Django上传'文件名'错误

Question

代码只是上传CVS文件并将数据转换为LIST。

当用户上传csv文件时，单个用户使用的文件名不会被强制执行。我希望能够处理用户使用csv文件的任何文件名。

我收到此错误：

File "views.py", line 35, in upload_file phone_list = handle_uploaded_file(request.FILES['phonelistfile'])

File "views.py", line 17, in handle_uploaded_file with open(settings.MEDIA_URL+'documents/'+str(filename),'wb') as destination:

NameError: name 'filename' is not defined

Views.py

from django.shortcuts import render 

# Create your views here. 

from django.http import HttpResponseRedirect 
from .models import Upload 
from .forms import UploadFileForm 
import csv 
import io 

# Imaginary function to handle an uploaded file. 
#from somewhere import handle_uploaded_file 

def handle_uploaded_file(f): 
    with open(settings.MEDIA_URL+'documents/'+str(filename),'wb') as destination: 
     for chunk in f.chunks(): 
      destination.write(chunk) 
     destination.close() 
    #csvfile = f 
    csvfile = io.TextIOWrapper(f) # Python 3 Only 
    #dialect = csv.sniffer().sniff(codecs.EncodedFile(csvfile, "utf-8").read(1024)) 
    dialect = csv.sniffer().sniff(csvfile.read(1024), delimiter=";,") 
    #csvfile.open() 
    csvfile.seek(0) 
    #csvreader = csv.reader(codecs.EncodedFile(csvfile, "utf-8"), delimiter=',', dialect=dialect) 
    csvreader = csv.reader(csvfile, dialect) 
    return list(csvreader) 

def upload_file(request): 
    if request.method == 'POST': 
     form = UploadFileForm(request.POST, request.FILES) 
     if form.is_valid(): 
      phone_list = handle_uploaded_file(request.FILES['phonelistfile']) 
      upload_phone_list = Upload() 
      upload_phone_list.name = request.name 
      upload_phone_list.phonelistfile = request.FILES['file'].file 
      upload_phone_list.phonelist = phone_list 
      #form.save() 
      upload_phone_list.save() 

      return HttpResponseRedirect('/success/url/') 
    else: 
     form = UploadFileForm() 
    return render(request, 'upload.html', {'form': form}) 

models.py

from django.db import models 
#from django.forms import ModelForm 
from django.db import models 
from django.contrib.postgres.fields import ArrayField 

class Upload(models.Model): 
    name = models.CharField(max_length=50) 
    phonelistfile = models.FileField("phonelistfile", upload_to="media/%Y/%m/%d/") 
    upload_date = models.DateTimeField(auto_now_add =True) 
    phonelist = ArrayField(models.TextField()) 

Forms.py

from django import forms 
from .models import Upload 
from django.forms import ModelForm 

# FileUpload form class. 
class UploadFileForm(ModelForm): 
    #name = forms.CharField(max_length=100) 
    #phonelistfile = forms.FileField("phonelistfile", allow_empty_file=True, required=False) 
    class Meta: 
     model = Upload 
     fields = ('name', 'phonelistfile') 

回答： 我已经能够找到答案。现在，我可以通过添加下面的代码来获取用户上传的CSV文件的文件名;由此使用折叠在views.py两个功能合为一体：

def upload_file(request): 
    if request.method == 'POST': 
     form = UploadFileForm(request.POST, request.FILES) 
     if form.is_valid(): 
      fil = request.FILES['phonelistfile'] 
      with open('f', 'wb+') as destination: 
       for chunk in fil.chunks(): 
        destination.write(chunk) 
        destination.close() 
      csvfile = io.TextIOWrapper(open('f', 'rb')) # Python 3 Only 
      #Do something with he file.... 

这是卓有成效.... - FIL = request.FILES [ 'phonelistfile']

Answer 1

在下面的行，则从不界定什么是filename：

with open(settings.MEDIA_URL+'documents/'+str(filename),'wb') as destination: 

的Django不知道那是什么。您需要将其设置为某个值。