Question

我的所有对象newmedia都以相同名称保存。我该如何解决这个问题？

#view.py
if request.method == 'POST':
    formset = MediaFormSet(request.POST, request.FILES)
    if formset.is_valid():
        Page_key = Page.objects.get(pk=1)
        slide = SlideshowComponent(page=Page_key, order=0, label="slideshow", x=0, y=0, width=0, height=0, viewport_type="simle_page", keywords="slideshow")
        slide.save()      
        for filename, file in request.FILES.iteritems():
            name = request.FILES[filename].name               
        for form in formset.forms:        
            file_type = file.content_type
            if file_type == "image/png" or file_type == "image/jpeg" or file_type == "image/gif":
                newmedia = formset.save(commit=False)
                newmedia.filename = name
                newmedia.content_type = "photos"
                newmedia.save()

Answer 1

我猜你省略了一个表格：

    for filename, file in request.FILES.iteritems():
        name = request.FILES[filename].name               
        for form in formset.forms:        
            file_type = file.content_type
            if file_type == "image/png" or file_type == "image/jpeg" or file_type == "image/gif":
                newmedia = formset.save(commit=False)
                newmedia.filename = name
                newmedia.content_type = "photos"
                newmedia.save()

否则它始终保留您处理的最后一个文件的名称。

编辑：事实上，我想这不是你想要的。您应该跟踪已经处理的formset.forms，然后指定与尚未处理的formset.forms一起的名称。

Answer 2

相反，我有2个，我只做一个。

for file in request.FILES.getlist('form-0-source'):
    file_type = file.content_type
    if file_type == "image/png" or file_type == "image/jpeg" or file_type == "image/gif":
        name= file.name.split('.')[0]
        newmedia = Media(source=file, filename=name, content_type = "photos", created='03/25/12')
        newmedia.save()                  
        image2 = ImageComponent(page=Page_key, order=0, label="imagem", x=0, y=0, width=0, height=0, viewport_type="simle_page", keywords="imagem", media=newmedia, is_slideshow='true')
        image2.save()
        slide.image.add(image2)                 
    else:
        return render_to_response('revista_digital/error.html', context_instance=RequestContext(request))

django formset中的多个文件名

2 个答案: