'tuple' 对象在更新图像对象时没有属性 '_committed' 错误？

Question

在这里，我试图更新特定产品的每个产品图像。但它不能正常工作。这里只有第一个对象的图像正在更新。

有一个模板，我们可以在其中一次更新产品和产品图片。与模型ProductImage有ManyToOne关系，Product因此在模板中可以有单个产品的多个图像对象。

更新产品模型工作正常，但在更新 ProductImage 对象时它不起作用。

除了压缩，还有其他方法可以一次更新多个图像对象吗？

编辑：如果我解压缩图像列表，则更新无法正常工作。例如，如果我更改一个对象的图像，则另一个对象的图像也会更改。 但是当我更改所有图像对象图像时，更新工作正常。当我只更改某些对象时，它无法正常工作。

当我压缩然后图像列表时，这是错误回溯。

Traceback (most recent call last):
  File "venv\lib\site-packages\django\core\handlers\exception.py", line 47, in inner
    response = get_response(request)
  File "venv\lib\site-packages\django\core\handlers\base.py", line 179, in _get_response
    response = wrapped_callback(request, *callback_args, **callback_kwargs)
  File "\venv\lib\site-packages\django\views\generic\base.py", line 73, in view
    return self.dispatch(request, *args, **kwargs)
  File "\venv\lib\site-packages\django\views\generic\base.py", line 101, in dispatch
    return handler(request, *args, **kwargs)
  File "dashboard\product\views\views.py", line 293, in post
    p.save()
  File "venv\lib\site-packages\django\db\models\base.py", line 751, in save
    force_update=force_update, update_fields=update_fields)
  File "venv\lib\site-packages\django\db\models\base.py", line 789, in save_base
    force_update, using, update_fields,
  File "venv\lib\site-packages\django\db\models\base.py", line 867, in _save_table
    for f in non_pks]
  File "lib\site-packages\django\db\models\base.py", line 867, in <listcomp>
    for f in non_pks]
  File "\venv\lib\site-packages\django\db\models\fields\files.py", line 303, in pre_save
    if file and not file._committed:
AttributeError: 'tuple' object has no attribute '_committed'

楷模

class ProductImage(models.Model):
    image = models.ImageField(upload_to='imgs',blank=True, null=True)
    product = models.ForeignKey(Product, on_delete=models.CASCADE)

模板

 <form method="post" enctype="multipart/form-data">
       {% csrf_token %}
       {{product_form}}
    <th>Current Image</th>
    <th>Change</th>
    {% for image in p_images %}
    <tr><td>{{image.pk}}</td>
    <td><img src="{{image.image.url}}" width="50" height="50"></td>
    <td><input type="file" name="image"></td>
    </tr>
    {% endfor %}
   <input type="submit">

意见

def post(self, request, *args, **kwargs):
    product = Product.objects.get(pk=kwargs['pk'])
    product_form = ProductForm(request.POST, instance=product)
    images = zip(request.FILES.getlist('image')) 
    p_images = ProductImage.objects.filter(product=product).order_by('pk')
    if product_form.is_valid():
        product_form.save()             
        
       # updating product images 
        for p, img in zip(p_images, images):  # error is here
           p.image = img
           p.save()

           # Tried this way too:

       for img in images:
           ProductImage.objects.filter(product=product).update(image=img)

Answer 1

首先，进行更改input name以能够识别哪些内容ProductImage已更新。

<!-- <td><input type="file" name="image"></td> -->
     <td><input type="file" name="image-{{image.pk}}"></td>

接下来，迭代input_nameinrequest.FILES并得到ProductImagePK。
然后，查找ProductImage p、更新image字段和save模型。

def post(self, request, *args, **kwargs):
    product = Product.objects.get(pk=kwargs['pk'])
    product_form = ProductForm(request.POST, instance=product)
    if product_form.is_valid():
        product_form.save()

        # Updating product images
        if request.FILES:
            p_images = ProductImage.objects.filter(product=product).order_by('pk')
            p_images_lookup = {p_image.pk: p_image for p_image in p_images}
            for input_name in request.FILES:
                p = p_images_lookup[int(input_name[len('image-'):])]
                p.image = request.FILES[input_name]
                p.save()