Django ImageField传递一个可以上传到upload_to的图像

问题描述

我正在尝试将自定义的upload_to函数传递给我的模型imageField，但我想将函数定义为模型函数....是可能的吗？

$ _ code class MyModel（models.Model）：
...
image = models.ImageField（upload_to = self.get_image_path）
...

def get_image_path（self，filename）：
...
return image_path

现在我知道我不能用'self'来引用它，因为在这一点上自己不存在...有没有办法呢？如果没有，那么定义该函数的最佳位置在哪里？

解决方案

所以只需删除@classmethod，Secator的代码将会工作

  class MyModel（models.Model）：
 
＃需要在字段前定义
 def get_image_path（self，filename）：
＃'self'将工作，因为Django明确地传递它。 
返回文件名
 
 image = models.ImageField（upload_to = get_image_path）
  

I'm trying to pass a custom upload_to function to my models imageField but I'd like to define the function as a model function....is that possible?

class MyModel(models.Model):
    ...
    image = models.ImageField(upload_to=self.get_image_path)
    ...

    def get_image_path(self, filename):
        ...
        return image_path

Now i know i can't reference it by 'self' since self doesn't exist at that point...is there a way to do this? If not - where is the best place to define that function?

解决方案

So Just remove "@classmethod" and Secator's code will work.

class MyModel(models.Model):

    # Need to be defined before the field    
    def get_image_path(self, filename): 
        # 'self' will work, because Django is explicitly passing it.
        return filename

    image = models.ImageField(upload_to=get_image_path)

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