我有两个类似的形式上传文件，但我想保存文件在不同的目录取决于形式。
让我解释一下，例如：
如果用户从Form_1上传文件->文件应保存在media/folder_file_1/file.csv中
如果用户从Form_2上传文件->文件应保存在media/folder_file_2/file.csv中
关于models.py, forms.py, views.py, urls.py，我只是使用了Django文档中的示例。
index.html：
<!DOCTYPE html>
{% load static %}
<body>
<div class="page_secondblock secondblock">
<div class="secondblock__container _container">
<h1 class="secondblock__title">
The file you want to update:
</h1>
<h2 class="secondblock__title">
The file from which you want to get information:
</h2>
</div>
</div>
<div class="page_thirdblock thirdblock">
<div class="thirdblock__container _container">
<form method="POST" enctype="multipart/form-data" class="upload1" id="upload_container">
{% csrf_token %}
{{form.as_p}}
<input type="submit" value="Submit">
</form>
<form method="POST" enctype="multipart/form-data" class="upload2" id="upload_container">
{% csrf_token %}
{{form.as_p}}
<input type="submit" value="Submit">
</form>
</div>
</div>
</body>models.py：
from django.db import models
class UploadFile(models.Model):
file = models.FileField(upload_to='working/')forms.py：
from django import forms
from .models import UploadFile
class UploadFileForm(forms.ModelForm):
class Meta:
model = UploadFile
fields = ['file']views.py：
from django.shortcuts import render
from django.http import HttpResponse, HttpResponseRedirect
from .forms import UploadFileForm
def index(request):
return render(request, "index.html")
def tools(request):
return render(request, "index.html")
def login(request):
return render(request, "index.html")
def upload_file(request):
if request.method == 'POST':
form = UploadFileForm(request.POST,request.FILES)
if form.is_valid():
form.save()
return HttpResponseRedirect('home')
else:
form = UploadFileForm()
context = {
'form':form,
}
return render(request, 'index.html', context)url.py：
from django.contrib import admin
from django.urls import path, re_path
from actualization import views
urlpatterns = [
re_path(r'^tools', views.tools, name='tools'),
re_path(r'^login', views.login, name='login'),
path('get_result/', views.get_result),
path('home/', views.upload_file, name='upload_file'),
]
1条答案
按热度按时间relj7zay1#
好的，我的做法是：
通过这种方式，它将文件存储在您在www.example.com中分配的文件夹中settings.py作为
MEDIA_ROOT。然后将其放入与您的用户名同名的文件夹中。然后是文件名。如果用户名作为文件夹名是一个很好的选择是肯定有争议的，但你得到的想法。
在这里找到另一个很好的例子：
FileField.upload_to