Question

我有一个用于上传文件的表单。我的views.py文件中的代码是：

def upload(request):
    # Handle file upload
    if request.method == 'POST':
        form = DocumentForm(request.POST, request.FILES)
        if form.is_valid():
            newdoc = Document(docfile=request.FILES['docfile'])
            newdoc.save()
            # Redirect to the document list after POST

            return HttpResponse("Success!")
    else:
        form = DocumentForm()  # A empty, unbound form

        # Load documents for the list page
        documents = Document.objects.all()

        # Render list page with the documents and the form
        return render_to_response(
             'upload.html',
             {'documents': documents, 'form': form},
             context_instance=RequestContext(request)
             )

如何了解正在创建的文件名？我正在使用Django v.1.9

Answer 1

在python中，你可以使用name属性获取它。，。

newdoc = request.FILES['docfile']
newdoc.name # file name

Andv如果您希望在模板中看到代码

{% for document in documents %}
{{ document.file.filename }}
{% endfor }}

在模型对象中

newdoc = Document.objects.get(id=1)
newdoc.docfile.url # url of the file
newdoc.docfile.path # exact path

从Django Document对象获取文件名

1 个答案: