Question

我想上传excel文件并将该文件保存到django中的特定位置，而无需为该文件创建模型。

我在这里试过我的forms.py文件

class UploadFileForm(forms.Form):

    file  = forms.FileField(label='Select a file',
        help_text='max. 42 megabytes')

my views.py

import xlrd  
from property.forms import UploadFileForm

def excel(request):

    if request.method == 'POST':

        form = UploadFileForm(request.POST, request.FILES)
        if form.is_valid():
            newdoc = handle_uploaded_file(request.FILES['file'])
            print newdoc
            print "you in"
            newdoc.save()

            return HttpResponseRedirect(reverse('upload.views.excel'))
    else:
        form = UploadFileForm() # A empty, unbound form
    #documents = Document.objects.all()
    return render_to_response('property/list.html',{'form': form},context_instance=RequestContext(request))

def handle_uploaded_file(f):
    destination = open('media/file/sheet.xls', 'wb+')
    for chunk in f.chunks():
        destination.write(chunk)
    destination.close()

因此，在尝试此操作时，我收到错误。

IOError at /property/excel/
[Errno 2] No such file or directory: 'media/file/sheet.xls'
Request Method: POST
Request URL:    http://127.0.0.1:8000/property/excel/
Django Version: 1.5
Exception Type: IOError
Exception Value:    
[Errno 2] No such file or directory: 'media/file/sheet.xls'
Exception Location: D:\Django_workspace\6thMarch\dtz\property\views.py in handle_uploaded_file, line 785

请帮帮我，handle_uploaded_file（）函数有问题。

Answer 1

如果你使用开放式（如开放式路径＆＃39;路径＆＃39;，＆＃39; wb +＆＃39;），则需要使用FULL路径。

你能做的是：

from django.conf import settings
destination = open(settings.MEDIA_ROOT + filename, 'wb+')

如何在不创建模型的情况下在django中保存文件

1 个答案: