我试图在保存之前更改表单的实例。我需要像这样在 View 中设置某些信息:

 class UploadedFile(models.Model):

     file = models.FileField(storage=s3store, upload_to=custom_upload_to)
     slug = models.SlugField(max_length=50, blank=True)
     bucket = models.ForeignKey(S3Bucket, blank=False)
     uploaded_by = models.ForeignKey(User, related_name='uploaded_by', blank=False)
     company = models.ForeignKey(Company, blank=False)

--

class UploadForm(ModelForm):

    class Meta:
        model = UploadedFile    

--

form = UploadForm(request.POST, request.FILES)
form.instance.company_id = r_user.company.id
form.instance.uploaded_by_id = r_user.id
form.instance.bucket_id = r_user.company.s3_bucket_id
if form.is_valid():
     form_object = form.save()

现在，我知道表单无效，因为 company/uploaded/bucket 是空的:

错误表:

但我确实设置了它们!我是否需要让它们空白=True，然后保存(commit=false)，更改它们，然后重新保存？如果是，为什么会这样？我确实更改了表格....

最佳答案

试试这个:

data = request.POST.copy()
data['company_id'] = r_user.company.id
data['uploaded_by_id'] = r_user.id
data['bucket_id'] = r_user.company.s3_bucket_id

form = UploadForm(data, request.FILES)
if form.is_valid():
     form_object = form.save()

通过这种方式，您可以在创建表单之前设置数据。

https://stackoverflow.com/questions/10723054/