employee模型，其中包括firstname， lastname和middlename领域。

在管理方面，可能在其他地方，我想显示为：

lastname, firstname middlename

对我而言，执行此操作的逻辑位置是在模型中通过创建一个这样的计算字段：

from django.db import models
from django.contrib import admin

class Employee(models.Model):
    lastname = models.CharField("Last", max_length=64)
    firstname = models.CharField("First", max_length=64)
    middlename = models.CharField("Middle", max_length=64)
    clocknumber = models.CharField(max_length=16)
    name = ''.join(
        [lastname.value_to_string(),
        ',',
         firstname.value_to_string(),
        ' ',
         middlename.value_to_string()])

    class Meta:
        ordering = ['lastname','firstname', 'middlename']

class EmployeeAdmin(admin.ModelAdmin):
    list_display = ('clocknumber','name')
    fieldsets = [("Name", {"fields":(("lastname", "firstname", "middlename"), "clocknumber")}),
        ]

admin.site.register(Employee, EmployeeAdmin)

最终，我认为我需要以字符串形式获取名称字段的值。我得到的错误是value_to_string() takes exactly 2 arguments (1 given)。字符串所需的值self, obj。我不知道这obj是什么意思。

一定有一种简单的方法可以做到这一点，我敢肯定我不是第一个想要这样做的人。

编辑：下面是我的代码修改为丹尼尔的答案。我得到的错误是：

django.core.exceptions.ImproperlyConfigured:
    EmployeeAdmin.list_display[1], 'name' is not a callable or an
    attribute of 'EmployeeAdmin' of found in the model 'Employee'.

from django.db import models
from django.contrib import admin

class Employee(models.Model):
    lastname = models.CharField("Last", max_length=64)
    firstname = models.CharField("First", max_length=64)
    middlename = models.CharField("Middle", max_length=64)
    clocknumber = models.CharField(max_length=16)

    @property
    def name(self):
        return ''.join(
            [self.lastname,' ,', self.firstname, ' ', self.middlename])

    class Meta:
        ordering = ['lastname','firstname', 'middlename']

class EmployeeAdmin(admin.ModelAdmin):
    list_display = ('clocknumber','name')
    fieldsets = [("Name", {"fields":(("lastname", "firstname", "middlename"), "clocknumber")}),
]

admin.site.register(Employee, EmployeeAdmin)