【发布时间】:2020-03-04 16:22:14
【问题描述】:
我的gradeScalesSettings 模型有一个ModelAdmin 子类:
@admin.register(gradeScalesSetting)
class gradeScalesSettingAdmin(admin.ModelAdmin):
list_display = ('configuration_select', 'NumberOfGrades', 'Rounding','Precision', 'Status',)
change_list_template = 'admin/Homepage/view.html'
实际结果
点击等级设置后:
如何将它连接到我的views.py? 这就是我想在我的views.py中编码的内容:
def gradescales(request):
gradeScalesSettings = gradeScalesSetting.objects.all()
configurations = configuration.objects.all()
rounding = gradeScalesSetting.objects.all().values_list('Rounding', flat=True).distinct()
print(rounding)
return render(request, 'Homepage/gradescale.html', {"rounding": rounding,"gradeScalesSetting":gradeScalesSettings,"configurations":configurations})
当我尝试这个时:
@admin.register(gradeScalesSetting)
class gradeScalesSettingAdmin(admin.ModelAdmin):
def new_NumberOfGrades(self, obj):
if obj.NumberOfGrades == 'Grade Scale Settings':
return '<a href="view.html" </a>' # this url will redirect to your
在我的ModelAdmin 子类中:
list_display = ('configuration_select', 'new_NumberOfGrades', 'Rounding','Precision', 'Status',)
有什么方法可以将它连接到我的views.py?
预期结果
这就是我想在 view.html 中显示的内容:
这就是为什么我想将它连接到我的views.py。
【问题讨论】:
-
view.html 和gradescale.html 有什么区别?
标签: python django django-admin overriding django-modeladmin