2

所以我试图上传一个没有任何外部插件的文件，但我遇到了一些错误。

                <form method="" action="" name='upload_form' id='upload_form' >
                    {% csrf_token %}
                   <input type='file' name='file' id='file' />
                   <input type='button' value='Upload' id='upload'/>
                </form>

                <script type='text/javascript'>
                $(document).ready(function() {
                    var csrf_token = $('input[name="csrfmiddlewaretoken"]').val();
                    $('#upload').click(function() {
                        $.ajax({
                            csrfmiddlewaretoken: csrf_token,
                            type: 'POST',
                            url : 'upload',
                            enctype: "multipart/form-data",
                            data  : {
                                'file': $('#file').val()
                            },
                            success: function(data) {
                                console.log(data)
                            }
                        })
                    })
                })
                </script>

我的服务器：

class ImageUploadView(LoginRequiredMixin, JSONResponseMixin, AjaxResponseMixin, CurrentUserIdMixin, View):

    @method_decorator(csrf_protect)
    def dispatch(self, *args, **kwargs):
        return super(ImageUploadView, self).dispatch(*args, **kwargs)

    def post_ajax(self, request, username):
                print request.POST.get('file', None)
                print request.FILES

        # id = request.POST['id']
        # path = 'pictures/'
        # f = request.FILES['picture']
        # destination = open(path, 'wb+')
        # for chunk in f.chunks():
        #   destination.write(chunk)
        # destination.close()
return HttpResponse("image uploaded")

我得到一个<MultiValueDict: {}>request.FILES

现在如何使用我的代码正确获取上传的文件？

4

2 回答 2

5

这是我使用javascript上传文件的方法，希望对您有所帮助！只需将您的 $('#file') 作为参数传递。 

function upload(field, upload_url) {
    if (field.files.length == 0) {
        return;
    }
    file = field.files[0];
    var formdata = new FormData();
    formdata.append('file_upload', file);
    $.ajax({
        url: upload_url,
        type: 'POST',
        data: formdata,
        processData: false,
        contentType: false,
        success: console.log('success!')
    });
}

[编辑]

这就是我在服务器端所做的（简化）：

def save_file(dest_path, f, filename):
    original_name, file_extension = os.path.splitext(f.name)
    filename = filename + '-' + datetime.datetime.now().strftime('%Y-%m-%d-%H-%M-%S') + file_extension
    url = '/' + dest_path + '/' + filename
    path = django_settings.MEDIA_ROOT + url
    destination = open(path, 'wb+')
    for chunk in f.chunks():
        destination.write(chunk)
    destination.close()
    return path

class FileUploadView(View):
    def post(self, request, *args, **kwargs):
        if request.FILES and request.FILES.get('file_upload'):
            path = save_file(UPLOAD_TO, 
                             request.FILES.get('file_upload'), 
                             FILENAME)
        return self.render_to_response({})
于 2013-05-14T14:28:39.130 回答
3

我按照本教程http://www.script-tutorials.com/pure-html5-file-upload/并在 php 部分替换为：

class UploadImageView(LoginRequiredMixin, CurrentUserIdMixin, View):

    @method_decorator(csrf_protect)
    def dispatch(self, *args, **kwargs):
        return super(UploadImageView, self).dispatch(*args, **kwargs)

    def post(self, request, username):
        path = 'myproject/media/pictures/guitar.jpg'
        f = request.FILES['image_file']
        destination = open(path, 'wb+')
        for chunk in f.chunks():
            destination.write(chunk)
            destination.close()

            return HttpResponse("image uploaded")

也改变了这一行

<form id="upload_form" enctype="multipart/form-data" method="post" action=".">
                                {% csrf_token %}
于 2013-05-14T10:57:17.640 回答