需要一个最小的Django文件上传示例

Question

作为Django的新手,我在Django 1.3中制作上传应用程序时遇到了困难.我找不到任何最新的示例/代码段.可能有人发布了一个最小但完整的(模型,视图,模板)示例代码吗？

Answer 1

Phew,Django文档确实没有这方面的好例子.我花了两个多小时来挖掘所有碎片,以了解它是如何工作的.凭借这些知识,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https://github.com/axelpale/minimal-django-file-upload-example或克隆它:

> git clone https://github.com/axelpale/minimal-django-file-upload-example.git

更新2013-01-30:除了1.3之外,GitHub上的源代码还实现了Django 1.4.即使几乎没有变化,下面的教程对1.4也很有用.

更新2013-05-10:在GitHub上实现Django 1.5.urls.py中重定向的微小更改以及list.html中url模板标记的使用.感谢hubert3的努力.

更新2013-12-07: GangHub支持Django 1.6.myapp/urls.py中的一个导入已更改.谢谢Arthedian.

更新2015-03-17:由于aronysidoro,GangHub支持Django 1.7 .

更新2015-09-04:由于nerogit,GangHub支持Django 1.8 .

更新2016-07-03:GangHub支持Django 1.9,感谢daavve和nerogit

项目树

一个基本的Django 1.3项目,包含单个应用程序和上传媒体/目录.

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

1.设置:myproject/settings.py

要上传和提供文件,您需要指定Django存储上传文件的位置以及Django为其提供的URL.MEDIA_ROOT和MEDIA_URL默认情况下在settings.py中,但它们是空的.有关详细信息,请参阅Django Managing Files中的第一行.记住还要设置数据库并将myapp添加到INSTALLED_APPS

...
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',
)

2.型号:myproject/myapp/models.py

接下来,您需要一个带有FileField的模型.该特定字段存储文件,例如基于当前日期和MEDIA_ROOT的媒体/文档/ 2011/12/24.请参见FileField参考.

# -*- coding: utf-8 -*-
from django.db import models

class Document(models.Model):
    docfile = models.FileField(upload_to='documents/%Y/%m/%d')

3.表单:myproject/myapp/forms.py

要很好地处理上传,您需要一个表单.这个表单只有一个字段,但这就足够了.有关详细信息,请参见Form FileField参考.

# -*- coding: utf-8 -*-
from django import forms

class DocumentForm(forms.Form):
    docfile = forms.FileField(
        label='Select a file',
        help_text='max. 42 megabytes'
    )

4.查看:myproject/myapp/views.py

所有魔法发生的视图.注意如何request.FILES处理.对我来说,很难发现request.FILES['docfile']可以保存到models.FileField 的事实.模型的save()自动处理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/urls.py

Django默认不提供MEDIA_ROOT.这在生产环境中会很危险.但在发展阶段,我们可以做空.注意最后一行.该行使Django能够从MEDIA_URL提供文件.这仅适用于开发阶段.

有关详细信息,请参阅django.conf.urls.static.static参考.另请参阅有关提供媒体文件的讨论.

# -*- coding: utf-8 -*-
from django.conf.urls import patterns, include, url
from django.conf import settings
from django.conf.urls.static import static

urlpatterns = patterns('',
    (r'^', include('myapp.urls')),
) + static(settings.MEDIA_URL, document_root=settings.MEDIA_ROOT)

6.应用网址:myproject/myapp/urls.py

要使视图可访问,您必须为其指定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:列表模板和下面的上传表单.表单必须将enctype-attribute设置为"multipart/form-data",将方法设置为"post"以使上传到Django成为可能.有关详细信息,请参阅文件上载文档

FileField有许多可以在模板中使用的属性.例如{{document.docfile.url}}和{{document.docfile.name}},如模板中所示.有关详细信息,请参阅模型中的文件文章和文件对象文档.

<!DOCTYPE html>
<html>
    <head>
        <meta charset="utf-8">
        <title>Minimal Django File Upload Example</title>   
    </head>
    <body>
    <!-- List of uploaded documents -->
    {% if documents %}
        <ul>
        {% for document in documents %}
            <li><a href="{{ document.docfile.url }}">{{ document.docfile.name }}</a></li>
        {% endfor %}
        </ul>
    {% else %}
        <p>No documents.</p>
    {% endif %}

        <!-- Upload form. Note enctype attribute! -->
        <form action="{% url 'list' %}" method="post" enctype="multipart/form-data">
            {% csrf_token %}
            <p>{{ form.non_field_errors }}</p>
            <p>{{ form.docfile.label_tag }} {{ form.docfile.help_text }}</p>
            <p>
                {{ form.docfile.errors }}
                {{ form.docfile }}
            </p>
            <p><input type="submit" value="Upload" /></p>
        </form>
    </body>
</html> 

8.初始化

只需运行syncdb和runserver.

> cd myproject
> python manage.py syncdb
> python manage.py runserver

结果

最后,一切准备就绪.在默认的Django开发环境中,可以看到上传文档的列表localhost:8000/list/.今天,文件上传到/ path/to/myproject/media/documents/2011/12/17 /,可以从列表中打开.

我希望这个答案可以帮助我,就像它会帮助我一样.

