如何在Django中上载文件？
作为Django的新手，我很难在Django 1.3中制作上传应用程序。我找不到任何最新的示例/代码片段。是否有人可以发布一个最小但完整的（模型、视图、模板）示例代码来执行此操作？一般来说，当您试图“获取一个工作示例”时，最好是“开始编写代码”。这里没有代码可以帮助您，因此它让我们在回答这个问题时要做更多的工作 如果你想抓取一个文件，你需要在html文件的某个地方有这样的东西：如何在Django中上载文件？,django,file,upload,Django,File,Upload,作为Django的新手，我很难在Django 1.3中制作上传应用程序。我找不到任何最新的示例/代码片段。是否有人可以发布一个最小但完整的（模型、视图、模板）示例代码来执行此操作？一般来说，当您试图“获取一个工作示例”时，最好是“开始编写代码”。这里没有代码可以帮助您，因此它让我们在回答这个问题时要做更多的工作 如果你想抓取一个文件，你需要在html文件的某个地方有这样的东西： <form method="post" enctype="multipart/form-data">
<form method="post" enctype="multipart/form-data">
<input type="file" name="myfile" />
<input type="submit" name="submit" value="Upload" />
</form>
您可以使用上载的文件对象从视图中调用此
处理上载的文件entry = form.save()
# save uploaded file
if request.FILES['myfile']:
entry.myfile.save(request.FILES['myfile']._name, request.FILES['myfile'], True)
哎呀，Django文档中确实没有关于这方面的好例子。我花了两个多小时来挖掘所有的碎片，以了解这是如何工作的。有了这些知识，我实现了一个项目，可以上传文件并将其显示为列表。要下载项目的源代码，请访问或克隆它：
> git clone https://github.com/axelpale/minimal-django-file-upload-example.git
minimal-django-file-upload-example/
src/
myproject/
database/
sqlite.db
media/
myapp/
templates/
myapp/
list.html
forms.py
models.py
urls.py
views.py
__init__.py
manage.py
settings.py
urls.py
...
import os
BASE_DIR = os.path.dirname(os.path.dirname(__file__))
...
DATABASES = {
'default': {
'ENGINE': 'django.db.backends.sqlite3',
'NAME': os.path.join(BASE_DIR, 'database.sqlite3'),
'USER': '',
'PASSWORD': '',
'HOST': '',
'PORT': '',
}
}
...
MEDIA_ROOT = os.path.join(BASE_DIR, 'media')
MEDIA_URL = '/media/'
...
INSTALLED_APPS = (
...
'myapp',
)
# -*- coding: utf-8 -*-
from django import forms
class DocumentForm(forms.Form):
docfile = forms.FileField(
label='Select a file',
help_text='max. 42 megabytes'
)
请求.文件的处理方式。对我来说，很难发现request.FILES['docfile']
可以像这样保存到models.FileField。模型的save（）自动将文件存储到文件系统
# -*- coding: utf-8 -*-
from django.shortcuts import render_to_response
from django.template import RequestContext
from django.http import HttpResponseRedirect
from django.core.urlresolvers import reverse
from myproject.myapp.models import Document
from myproject.myapp.forms import DocumentForm
def list(request):
# Handle file upload
if request.method == 'POST':
form = DocumentForm(request.POST, request.FILES)
if form.is_valid():
newdoc = Document(docfile = request.FILES['docfile'])
newdoc.save()
# Redirect to the document list after POST
return HttpResponseRedirect(reverse('myapp.views.list'))
else:
form = DocumentForm() # A empty, unbound form
# Load documents for the list page
documents = Document.objects.all()
# Render list page with the documents and the form
return render_to_response(
'myapp/list.html',
{'documents': documents, 'form': form},
context_instance=RequestContext(request)
)
5.项目URL:myproject/url.py
默认情况下，Django不提供媒体根目录。这在生产环境中是危险的。但在开发阶段，我们可以缩短。注意最后一行。该行使Django能够从媒体URL提供文件。这只适用于开发阶段
有关详细信息，请参阅。另见
6.应用程序URL:myproject/myapp/URL.py
要使视图可访问，必须为其指定URL。这里没什么特别的
# -*- coding: utf-8 -*-
from django.conf.urls import patterns, url
urlpatterns = patterns('myapp.views',
url(r'^list/$', 'list', name='list'),
)
7.模板：myproject/myapp/templates/myapp/list.html
