python - 如何在 Django 中输出 FilePathField 的完整路径？

Question

考虑以下模型:

STOCK_IMAGE_DIR = os.path.join(settings.MEDIA_ROOT, 'stock_images')

class Product(models.Model):
    stock_image = models.FilePathField(path=STOCK_IMAGE_DIR, default='image.png')

访问产品的stock_image时，仅返回图像名称:

>>> p = Product.objects.first()
>>> p.stock_image
'image.png'

如何输出文件的整个路径？

Answer 1

import os
p = Product.objects.first()
print(os.path.join(Product._meta.get_field('stock_image').path, p.stock_image))

或者

print(os.path.join(STOCK_IMAGE_DIR, p.stock_image))

或者，如果您将路径存储在类中，您可以添加一个属性来获取完整路径

class Product(models.Model):

    STOCK_IMAGE_DIR = os.path.join(settings.MEDIA_ROOT, 'stock_images')

    stock_image = models.FilePathField(path=STOCK_IMAGE_DIR, default='image.png')

    @property
    def stock_image_path(self):
        return os.path.join(self.STOCK_IMAGE_DIR, self.stock_image)

然后你就可以使用该属性

p = Product.objects.first()
print(p.stock_image_path)