Django：如何将 upload_to=function 与 ModelForm 一起使用

Question

目标是动态更新 upload_to ，以便用户上传的文件存储在取决于用户的目录位置。网上有几个这样的例子，但没有一个使用 ModelForm。请参阅两个问题的代码片段，其中一个是我得到的 instance.user 值为空字符串，当我尝试修复该问题时，表单无效。

# models.py

def get_file_path( instance, filename ):
    # make the filepath include the signed in username
    print "inst: %s" % instance.__dict__.keys()
    print "inst:user:%s" % instance.user   # <-- This is empty string!!
    print "file: %s" % filename
    return "%s/myapp/%s/%s" % ( settings.MEDIA_ROOT, instance.user, filename )

class trimrcUpload(models.Model):
    user = models.CharField( max_length = 20 )
    inputFile = models.FileField( upload_to = get_file_path )


# forms. py

class trimrcUploadForm(ModelForm):

    class Meta:
        model = trimrcUpload
        exclude = ( 'resultFile', 'numTimesProcessed' )

# views.py

def myapp_upload( request, username, template_name="myapp/myapptemplate.html" ):

    dummy = trimrcUpload( user=username )
    if request.POST:
        form = trimrcUploadForm( request.POST, request.FILES, instance=dummy )
        if form.is_valid():
            success = form.save()
            success.save()

    # form is not valid, user is field is "required"
    # the user field is not displayed in the template by design,
    # it is to be populated by the view (above).

# http://docs.djangoproject.com/en/1.0/topics/forms/modelforms/
# about halfway down there is a "Note" section describing the use of dummy.

Answer 1

我想你的问题来自于尝试用用户名填充模型的用户属性。如果上传表单始终由登录用户使用，您可以使用以下选项：

dummy = trimrcUpload( user=request.user )

否则，如果您仍然想像现在一样传递用户名，您可以尝试以下操作：

try:
    user = User.objects.get(username=username)
    dummy = trimrcUpload( user=user )
except User.DoesNotExist:
    # Probably have to set some kind of form error

我建议使用第一个选项这将使您不必将用户名传递给视图。

Answer 2

问题中的原始代码实际上有效。