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Django文件上载未保存到文件系统

django-file-upload python-3.x django

0

Jahmul14 · 技术社区 · 7 年前

设置文件包括:

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

模型:

 from django.db import models 

 class Document(models.Model):
    description = models.CharField(max_length=255, blank=True)
    document = models.FileField(upload_to='documents/')
    uploaded_at = models.DateTimeField(auto_now_add=True)

形式:

from django import forms
from mysite.uploads.models import Document

class DocumentForm(forms.ModelForm):
    class Meta:
        model = Document
        fields = ['description', 'document']

 from django.shortcuts import render
 from django.http import HttpResponseRedirect
 from django.core.urlresolvers import reverse

 from mysite.uploads.models import Document
 from mysite.uploads.forms import DocumentForm

def model_upload(request):
    if request.method == 'POST':
        form = DocumentForm(request.POST, request.FILES)
        if form.is_valid():
            form.save(commit=True)
            return HttpResponseRedirect(reverse('uploadindex'))
   else:
       form = DocumentForm()
   return render(request, 'uploads/index.html', {
    'form': form
})

URL:

from django.conf.urls import url

from mysite.uploads  import views

urlpatterns = [
    url(r'^$', views.model_upload, name='uploadindex'),
]

 <html>
    <body>
    <form action="{% url "uploadindex" %}" 
       method="post"encytpe="multipart/form-data">
       {% csrf_token %}
       {{ form.as_p }}
       <button type="submit">Upload</button>
    </form>
    </body>
 </html>

2 回复 | 直到 7 年前

1

4

Casey Kinsey 7 年前

我猜你的文档表单无效,你在 document 字段,但不在模板中呈现任何反馈以表明这一点。

检查是否已在表单标记上设置了适当的编码。这是通过Django表单处理文件上传时最常见的问题:

请注意该请求。仅当请求方法为POST并且发布请求的具有属性enctype=“多部分/表单数据”时,文件才会包含数据

https://docs.djangoproject.com/en/1.11/topics/http/file-uploads/#basic-file-uploads

2

0

Arun V Jose 7 年前

...
if form.is_valid():
        form.save(commit=True)
        return HttpResponseRedirect(reverse('uploadindex'))
else:
    print form.errors
...

它将显示错误,以防 document