Django模型：从外键使用字段 [英] Django-models: use field from foreign key

问题描述

我正在为DJango制作一个图像库项目，只是为了它的缘故。而且，我有一个类名为 Gallery 和一个名为 ImageGallery 的类。

名为gallery的类将如下所示：

  （models.Model）：
 gallery = models.ForeignKey（Gallery，related_name =parent_gallery）
 title = models.CharField（max_length = 200）
 folder = models.CharField（max_length = 200）＃ex：images / gallery / slugify（self.title）
 
 class ImageGallery（models.Model）：
 gallery = models.ForeignKey（Gallery，related_name =parent_gallery）
 title = models.CharField（max_length = 200）
 image = models.ImageField（upload_to = self.gallery.folder）
  

嗯，最后一行代码是我想知道的，如果它可能或任何其他精细替换。

在 DJango-admin 中，我想添加表 ImageGallery ，而在选择画廊时，我想使用，将图像保存到画廊中指定的文件夹。文件夹字段。

有什么最好的方法来解决这个问题？我没有完成两个班的写作，但我怀疑他们会这样工作。先谢谢你。 :-)

解决方案

FileField.upload_to 定义如下

此属性提供了一种设置上传目录和文件
名称的方式，可以通过两种方式进行设置。在这两种情况下，值将
传递给Storage.save（）方法。 ... upload_to也可以是一个可调用的
，如函数。这将被调用来获取上传路径，
包括文件名。这个可调用必须接受两个参数，
返回一个Unix样式的路径（带有斜杠），以传递到
存储系统。两个论点是：

但是 self.gallery.folder 不是可打电话您需要的是沿着该示例中给出的线设置函数

  def get_upload_path（instance，filename）：
 return'{0} / {1}'。format（instance.gallery.folder，filename）
  

您的模型将更改为

  image = models.ImageField（upload_to = get_upload_path）
  

I'm working on an image gallery project in DJango, just for the sake of it. And well, I have a class named Gallery and a class named ImageGallery.

The class named gallery would look like this:

class Gallery(models.Model):
 gallery = models.ForeignKey(Gallery, related_name="parent_gallery")
 title = models.CharField(max_length=200)
 folder = models.CharField(max_length=200) # ex: images/galleries/slugify(self.title)

class ImageGallery(models.Model):
 gallery = models.ForeignKey(Gallery, related_name="parent_gallery")
 title = models.CharField(max_length=200)
 image = models.ImageField(upload_to=self.gallery.folder)

Well, the last line of code is what I want to know if its possible or any other fine replacement for it.

In the DJango-admin I want to be able to add records for the table ImageGallery and when selecting the Gallery I would like to use, the images to be saved to the folder specified in gallery.folder field.

What's the best way to go around this? I haven't completed writing the two classes but I doubt they're gonna work like this. Thank you in advance. :-)

解决方案

The FileField.upload_to is defined as follows

This attribute provides a way of setting the upload directory and file name, and can be set in two ways. In both cases, the value is passed to the Storage.save() method. ... upload_to may also be a callable, such as a function. This will be called to obtain the upload path, including the filename. This callable must accept two arguments and return a Unix-style path (with forward slashes) to be passed along to the storage system. The two arguments are:

But self.gallery.folder isn't a callable. What you need is to setup a function along the lines given in that example

def get_upload_path(instance, filename):
    return '{0}/{1}'.format(instance.gallery.folder, filename)

And your model will change to

image = models.ImageField(upload_to=get_upload_path)

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