Question

我有一个django项目，并且在此项目中创建了一个后期操作，因此用户将可以共享多个图片后期。但是我有一个问题。我无法编写多张图片上传功能。我看了很多内容，但是没有用。它报告了一个问题，或者contex没有发送到我想要的html页面。请帮我。我想要的多图片功能应该在CreateView下，并且应该与它放在同一模板中。另外，应该有4个图片上传按钮，最后一个应该分配多个功能（以html标记）

models.py：

class Photo(models.Model):
post= models.ForeignKey(Person, on_delete=models.CASCADE)

image = models.ImageField(upload_to=get_image_filename)
uploaded_at = models.DateTimeField(auto_now_add=True)

views.py：

class PersonCreateView(CreateView):
model = Person
form_class = PersonForm
success_url = reverse_lazy('person_changelist')

forms.py：

    class PhotoForm(forms.ModelForm):
    image = forms.ImageField(label='Image')
    class Meta:
        model = Photo
        fields = ('image', )


class PersonForm(forms.ModelForm):
    title = forms.CharField(max_length=128)
    body = forms.CharField(max_length=245, label="Item Description.")
    file_field = forms.FileField(widget=forms.ClearableFileInput(attrs={'multiple': True}))

    class Meta:
        model = Person
        widgets = {
            'valyuta': forms.RadioSelect,
            'barter': forms.CheckboxInput,

        }
        fields = ('country', 'city', 'ban', 'yurus', 'reng', 'qiymet', 'valyuta', 'yanacaqnovu', 'oturucu', 'squtu', 'buraxilisili', 'hecm', 'seher', 'barter', 'metin')

    def __init__(self, *args, **kwargs):
        super().__init__(*args, **kwargs)
        self.fields['city'].queryset = City.objects.none()


        if 'country' in self.data:
            try:
                country_id = int(self.data.get('country'))
                self.fields['city'].queryset = City.objects.filter(country_id=country_id).order_by('name')
            except (ValueError, TypeError):
                pass  # invalid input from the client; ignore and fallback to empty City queryset
        elif self.instance.pk:
            self.fields['city'].queryset = self.instance.country.city_set.order_by('name')

Answer 1

查看是否可行。创建一个简单的HTML表单，而不是Django的模型表单

<form enctype="multipart/form-data" action="" method="post">
    <input type="file" name="uploaded_images" accept="image/*" multiple>
    <input type="submit" name="upload" value="Upload">
</form>

您认为：

if request.method == 'POST':
    for afile in request.FILES.getlist('uploaded_images'):
        #Save images in respective models with the links to other models
        return redirect('back/to/same/url')

这将使用户能够一次上传多张图像，并且您将能够轻松处理单个图像。

Answer 2

我一直在努力解决同一问题，但最终我成功了。所以我想分享我的解决方案。首先，我们需要创建2个模型，其中一个模型放置ImageField并使用ForeignKey与第一个模型连接。

models.py

class Item(models.Model):
    user = models.ForeignKey(settings.AUTH_USER_MODEL, on_delete=models.CASCADE, related_name="items")
    name = models.CharField(max_length=100)


class ItemImage(models.Model):
    item = models.ForeignKey(Item, on_delete=models.CASCADE)
    img = models.ImageField(default="store/default_noitem.jpg", upload_to=get_image_dir)

然后使用inlineformset_factory创建表单。在通过外键处理相关对象的情况下尤其有用。

forms.py

from django import forms
from django.forms.models import inlineformset_factory

from .models import Item, ItemImage

class  ItemImageForm(forms.ModelForm):
    class Meta:
        model = ItemImage
        exclude = ()

class ItemForm(forms.ModelForm):
    class Meta:
        model = Item
        fields = ["name",]

ItemImageFormSet = inlineformset_factory(
    Item, ItemImage, form=ItemImageForm,
    fields=['img'], extra=3, can_delete=True  # <- place where you can enter the nr of img
)

最后

views.py

class ItemCreateView(LoginRequiredMixin, SuccessMessageMixin, CreateView):
    template_name = "items/add_item_form.html"
    success_message = 'Item successfully added!'
    form_class = ItemForm

    def get_context_data(self, **kwargs):
        data = super(ItemCreateView, self).get_context_data(**kwargs)
        data['form_images'] = ItemImageFormSet()
        if self.request.POST:
            data['form_images'] = ItemImageFormSet(self.request.POST, self.request.FILES)
        else:
            data['form_images'] = ItemImageFormSet()
        return data

    def form_valid(self, form):
        context = self.get_context_data()
        form_img = context['form_images']
        with transaction.atomic():
            form.instance.user = self.request.user
            self.object = form.save()
            if form_img.is_valid():
                form_img.instance = self.object
                form_img.save()
        return super(ItemCreateView, self).form_valid(form)

请记住在(self.request.POST, self.request.FILES)标记所在的HTML格式集内添加enctype="multipart/form-data"，在html内添加<form>。

如何在Django CreateView中创建上传多个图像？

2 个答案: