将参数传递给Django中的upload_to函数 [英] Passing parameter to upload_to function in Django

问题描述

我尝试将一个实例的多个图像上传到不同的子文件夹.但是我还需要重命名每个上载的文件，因此我实现了一个针对upload_to字段的函数，如下所示.

I try to upload multiple images of an instance to different sub folders. But I need to rename each uploaded file as well so I implement a function for upload_to field as below.

class MyModel(models.Model):
    code = models.CharField()
    logo = models.FileField(upload_to=get_path)
    cover = models.FileField(upload_to=get_path)

def get_path(instance, filename):
    ext = filename.split('.')[-1]
    new_name = "%s.%s" % (slughifi(filename), ext)
    return new_name

但是，我不确定如何将图像划分为子文件夹，例如logos和cover_images 最好将参数传递给像这样的get_path函数

However, I'm not sure how I can divide images to subfolders like logos and cover_images It'd better if I pass a parameter to get_path function like

    ...
    logo = models.FileField(upload_to=get_path("logos/"))
    cover = models.FileField(upload_to=get_path("cover_images/"))
    ...

我需要为每个文件字段编写不同的upload_to函数吗?

Do I need to write different upload_to functions for each file field?

推荐答案

我认为不，

您可以这样使用它:

upload_to=lambda s,f: MyModel.upload_to(s,f,your_custom_path)


@classmethod
def upload_to(cls, obj, filename, custom_path):
    name,extension = os.path.splitext(filename)
    new_name = "%s.%s" % (slughifi(filename), extension)
    return "{}{}".format(custom_path, new_name)

这篇关于将参数传递给Django中的upload_to函数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！