Question

我想通过Django Forms更新多个文件，并将上传文件的路径保存到数据库表中。但是每当我使用这种方法更新文件数据时，都会将文件数据存储到数据库中，但在数据库表中会为多次上传的每个文件创建不同的行。

我尝试了以下方法

多个文件是通过other_files上传的

forms.py

from django import forms
from categories.models import Categories
from .models import Process

class ProcessingForm(forms.ModelForm):
    files = forms.ModelChoiceField(queryset=Categories.objects.all())
    class Meta:
        model = Process
        fields = ('category', 'files', 'lookup_files', 'other_files')

models.py

from django.db import models

class Process(models.Model):
    category = models.CharField(max_length=25, choices=CATEGORIES, null=True, 
              blank=True)
    files = models.CharField(max_length=50, null=True, blank=True)
    lookup_files = models.FileField(upload_to='processing/%Y/%m/%d/Lookup', 
    verbose_name=('Lookup File [For Tagging]'), validators=
     [validate_file_extension])
    other_files =models.FileField(upload_to='processing/%Y/%m/%d/other_files', 
    null=True, blank=True)

views.py

from django.shortcuts import render, redirect
from categories.models import Categories
from django.contrib import messages
from .models import Process

def process(request):
    if request.method == 'POST':
        form = ProcessingForm(request.POST, request.FILES)        
        # print(form)
        if form.is_valid():
            category = form.cleaned_data['category']
            files = form.cleaned_data['files']
            lookup_files = request.FILES['lookup_files']
            other_files =  request.FILES.getlist('other_files')
            print(len(other_files))
            # file_list = []
            for oth in other_files:
                print(oth)
                # file_list.append(oth)
                form = Process(category=category, files=files, 
                       lookup_files=lookup_files, other_files=oth)
                form.save()
            messages.success(request, 'File uploaded succcessfully')
            return redirect('process')
        else:
            print("Form is not valid")
            messages.warning(request, 'File extension is not valid')
            return redirect('process')

    else:
        form = ProcessingForm()
        return render(request, 'process/process.html', {'form':form})

需要将所有这些行都排成一行

Answer 1

您应该更改策略（这是最好的方法，恕我直言）。您仅需要一个用于存储文件以及进程表和文件表之间关系的表。为此使用ForeignKey关系。您将拥有一个进程和许多文件（OneToMany）。

class OtherFile(models.Model):
    other_file = models.FileField(upload_to='processing/%Y/%m/%d/Lookup', verbose_name=('Lookup File [For Tagging]'), validators=[validate_file_extension])
    process = models.ForeignKey(to=Process, on_delete=models.PROTECT) # this is process_id

Answer 2

如果您不想使用外键，请使用任何分隔符（如 '---' AllLinks = [oth,]）连接文件名，您可以将其保存在一行中。当您必须获取数据时，您必须使用 AllLinks.split('---') 再次拆分名称以获取文件链接。

为了避免这种情况，最好建立外键关系。

Django上传多个文件

2 个答案: