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。但是，在Django中最快速的开发方式是利用class-based views来进行开发。所以，我自己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。
model
既然是数据驱动的web，自然先要有model。
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，对应每个FileField都提供不同的上传路径。因为upload_to可以接受一个callable的对象，所以我尝试过把lambda赋值给upload_to，但是在测试中发现，给upload_to赋值为lambda表达式是会报错的ValueError: Cannot serialize function: lambda。应该可以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。
经过后续的测试发现，闭包也是不支持的
def _roles_subdir(roles_path, subdir):
def wrapper(instance, filename):
return os.path.join(roles_path, instance.name, subdir, filename)
return wrapperPlease 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-values
view
本例中使用Django提供的CreateView。在实际的使用中，可以针对于自己的应用场景选择CreateView、UpdateView。
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：
url(r'^upload/$', views.UploadRolesFormView.as_view(), name='rolesUpload'),这样，就可以利用Django最方便的class-based views开处理文件上传的场景了。
python
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