Question

当我添加新工作时，我有2个模型Worker和Position应该有一个选择框，其中Positions模型中列出的位置。

我尝试使用Select小部件，但django因no such property例外而失败。

模型


# Position model
class Position(models.Model):
    p_name   = models.CharField(max_length = 40, unique = True, verbose_name = 'Position')
    p_salary = models.IntegerField(max_length = 11, verbose_name = 'Salary')

    fields = ('p_name', 'p_salary')


# Worker model
class Worker(models.Model):
    w_name     = models.CharField(max_length = 40, verbose_name = 'Name')
    w_lastname = models.CharField(max_length = 40, verbose_name = 'Lastname')
    w_position = models.ForeignKey(to = Position, to_field = 'p_name', verbose_name = 'Position')
    w_dept     = models.ForeignKey(to = Department, to_field = 'd_name' verbose_name = 'Department')

表格


class WorkerForm(ModelForm):
    class Meta:
        model = Worker

        widgets = {
            'w_name'    : TextInput(attrs = {'class': 'e name'}),
            'w_lastame' : TextInput(attrs = {'class': 'e lastname'}),
            'w_position': Select(attrs = {'class': 'e position'}),
            'w_dept'    : Select(attrs = {'class': 'e department'}),
        }

在admin.py中定义


class WorkerAdmin(admin.ModelAdmin):
    form = WorkerForm

list_display = ('w_name', 'w_lastname', 'w_position', 'w_dept')

list_display = ('w_name', 'w_lastname', 'w_position', 'w_dept')

如何根据外键显示选择框会出现什么问题？

苏丹

Answer 1

您可以尝试重新声明表单内部表单的字段（确保使用与模型中相同的名称）并在其中设置窗口小部件和attrs？

Django：模型外键和选择框小部件

1 个答案: