Question

我指的是这个youtube video，了解如何使用ImageField上传图片。他已经解释了如何在保存图像的同时使用instance.id。我试过了，但instance.id正在返回None。而对他来说它完美无缺。以下是代码：

#models.py
import os

def get_image_path(instance, filename):
    return os.path.join(str(instance.id), filename)

class AdProfile(models.Model):
    name = models.CharField(max_length=100)
    profile_image = models.ImageField(upload_to=get_image_path, blank=True, null=True)

每当保存文件时，保存为None/filename。

即使这个link通知也是如此。我正在使用Django 10.5和MySQL数据库。

可能是什么问题？

Answer 1

Django admin以某种方式调用了get_image_path函数而没有将模型保存到数据库，因此id为None。我们可以使用save方法覆盖django模型，并确保保存图像，get_image_path获取id为

的实例

class AdProfile(models.Model):
    name = models.CharField(max_length=100)
    profile_image = models.ImageField(upload_to=get_image_path, blank=True, null=True)

    # Model Save override 
    def save(self, *args, **kwargs):
        if self.id is None:
            saved_image = self.profile_image
            self.profile_image = None
            super(AdProfile, self).save(*args, **kwargs)
            self.profile_image = saved_image
            if 'force_insert' in kwargs:
                kwargs.pop('force_insert')

        super(AdProfile, self).save(*args, **kwargs)

Answer 2

使用Raja Simon的答案，可以使用模型来处理模型中的所有FileField

class MyModel(models.Model):

    file_field = models.FileField(upload_to=upload_to, blank=True, null=True)

    def save(self, *args, **kwargs):
        if self.id is None:
            saved = []
            for f in self.__class__._meta.get_fields():
                if isinstance(f, models.FileField):
                    saved.append((f.name, getattr(self, f.name)))
                    setattr(self, f.name, None)

            super(self.__class__, self).save(*args, **kwargs)

            for name, val in saved:
                setattr(self, name, val)
        super(self.__class__, self).save(*args, **kwargs)

此外，我们可以使文件位置动态化，即不仅基于self.id，而且还基于外键的id或其他任何值。只需遍历字段并检查路径是否已更改。

def upload_to(o, fn):
    if o.parent and o.parent.id:
        return parent_upload_to(o.parent, fn)

    return "my_temp_dir/{}/{}".format(o.id, fn)


class MyModel(models.Model):

    parent = models.ForeignKey(Parent)

    def save(self, *args, **kwargs):

        # .... code from save() above here

        for f in [f for f in self.__class__._meta.get_fields() if isinstance(f, models.FileField)]:

            upload_to = f.upload_to

            f = getattr(self, f.name)  # f is FileField now

            if f and callable(upload_to):
                _, fn = os.path.split(f.name)
                old_name = os.path.normpath(f.name)
                new_name = os.path.normpath(upload_to(self, fn))

                if old_name != new_name:

                    old_path = os.path.join(settings.MEDIA_ROOT, old_name)
                    new_path = os.path.join(settings.MEDIA_ROOT, new_name)

                    new_dir, _ = os.path.split(new_path)
                    if not os.path.exists(new_dir):
                        print "Making  dir {}", new_dir
                        os.makedirs(new_dir)

                    print "Moving {} to {}".format(old_path, new_path)
                    try:
                        os.rename(old_path, new_path)
                        f.name = new_name

                    except WindowsError as e:
                        print "Can not move file, WindowsError: {}".format(e)

        super(self.__class__, self).save(*args, **kwargs)

Django - 在上传图片时使用instance.id.

2 个答案: