使用当前模型ID的Django管理文件上传

Question

我正在尝试使用默认的Django
admin创建一个简单的相册。我想为每个画廊保存一张样本照片，但是我不想保留filname。除了文件名，我想保存模型的ID（N.jpg）。但是，第一次我要保存该对象时，该ID不存在。我如何知道模型中的下一个自动增量，或者以某种方式保存模型数据（在存在super.save文件时上传之前和之后）self.id？有没有很酷的解决方案？

像这样：

def upload_path_handler(instance, filename):
    ext = filename extension
    return "site_media/images/gallery/{id}.{ext}".format(id=instance.nextincrement, ext=ext)

class Gallery(models.Model):
    name  = models.TextField()
    image = models.FileField(upload_to=upload_path_handler)

也许将文件名存储在其他字段中。

Answer 1

图像文件将在Gallery实例之前保存。因此，您必须通过使用带有状态的Gallery实例本身的信号将保存分为两个阶段：

from django.db.models.signals import post_save, pre_save
from django.dispatch import receiver

_UNSAVED_FILEFIELD = 'unsaved_filefield'

@receiver(pre_save, sender=Image)
def skip_saving_file(sender, instance, **kwargs):
    if not instance.pk and not hasattr(instance, _UNSAVED_FILEFIELD):
        setattr(instance, _UNSAVED_FILEFIELD, instance.image)
        instance.image = None

@receiver(post_save, sender=Image)
def save_file(sender, instance, created, **kwargs):
    if created and hasattr(instance, _UNSAVED_FILEFIELD):
        instance.image = getattr(instance, _UNSAVED_FILEFIELD)
        instance.save()        
        # delete it if you feel uncomfortable...
        # instance.__dict__.pop(_UNSAVED_FILEFIELD)

upload_path_handler看起来像

def upload_path_handler(instance, filename):
    import os.path
    fn, ext = os.path.splitext(filename)
    return "site_media/images/gallery/{id}{ext}".format(id=instance.pk, ext=ext)

如果字段仅用于图像上传，我建议使用ImageField而不是FileField进行类型检查。另外，您可能希望规范化文件名扩展名（由于mimetype而不需要），例如

def normalize_ext(image_field):
    try:
        from PIL import Image
    except ImportError:
        import Image
    ext = Image.open(image_field).format
    if hasattr(image_field, 'seek') and callable(image_field.seek):
       image_field.seek(0)
    ext = ext.lower()
    if ext == 'jpeg':
        ext = 'jpg'
    return '.' + ext