相关疑难解决方法(0)

我想做的事情如下:

变量p来自test.py,它是一个列表['a','b','c','d']

{% for i in p %}
{{variable++}}
{{variable}}

结果输出为:1 2 3 4

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

在管理员方面,可能在其他地方,我想将其显示为:

lastname, firstname middlename

对我来说,这样做的合理位置是在模型中通过创建一个计算字段:

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)

最终我认为我需要的是将名称字段的值作为字符串.我得到的错误是value_to_string() takes exactly 2 arguments (1 given).字符串想要的值self, obj …