最后一部分：列表模板及其下方的上传表单。表单必须将enctype属性设置为“multipart/form data”，并将method设置为“post”，才能上载到Django。有关详细信息，请参阅
# -*- coding: utf-8 -*-
from django import forms
class DocumentForm(forms.Form):
docfile = forms.FileField(
label='Select a file',
help_text='max. 42 megabytes'
)
FileField有许多可在模板中使用的属性。例如{document.docfile.url}}和{{document.docfile.name}}在模板中。在和中查看有关这些的更多信息
结果
最后，一切都准备好了。在默认的Django开发环境中，可以在localhost:8000/list/
上看到上载文档的列表。今天，这些文件被上传到/path/to/myproject/media/documents/2011/12/17/并可以从列表中打开
我希望这个答案能像帮助我一样帮助别人。我必须说，我觉得django的文档很混乱。
还有一个最简单的例子，为什么要提到表单？
我在views.py中使用的示例是：-
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文件如下面的代码所示，但本示例仅上载一个文件，保存文件的代码处理多个文件：-
<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>
{%csrf\u令牌%}
文件名：
这些示例不是我的代码，它们是从我发现的另外两个示例中获得的。
我是django的相对初学者，因此我很可能遗漏了一些关键点。这里它可能会帮助您：
在models.py中创建文件字段
用于上载文件（在admin.py中）：
在模板中也使用该字段。我遇到了类似的问题，django管理站点解决了这个问题
# 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)
请参见，与Django 3一起使用
一个最小的Django文件上载示例
1.创建django项目
运行startproject:：
$ django-admin.py startproject sample
现在创建了一个文件夹（示例）
2.创建一个应用程序
创建应用程序：：
$ cd sample
$ python manage.py startapp uploader
现在创建了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上
将'uploader'
添加到已安装的应用程序
，并添加媒体根目录
和媒体URL
<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>
def save_model(self, request, obj, form, change):
url = "http://img.youtube.com/vi/%s/hqdefault.jpg" %(obj.video)
url = str(url)
if url:
temp_img = NamedTemporaryFile(delete=True)
temp_img.write(urllib2.urlopen(url).read())
temp_img.flush()
filename_img = urlparse(url).path.split('/')[-1]
obj.image.save(filename_img,File(temp_img)
# 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)
$ django-admin.py startproject sample
$ cd sample
$ python manage.py startapp uploader
uploader/
__init__.py
admin.py
app.py
models.py
tests.py
views.py
migrations/
__init__.py
INSTALLED_APPS = [
'uploader',
...<other apps>...
]
MEDIA_ROOT = os.path.join(BASE_DIR, 'media')
MEDIA_URL = '/media/'
...<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)
from django.db import models
class Upload(models.Model):
upload_file = models.FileField()
upload_date = models.DateTimeField(auto_now_add =True)
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
<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>
$ python manage.py makemigrations
$ python manage.py migrate
$ python manage.py runserver
if request.method == 'POST':
file1 = request.FILES['file']
contentOfFile = file1.read()
if file1:
return render(request, 'blogapp/Statistics.html', {'file': file1, 'contentOfFile': contentOfFile})
{% block content %}
<h1>File content</h1>
<form action="{% url 'blogapp:uploadComplete'%}" method="post" enctype="multipart/form-data">
{% csrf_token %}
<input id="uploadbutton" type="file" value="Browse" name="file" accept="text/csv" />
<input type="submit" value="Upload" />
</form>
{% endblock %}
{% block content %}
<h3>File uploaded successfully</h3>
{{file.name}}
</br>content = {{contentOfFile}}
{% endblock %}
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'
}))
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)