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': $('#file').val()
},success: function(data) {
console.log(data)
}
})
})
})
</script>
我的服务器：@H_403_5@
class ImageUploadView(LoginrequiredMixin,JSONResponseMixin,AjaxResponseMixin,CurrentUserIdMixin,View): @method_decorator(csrf_protect) def dispatch(self,*args,**kwargs): return super(ImageUploadView,self).dispatch(*args,**kwargs) def post_ajax(self,request,username): print request.POST.get('file',None) print request.FILES # id = request.POST['id'] # path = 'pictures/' # f = request.FILES['picture'] # destination = open(path,'wb+') # for chunk in f.chunks(): # destination.write(chunk) # destination.close() return HttpResponse("image uploaded")
我得到一个< MultiValueDict：{}>对于request.FILES@H_403_5@
解决方法
这是我使用javascript上传文件的内容,希望这有帮助！只需将$(‘#file’)作为参数传递即可.
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,data: formdata,processData: false,contentType: false,success: console.log('success!')
});
}
[编辑]@H_403_5@
这就是我在服务器端所做的事情(简化)：@H_403_5@
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,**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({})
