在 django 中处理图像上传表单：何时使用 save()、chunks() 和 clean_data？

发布于 2024-09-14 01:49:11 字数 1868 浏览 10 评论 0原文

我已使用以下代码成功上传了图像：

views.py

from django.conf.urls.defaults import *
from django.http import HttpResponse, HttpResponseRedirect
from django.shortcuts import render_to_response
from django import template
from django.template import RequestContext

from mysite.uploadr.forms import UploadFileForm

def upload_file(request):
    if request.method == 'POST':
        form = UploadFileForm(request.POST, request.FILES)
        if form.is_valid():
            form.handle_uploaded_file(request.FILES['file'])
            return HttpResponse(template.Template('''
                        <html><head><title>Uploaded</title></head> <body>
                        <h1>Uploaded</h1>
                        </body></html>
                        '''
                        ).render( template.Context({}))
                    )
    else:  
        form = UploadFileForm()
    return render_to_response('upload.html', {'form': form}, context_instance=RequestContext(request))

forms.py

from django import forms
from settings import MEDIA_ROOT

class UploadFileForm(forms.Form):
    title = forms.CharField(max_length = 50)
    file = forms.FileField()

    def handle_uploaded_file(self,file):
        #print type(file), "file.name=",file.name
        #print dir(file)
        destination = open(MEDIA_ROOT + '/images/'+file.name, 'wb+')
        for chunk in file.chunks():
            destination.write(chunk)

我想更进一步，将图像与正在上传的用户关联起来。我看过一些例子，并且喜欢这篇文章中的技术：https://stackoverflow .com/questions/3348013/django-image-file-uploads。

我注意到他们的代码使用了 save() 和 clean_data。是否不需要像文档中的示例那样迭代块并写入目标文件夹？我必须使用cleaned_data吗？只是想找出最有效的上传文件的方法，我已经看到了很多不同的方法。非常感谢您的帮助。

原文

I have successfully uploaded an image using the following code:

views.py

from django.conf.urls.defaults import *
from django.http import HttpResponse, HttpResponseRedirect
from django.shortcuts import render_to_response
from django import template
from django.template import RequestContext

from mysite.uploadr.forms import UploadFileForm

def upload_file(request):
    if request.method == 'POST':
        form = UploadFileForm(request.POST, request.FILES)
        if form.is_valid():
            form.handle_uploaded_file(request.FILES['file'])
            return HttpResponse(template.Template('''
                        <html><head><title>Uploaded</title></head> <body>
                        <h1>Uploaded</h1>
                        </body></html>
                        '''
                        ).render( template.Context({}))
                    )
    else:  
        form = UploadFileForm()
    return render_to_response('upload.html', {'form': form}, context_instance=RequestContext(request))

forms.py

from django import forms
from settings import MEDIA_ROOT

class UploadFileForm(forms.Form):
    title = forms.CharField(max_length = 50)
    file = forms.FileField()

    def handle_uploaded_file(self,file):
        #print type(file), "file.name=",file.name
        #print dir(file)
        destination = open(MEDIA_ROOT + '/images/'+file.name, 'wb+')
        for chunk in file.chunks():
            destination.write(chunk)

I'd like to go one step further and associate an image with the user who is uploading. I've seen a few examples and have taken a liking to the technique in this post: https://stackoverflow.com/questions/3348013/django-image-file-uploads.

I noticed that their code uses save() and cleaned_data. Is it not necessary to iterate thru the chunks and write to the destination folder like the examples in the documentation? Do I have to use cleaned_data? Just trying to figure out the most efficient means of uploading files, I 've seen so many different ways of doing this. Your help you be greatly appreciated.

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

假扮的天使 2024-09-21 01:49:11

当文件大于 settings.FILE_UPLOAD_MAX_MEMORY_SIZE 时需要分块（django 1.2 中默认为 2.5M）

请查看 django.core.files.storage.FileSystemStorage 类。它的 save() 方法为您执行按块保存的工作并执行正确的文件锁定。

storage = FileSystemStorage(
                    location = '/var/www/site/upfiles', 
                    base_url = '/upfiles'
                  )


content = request.FILES['the_file']
name = storage.save(None, content) #you can use some suggested name instead of
                                   #None. content.name will be used with None
url = storage.url(name)   #<-- get url for the saved file

在旧版本的 django 中（例如 1.0），文件名的生成存在缺陷。它不断在文件名中添加_，并且如果重复上传相同的文件，上传的文件名会变得越来越长。这似乎在 1.2 版本中得到了修复。

Chunking is needed when file is bigger than settings.FILE_UPLOAD_MAX_MEMORY_SIZE (default 2.5M in django 1.2)

Take a look at django.core.files.storage.FileSystemStorage class. It's save() method does the chunk-wise saving job for you and does the proper file locking.

storage = FileSystemStorage(
                    location = '/var/www/site/upfiles', 
                    base_url = '/upfiles'
                  )


content = request.FILES['the_file']
name = storage.save(None, content) #you can use some suggested name instead of
                                   #None. content.name will be used with None
url = storage.url(name)   #<-- get url for the saved file

In the older versions of django (e.g in 1.0) there was a defect in generation of file names. It kept adding _ to file names and the uploaded file name got longer and longer if you upload the same file repeatedly. This seems to be fixed in version 1.2.

回复收藏 0 原文

鲜肉鲜肉永远不皱 2024-09-21 01:49:11

通过直接访问 request.FILES['file']，您将绕过 UploadFileForm 正在执行的任何处理（您甚至不需要表单类来处理此类文件）。 form.cleaned_data['file'] 将访问已处理（如果添加了 clean 方法，则已清理）的表单数据。您还可以直接访问 request.POST 字典，而不是表单数据。除非您有充分的理由，否则最好使用清理后的表单数据。

在您提供的示例中，还使用了一个模型（这就是调用 save() 方法的模型），并且该模型的字段正在处理文件访问。如果您想将有关上传文件的信息保存在数据库中，那么就可以这样做。

您可以使用内置文件存储 API 来保存文件： http://docs.djangoproject.com/en/dev/ref/files/storage/#ref-files-storage。

此外，简单地使用用户指定的文件名调用 open(MEDIA_ROOT + '/images/'+file.name, 'wb+') 也不是一个好主意。这只是要求目录遍历或其他问题。