在Django 1.3中,用户上传一个“hello world”应用程序所需的最少示例代码是什么?
当前回答
你可以参考Fine Uploader中的服务器示例,它有django版本。 https://github.com/FineUploader/server-examples/tree/master/python/django-fine-uploader
它非常优雅,最重要的是,它提供了特色的js lib。Template不包含在服务器示例中,但是你可以在它的网站上找到demo。 精美上传:http://fineuploader.com/demos.html
django-fine-uploader
views.py
UploadView将post和delete请求分派给各自的处理程序。
class UploadView(View):
@csrf_exempt
def dispatch(self, *args, **kwargs):
return super(UploadView, self).dispatch(*args, **kwargs)
def post(self, request, *args, **kwargs):
"""A POST request. Validate the form and then handle the upload
based ont the POSTed data. Does not handle extra parameters yet.
"""
form = UploadFileForm(request.POST, request.FILES)
if form.is_valid():
handle_upload(request.FILES['qqfile'], form.cleaned_data)
return make_response(content=json.dumps({ 'success': True }))
else:
return make_response(status=400,
content=json.dumps({
'success': False,
'error': '%s' % repr(form.errors)
}))
def delete(self, request, *args, **kwargs):
"""A DELETE request. If found, deletes a file with the corresponding
UUID from the server's filesystem.
"""
qquuid = kwargs.get('qquuid', '')
if qquuid:
try:
handle_deleted_file(qquuid)
return make_response(content=json.dumps({ 'success': True }))
except Exception, e:
return make_response(status=400,
content=json.dumps({
'success': False,
'error': '%s' % repr(e)
}))
return make_response(status=404,
content=json.dumps({
'success': False,
'error': 'File not present'
}))
forms.py
class UploadFileForm(forms.Form):
""" This form represents a basic request from Fine Uploader.
The required fields will **always** be sent, the other fields are optional
based on your setup.
Edit this if you want to add custom parameters in the body of the POST
request.
"""
qqfile = forms.FileField()
qquuid = forms.CharField()
qqfilename = forms.CharField()
qqpartindex = forms.IntegerField(required=False)
qqchunksize = forms.IntegerField(required=False)
qqpartbyteoffset = forms.IntegerField(required=False)
qqtotalfilesize = forms.IntegerField(required=False)
qqtotalparts = forms.IntegerField(required=False)
其他回答
不确定这种方法是否有任何缺点,但更小,在views.py中:
entry = form.save()
# save uploaded file
if request.FILES['myfile']:
entry.myfile.save(request.FILES['myfile']._name, request.FILES['myfile'], True)
我也遇到过类似的问题,由django管理站点解决了。
# models
class Document(models.Model):
docfile = models.FileField(upload_to='documents/Temp/%Y/%m/%d')
def doc_name(self):
return self.docfile.name.split('/')[-1] # only the name, not full path
# admin
from myapp.models import Document
class DocumentAdmin(admin.ModelAdmin):
list_display = ('doc_name',)
admin.site.register(Document, DocumentAdmin)
以亨利为例:
import tempfile
import shutil
FILE_UPLOAD_DIR = '/home/imran/uploads'
def handle_uploaded_file(source):
fd, filepath = tempfile.mkstemp(prefix=source.name, dir=FILE_UPLOAD_DIR)
with open(filepath, 'wb') as dest:
shutil.copyfileobj(source, dest)
return filepath
你可以在上传文件对象的视图中调用handle_uploaded_file函数。这将在文件系统中保存具有唯一名称的文件(以原始上传文件的文件名作为前缀),并返回所保存文件的完整路径。您可以将该路径保存在数据库中,稍后对该文件进行处理。
Demo
参见github repo,适用于Django 3
一个简单的Django文件上传示例
1. 创建一个django项目
运行startproject::
$ django-admin.py startproject sample
现在创建了一个文件夹(示例)。
2. 创建应用程序
创建一个应用::
$ cd sample
$ python manage.py startapp uploader
现在创建一个包含这些文件的文件夹(uploader):
uploader/
__init__.py
admin.py
app.py
models.py
tests.py
views.py
migrations/
__init__.py
3.更新settings.py
在sample/settings.py中添加'uploader'到INSTALLED_APPS,并添加MEDIA_ROOT和MEDIA_URL,即::
INSTALLED_APPS = [
'uploader',
...<other apps>...
]
MEDIA_ROOT = os.path.join(BASE_DIR, 'media')
MEDIA_URL = '/media/'
4. 更新urls . py
在sample/urls.py中添加::
...<other imports>...
from django.conf import settings
from django.conf.urls.static import static
from uploader import views as uploader_views
urlpatterns = [
...<other url patterns>...
path('', uploader_views.UploadView.as_view(), name='fileupload'),
]+ static(settings.MEDIA_URL, document_root=settings.MEDIA_ROOT)
5. 更新models.py
更新上传/ models.py::
from django.db import models
class Upload(models.Model):
upload_file = models.FileField()
upload_date = models.DateTimeField(auto_now_add =True)
6. 更新views.py
更新上传/ views.py::
from django.views.generic.edit import CreateView
from django.urls import reverse_lazy
from .models import Upload
class UploadView(CreateView):
model = Upload
fields = ['upload_file', ]
success_url = reverse_lazy('fileupload')
def get_context_data(self, **kwargs):
context = super().get_context_data(**kwargs)
context['documents'] = Upload.objects.all()
return context
7. 创建模板
创建文件夹sample/uploader/templates/uploader
创建文件upload_form.html ie sample/uploader/templates/uploader/upload_form.html::
<div style="padding:40px;margin:40px;border:1px solid #ccc">
<h1>Django File Upload</h1>
<form method="post" enctype="multipart/form-data">
{% csrf_token %}
{{ form.as_p }}
<button type="submit">Submit</button>
</form><hr>
<ul>
{% for document in documents %}
<li>
<a href="{{ document.upload_file.url }}">{{ document.upload_file.name }}</a>
<small>({{ document.upload_file.size|filesizeformat }}) - {{document.upload_date}}</small>
</li>
{% endfor %}
</ul>
</div>
8. Syncronize数据库
同步数据库和runserver::
$ python manage.py makemigrations
$ python manage.py migrate
$ python manage.py runserver
访问http://localhost: 8000 /
我必须说我发现django的文档令人困惑。 还有一个最简单的例子,为什么要提到表单? 我在views.py中工作的例子是:-
for key, file in request.FILES.items():
path = file.name
dest = open(path, 'w')
if file.multiple_chunks:
for c in file.chunks():
dest.write(c)
else:
dest.write(file.read())
dest.close()
html文件看起来像下面的代码,尽管这个例子只上传了一个文件,保存文件的代码处理许多:-
<form action="/upload_file/" method="post" enctype="multipart/form-data">{% csrf_token %}
<label for="file">Filename:</label>
<input type="file" name="file" id="file" />
<br />
<input type="submit" name="submit" value="Submit" />
</form>
这些示例不是我的代码,它们是从我找到的其他两个示例中选取的。 我是一个相对初学者django,所以很可能我错过了一些关键点。
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