Django中的文件上传(利用class-based view)

时间 2019-11-10

标签 django 文件上传利用 class based view 栏目 Python 繁體版

原文原文链接

背景介绍

在Django的官网上有专门介绍如何处理文件上传的文档，其中说到了如何利用model来处理文件上传的场景。可是，在Django中最快速的开发方式是利用class-based views来进行开发。因此，我本身整理了一下如何利用class-based views来处理文件上传的场景，特此记录。html

model

既然是数据驱动的web，天然先要有model。web

from django.db import models
from django.contrib.auth.models import User
from django.conf import settings
import os

_roles_path = os.path.join(_base_path, 'roles')


def var_dir(instance, filename):
    return os.path.join(_roles_path, instance.name, 'vars', filename)


def task_dir(instance, filename):
    return os.path.join(_roles_path, instance.name, 'tasks', filename)


class Roles(models.Model):
    id = models.AutoField(primary_key=True)
    name = models.CharField(max_length=50)
    creator = models.ForeignKey(User)
    createDatetime = models.DateTimeField(auto_now_add=True)
    directory = models.FilePathField(path=_roles_path, match='*.yml', recursive=True, max_length=200)
    tasks = models.FileField(upload_to=task_dir, blank=False)
    vars = models.FileField(upload_to=var_dir)

    def __unicode__(self):
        return u'%s' % self.name

上面用到了动态的upload_to，对应每一个FileField都提供不一样的上传路径。由于upload_to能够接受一个callable的对象，因此我尝试过把lambda赋值给upload_to，可是在测试中发现，给upload_to赋值为lambda表达式是会报错的ValueError: Cannot serialize function: lambda。应该能够尝试利用闭包的方式来给upload_to赋值，以解决多种动态路径的需求。django

通过后续的测试发现，闭包也是不支持的闭包

def _roles_subdir(roles_path, subdir):
    def wrapper(instance, filename):
        return os.path.join(roles_path, instance.name, subdir, filename)
    return wrapper

Please note that due to Python 2 limitations, you cannot serialize unbound method functions (e.g. a method declared
and used in the same class body). Please move the function into the main module body to use migrations.
For more information, see
https://docs.djangoproject.com/en/1.7/topics/migrations/#serializing-valuesapp

view

本例中使用Django提供的CreateView。在实际的使用中，能够针对于本身的应用场景选择CreateView、UpdateView。ide

from django.views.generic.edit import FormView, CreateView
from django.views.decorators.csrf import csrf_exempt
from django.core.urlresolvers import reverse_lazy

class UploadRolesFormView(CreateView):
    template_name = 'app/upload_roles.html'
    model = Roles
    fields = ['name', 'tasks', 'vars']
    success_url = reverse_lazy('app:index')

    #临时去掉CSRF保护，千万别学我！
    @csrf_exempt
    def dispatch(self, request, *args, **kwargs):
        return super(UploadRolesFormView, self).dispatch(request, *args, **kwargs)

    #override
    def form_valid(self, form):
        #在form中加入user对象存入model
        form.instance.creator = self.request.user
        return super(UploadRolesFormView, self).form_valid(form)

template

<html>
    <head>
        <title>upload</title>
        <meta http-equiv="description" content="this is my page">
        <meta http-equiv="content-type" content="text/html; charset=GB18030">
    </head>

    <body>
        <form action="{% url 'app:rolesUpload' %}" method="post" enctype="multipart/form-data">
            <input type="text" name="name" />
            <input type="file" name="tasks" />
            <input type="file" name="vars" />
            <input type="submit" value="上传" />
        </form>
    </body>
</html>

url

在app的urls.py中加入一条对应的url规则：post

url(r'^upload/$', views.UploadRolesFormView.as_view(), name='rolesUpload'),

这样，就能够利用Django最方便的class-based views开处理文件上传的场景了。测试