django表单有多个文件字段

How do you handle multiple file fields in Django. For example if I had only one form field i would pass the request.FILES['file'] to a handling function. But what is the way to go when there are more files?

你如何处理Django中的多个文件字段。例如，如果我只有一个表单字段，我会将request.FILES ['file']传递给处理函数。但是当有更多文件时，该怎么办？

3 个解决方案

#1

I'm late to the party, but I've been trying to figure this out for a while and finally have a solution. Have a look at the code used here: https://code.djangoproject.com/ticket/12446

我迟到了，但我一直试图弄清楚这一点，最后有一个解决方案。看看这里使用的代码：https：//code.djangoproject.com/ticket/12446

You can access multipart values with getlist. If my HTML form was:

您可以使用getlist访问多部分值。如果我的HTML表单是：

<form enctype="multipart/form-data" action="" method="post">
<input type="file" name="myfiles" multiple>
<input type="submit" name="upload" value="Upload">
</form>

My django code to process it would look like:

我的django代码处理它看起来像：

for afile in request.FILES.getlist('myfiles'):
    # do something with afile

Writing a form field/widget to handle this properly is my next step. I'm still rather new to using Django, so I'm learning as I go.

编写表单字段/窗口小部件以正确处理此问题是我的下一步。我仍然是使用Django的新手，所以我正在学习。

#2

request.FILES.get('filename', None) responds to the existence of a form-named field like this:

request.FILES.get（'filename'，None）响应存在一个表单命名的字段，如下所示：

<input type="file" name="filename"></input>

If you had two such fields:

如果您有两个这样的字段：

<input type="file" name="file1"></input>
<input type="file" name="file2"></input>

Then request.FILES.get('file1', None) and request.FILES.get('file2', None) should give you those files respectively.

然后request.FILES.get（'file1'，None）和request.FILES.get（'file2'，None）应该分别给你那些文件。

The reason for this is multipart mime. The three parts (form data, file1, file2) should be uploaded and Django's UploadFileHandler splits this apart into request.POST and request.FILES respectively.

原因是多部分哑剧。应该上传这三个部分（表单数据，文件1，文件2），Django的UploadFileHandler分别将它拆分为request.POST和request.FILES。

#3

Here's is a good link for this answer: https://github.com/Chive/django-multiupload. However, since I was not using ModelForm I had to make few changes. In view I have written below code and saved files to disk:

这是这个答案的一个很好的链接：https：//github.com/Chive/django-multiupload。但是，由于我没有使用ModelForm，我不得不做一些改动。在视图中，我已将下面的代码和保存的文件写入磁盘：

for each in form.cleaned_data['attachments']:
    handle_uploaded_file(each)

def uploadMyWork(request):
    from .forms import UploadFileForm, handle_uploaded_file
    print 'in uploadMyWork'

    if request.method == 'GET':
        print 'in uploadMyWork : GET'
        form = UploadFileForm()
    else:
        form = UploadFileForm(request.POST, request.FILES)
        print 'in uploadMyWork : POST'
        #for each in form.cleaned_data['attachments']:
        #    handle_uploaded_file(each)
        #return render(request, 'stitchme/uploadMyWork.html', {'msg': "file uploaded successfully"})

        if form.is_valid():
             print 'inside form valid'
             for each in form.cleaned_data['attachments']:
                 handle_uploaded_file(each)
             return render(request, 'stitchme/uploadMyWork.html', {'msg': "file uploaded successfully"})

    print 'returning to uploadmywork'
    return render(request, 'stitchme/uploadMyWork.html', {'form': form, 'msg':'hi'})

#1