修改unique_together結合的形式Django的

Question

我想修改我的數據庫的現有行，這是一個名爲「CharacterSkill」具有獨特的約束在一起的中介表：

models.py

class CharacterSkill(models.Model): 
    character = models.ForeignKey(Character, on_delete=models.CASCADE) 
    level = models.IntegerField(default=0) 
    skill = models.ForeignKey(Skill, on_delete=models.CASCADE) 
    class Meta: 
     unique_together = ("character","skill") 

我做了一個表格來更改水平，但我無法保存表單，我有兩個領域「字符」 &「技能」兩個錯誤消息：

Select a valid choice. That choice is not one of the available choices. 

燦你幫忙嗎？ :)

我的形式：

class SkillCreateForm(forms.ModelForm): 
    class Meta: 
     model = CharacterSkill 
     fields = ('skill','level','character',) 

我的觀點：

def skill_update(request,skillpk,instancepk): 
    form = SkillCreateForm(request.POST) 
    user = User.objects.get (id = request.user.id) 
    instance = Character.objects.get (id = instancepk) 
    skill = CharacterSkill.objects.get(id = skillpk) 
    data = {'character' : instance, 
     'skill' : skill.skill, 
     'level' : skill.level, 
     } 
    if form.is_valid(): 
     form.save() 
     return redirect('persomaker:skill_list', instance.id) 
    else: 
     form = SkillCreateForm(data) 
     #form.fields['skill'].widget = HiddenInput() 
     #form.fields['character'].widget = HiddenInput() 
    return render(request, 'character/create_skill.html', 
    {'instance':instance, 
    'skill':skill, 
    'form': form,}) 

Answer 1

基於一本書，我改變了我圍成的形式使用實例的方式：

def skill_update(request,skillpk,instancepk): 

    user = User.objects.get (id = request.user.id) 
    instance = Character.objects.get (id = instancepk) 
    skill = CharacterSkill.objects.get(id = skillpk) 
    if request.method == "POST": 
     form = SkillModifyForm(request.POST,instance = skill,) 
     print (skill) 
     if form.is_valid(): 
      form.save() 
      return redirect('persomaker:skill_list', instance.id) 
    else: 
     form = SkillModifyForm(instance = skill,) 
    return render(request, 'character/create_skill.html', 
    {'instance':instance, 
    'skill':skill, 
    'form': form,})