将文件从一个模型复制到另一个模型 [英] copy file from one model to another

问题描述

我有两个简单的模型：

  class UploadImage（models.Model）：
 Image = models.ImageField （upload_to =temp /）
 
 class RealImage（models.Model）：
 Image = models.ImageField（upload_to =real /）
  
 
 
 
和一种形式
  class RealImageForm ModelForm）：
 class Meta：
 model = RealImage 
  
文件从UploadImage到RealImage。我怎么可能这样做。 
以下代码不起作用
  realform.Image = UploadImage.objects.get（id = image_id）.Image 
 realform.save（）
  
 Tnx寻求帮助。
解决方案
启发Gerard的解决方案我想出了以下代码：
  from django.core.files.base import ContentFile 
 
＃... 
 
类示例（models.Model）：
 file = models.FileField （）
 
 def duplicate（self）：

复制此对象包括复制文件

 new_example = 
 new_file = ContentFile（self.file.read（））
 new_file.name = self.file.name 
 new_example.file = new_file 
 new_example.save（）
  
这实际上会通过向文件名添加一个_1来重命名文件，文件，该文件的新副本可以同时存在磁盘上。
 
I have 2 simple models:
class UploadImage(models.Model):
   Image = models.ImageField(upload_to="temp/")

class RealImage(models.Model):
   Image = models.ImageField(upload_to="real/")
And one form
class RealImageForm(ModelForm):
    class Meta:
        model = RealImage 
I need to save file from UploadImage into RealImage. How could i do this.
Below code doesn't work
realform.Image=UploadImage.objects.get(id=image_id).Image 
realform.save()
Tnx for help.
解决方案
Inspired by Gerard's solution I came up with the following code:
from django.core.files.base import ContentFile

#...

class Example(models.Model):
    file = models.FileField()

    def duplicate(self):
        """
        Duplicating this object including copying the file
        """
        new_example = Example()
        new_file = ContentFile(self.file.read())
        new_file.name = self.file.name
        new_example.file = new_file
        new_example.save()
This will actually go as far as renaming the file by adding a "_1" to the filename so that both the original file and this new copy of the file can exist on disk at the same time.

                        
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