Answer 2

一般来说,当你试图"只是得到一个有效的例子"时,最好"开始编写代码".这里没有任何代码可以帮助您,因此它可以让我们更多地回答这个问题.

如果你想获取一个文件,你需要在某个地方的html文件中这样的东西:

<form method="post" enctype="multipart/form-data">
    <input type="file" name="myfile" />
    <input type="submit" name="submit" value="Upload" />
</form>

这将为您提供浏览按钮,上传按钮以启动操作(提交表单)并记下enctype,以便Django知道给你 request.FILES

在某个地方您可以使用访问该文件的视图

def myview(request):
    request.FILES['myfile'] # this is my file

文件上传文档中有大量信息

我建议您彻底阅读该页面,然后开始编写代码 - 然后在不起作用时返回示例和堆栈跟踪.

Answer 3

演示

更新AkseliPalén的回答.看到github repo,适用于Django 2

一个最小的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(样本)::

sample/
  manage.py
  sample/
    __init__.py
    settings.py
    urls.py
    wsgi.py 

2.创建一个应用程序

创建一个应用::

$ cd sample
$ python manage.py startapp uploader

现在创建一个uploader包含这些文件的文件夹()::

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添加'uploader.apps.UploaderConfig'到INSTALLED_APPS,并添加MEDIA_ROOT和MEDIA_URL,即::

INSTALLED_APPS = [
    ...<other apps>...
    'uploader.apps.UploaderConfig',
]

MEDIA_ROOT = os.path.join(BASE_DIR, 'media')
MEDIA_URL = '/media/'

4.更新urls.py

在sample/urls.py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.home, name='imageupload'),
]+ static(settings.MEDIA_URL, document_root=settings.MEDIA_ROOT)

5.更新models.py

更新uploader/models.py::

from django.db import models
from django.forms import ModelForm

class Upload(models.Model):
    pic = models.FileField(upload_to="images/")    
    upload_date=models.DateTimeField(auto_now_add =True)

# FileUpload form class.
class UploadForm(ModelForm):
    class Meta:
        model = Upload
        fields = ('pic',)

6.更新views.py

更新uploader/views.py::

from django.shortcuts import render
from uploader.models import UploadForm,Upload
from django.http import HttpResponseRedirect
from django.urls import reverse
# Create your views here.
def home(request):
    if request.method=="POST":
        img = UploadForm(request.POST, request.FILES)       
        if img.is_valid():
            img.save()  
            return HttpResponseRedirect(reverse('imageupload'))
    else:
        img=UploadForm()
    images=Upload.objects.all().order_by('-upload_date')
    return render(request,'home.html',{'form':img,'images':images})

7.创建模板

在文件夹上传器中创建文件夹模板,然后创建一个文件home.html,即::sample/uploader/templates/home.html

<div style="padding:40px;margin:40px;border:1px solid #ccc">
    <h1>picture</h1>
    <form action="#" method="post" enctype="multipart/form-data">
        {% csrf_token %} {{form}} 
        <input type="submit" value="Upload" />
    </form>
    {% for img in images %}
        {{forloop.counter}}.<a href="{{ img.pic.url }}">{{ img.pic.name }}</a>
        ({{img.upload_date}})<hr />
    {% endfor %}
</div>

8. Syncronize数据库

Syncronize数据库和runserver ::

$ python manage.py makemigrations
$ python manage.py migrate
$ python manage.py runserver

访问http://localhost.com:8000

Answer 4

我必须说我发现django的文档令人困惑.另外,对于最简单的例子,为什么要提到表格？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,虽然这个例子只上传一个文件,保存文件的代码处理很多:

<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>

这些例子不是我的代码,它们来自我发现的另外两个例子.我是django的初学者,所以我很可能错过了一些关键点.

Answer 5

延伸亨利的例子:

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

您可以handle_uploaded_file使用上传的文件对象从视图中调用此函数.这将在文件系统中使用唯一名称(带有原始上载文件的文件名)保存文件,并返回已保存文件的完整路径.您可以将路径保存在数据库中,稍后对该文件执行某些操作.

Answer 6

我也有类似的要求.网上的大多数例子都要求创建模型并创建我不想使用的表单.这是我的最终代码.

if request.method == 'POST':
    file1 = request.FILES['file']
    contentOfFile = file1.read()
    if file1:
        return render(request, 'blogapp/Statistics.html', {'file': file1, 'contentOfFile': contentOfFile})

并在HTML上传我写道:

{% 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 %}

以下是显示文件内容的HTML:

{% block content %}
    <h3>File uploaded successfully</h3>
    {{file.name}}
    </br>content = {{contentOfFile}}
{% endblock %}

Answer 7

在这里它可以帮助您:在models.py中创建一个文件字段

要上传文件(在您的admin.py中):

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)

并在模板中使用该字段.

Answer 8

您可以参考具有django版本的Fine Uploader中的服务器示例. 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.模板不包含在服务器示例中,但您可以在其网站上找到演示.Fine Uploader:http://fineuploader.com/demos.html

Django的精细上传

views.py

UploadView将post和delete请求分派给各个处理程序.

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)

Answer 9

不确定这种方法是否有任何缺点,但在views.py中更为简单:

entry = form.save()

# save uploaded file
if request.FILES['myfile']:
    entry.myfile.save(request.FILES['myfile']._name, request.FILES['myfile'], True)