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Django-,以模型形式显示和过滤外键中的特定字段

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我正在使用模型表单,并且该字段是外键,对于另一个模型,外键模型包含我想要基于用户组过滤的选择字段(状态),这样如果用户属于一个组,例如,如果组EEER中的用户,可用的状态选项将是“每周任务”和“每日任务”

models.py

class ScrumyGoals(models.Model):
     user_name=models.ForeignKey('ScrumyUser', on_delete= models.CASCADE, null= True)
     status_id = models.ForeignKey('GoalStatus', on_delete=models.CASCADE,null=True)
     goal_type = models.CharField(choices=GOAL_TYPE, max_length=64, default='DT')
     def __str__(self):
         return '{},{}'.format(self.user_name,self.status_id)

class GoalStatus(models.Model):

    GOALS_TYPE= (('DT','Daily Task'),
        ('WT','Weekly Task'),
        ('V','Verified'),
        ('D','Done'),
)
    title = models.CharField(max_length=254, null=True) 
    task_id=models.IntegerField(default=1,null=False)
    description =models.CharField(max_length=254)
    verified_by=models.ForeignKey('ScrumyUser', on_delete= models.CASCADE, null=True)
    status=models.CharField(choices=GOALS_TYPE, max_length=2, default='DT')

forms.py

class ChangeTaskForm(forms.ModelForm):
    class Meta:
        model = ScrumyGoals
        fields = ['status_id']
    def __init__(self, *args, **kwargs):
    user = kwargs.pop('user', None)
    super(ChangeTaskForm, self).__init__(*args, **kwargs)
    if user.groups.filter(name ='DEVELOPER').exists():
        self.fields['status'].choices = (('WT','Weekly Task'), ('DT','Daily Task'),)

views.py

def changetask(request):
        if request.method == 'POST':
            form = ChangeTaskForm(request.POST, user=request.user)
        if form.is_valid():
            return HttpResponseRedirect('/index/')
        else:
           form = ChangeTaskForm(user=request.user)
        return render(request, 'oderascrumy/changetask.html', {'form': form})

表单目前仅显示status_id,它不显示选项(即“状态”)

所以本质上我想要的是如果登录用户在群组中说“开发者”,用户只能将目标(即“status_id”)选择(即“状态”)从每周任务改为日常任务和表格保存

1 回答

  • 0

    如果从forms.py中删除 if ,它是否有效,即您是否为名为'DEVELOPER'的组存在,并且您正在测试的用户实际上在该组中?

    另一种方法可能是始终在 fields 中包含 status ,然后只是有条件地在 if user.groups in ['DEVELOPER'] 模板中显示该字段 .

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