我有两种模式:
class Post(models.Model):
image = models.ImageField(upload_to='%Y/%m/%d')
title = models.CharField(max_length=200)
class Addimg(models.Model):
addimages = models.ForeignKey('Post', null=True)
addimg = models.ImageField(upload_to='images')
我希望使用“Addimg”模型将图像添加到我的Post模型中,该模型到目前为止工作得很好,但现在我想要在编辑我的父模型(Post)时,所有附加的“Addimg”模型也出现在形成。我怎么能这样做?什么是最简单的解决方案?
这是我处理“父”形式的视图:
def edit(request, pk):
post = get_object_or_404(Post, pk=pk)
if request.method == "POST":
form = PostForm(request.POST, request.FILES, instance=post)
if form.is_valid():
post = form.save(commit=False)
post.save()
return redirect('blog.views.detail', pk=post.pk)
else:
form = PostForm(instance=post)
return render(request, 'blog/edit.html', {'form': form})
和我的forms.py:
class PostForm(forms.ModelForm):
class Meta:
model = Post
fields = ('image', 'title',)
class AddimgForm(forms.ModelForm):
class Meta:
model = Addimg
fields = ('addimages', 'addimg',)
在我的表单模板中,我有:
<form enctype="multipart/form-data" method="POST" class="post-form">
{% csrf_token %}
{{ form.as_p }}
<button type="submit" >Save</button>
</form>
我会非常高兴任何提示或有用的链接,因为我是django /编程的新手我甚至不知道正确的搜索关键字。感谢
答案 0 :(得分:1)
答案 1 :(得分:0)
好的,对于django来说很简单:
form.py中的:
from django.forms.models import inlineformset_factory
MyFormSet = inlineformset_factory(Post, Addimg, extra=1, fields = ('addimages', 'addimg',))
在views.py中:
from .forms import PostForm, MyFormSet
def manageimages(request, pk):
post = get_object_or_404(Post, pk=pk)
if request.method == "POST":
formset = MyFormSet(request.POST, request.FILES, instance=post)
if formset.is_valid():
formset.save()
post.save()
return redirect('blog.views.someview')
else:
formset = MyFormSet(instance=post)
return render(request, 'blog/myformsettemplate.html', {'formset': formset})
和myformsettemplate.html:
<form enctype="multipart/form-data" method="POST" class="post-form">
{% csrf_token %}
{{ formset.management_form }}
{% for form in formset %}
<div class="third">
{{ form.as_p }}
</div>
{% endfor %}
<button type="submit" >Save</button>
</form>