如何在Django imagekit中指定字段定义之外的图像规格?
我有以下模型定义:如何在Django imagekit中指定字段定义之外的图像规格?,django,django-imagekit,Django,Django Imagekit,我有以下模型定义: class Photo(models.Model) : def get_gallery_path(self, filename): ext = filename.split('.')[-1] filename = "%s.%s" % (uuid.uuid4(), ext) return 'static/uploads/images/gallery/' + filename uploader = models.F
class Photo(models.Model) :
def get_gallery_path(self, filename):
ext = filename.split('.')[-1]
filename = "%s.%s" % (uuid.uuid4(), ext)
return 'static/uploads/images/gallery/' + filename
uploader = models.ForeignKey(User, on_delete=models.CASCADE)
date = models.DateField(default=date.today)
image = ProcessedImageField(default='', verbose_name='Image', upload_to=get_gallery_path, processors=[ResizeToFill(100, 50)], format='JPEG', options={'quality': 60})
caption = models.CharField(max_length=200, null=True, blank=True)
def __str__(self):
return str(self.id) + str(self.uploader.username) + str(self.date)
class Meta:
verbose_name_plural = "Photos"
verbose_name = "Photo"
我使用django imagekit修改上传的图像。但正如您所看到的,我必须在image
字段中内联声明所有属性和选项。当我指定许多属性时,这使得代码很难阅读
是否有一种方法可以在
models.py
文件中分别定义属性和选项,以便我可以将其用于所有图像字段。您只需利用字典解包即可
例如,定义字典已处理的\u图像\u字段\u规格
:
def get_gallery_path(filename):
ext = filename.split('.')[-1]
filename = "%s.%s" % (uuid.uuid4(), ext)
return 'static/uploads/images/gallery/' + filename
processed_image_field_specs = {
'default': '',
'verbose_name': 'Image',
'upload_to': get_gallery_path,
'processors': [ResizeToFill(100, 50)],
'format': 'JPEG',
'options': {'quality': 60}
}
现在打开字典,为ProcessedImageField
提供命名参数:
class Photo(models.Model) :
...
image = ProcessedImageField(**processed_image_field_specs)
...
如果有人能告诉我一种在不影响图像质量的情况下压缩图像的方法,那将是一个很大的帮助。简单有效:)