访问ForeignKey属性

Question

我有两个模型。基本上，我是从这个讨论扩展而来的：每个模型多个图像我想获得设计模型的machine_name的值，并将其用于在DesignImage模型中创建ImageField upload_to的路径。某种效果：image = models.ImageField(upload_to='designs/' + design.get_machine_name + '/images')

这将产生以下错误：AttributeError: 'ForeignKey' object has no attribute 'get_machine_name'。

我希望下面的代码是清晰的。

from django.db import models
from django.db.models.fields.related import ManyToManyField
from django.template.defaultfilters import slugify
class Design(models.Model):
    name = models.CharField(max_length=30, help_text='Name of the design.')
    machine_name = models.SlugField(editable=False)
    def save(self, *args, **kwargs):
        self.machine_name = slugify(self.name)
        super(Design, self).save(*args, **kwargs)
    def get_machine_name(self):
        return self.machine_name
    def __unicode__(self):
        return self.name
class DesignImage(models.Model):
    def get_upload_dir(self):
        return 'designs/' + self.design.machine_name + '/images'
    design = models.ForeignKey(Design, related_name='images')
    caption = models.CharField(max_length=100)
    image = models.ImageField(upload_to='designs/' + design.get_machine_name + '/images')
    def __unicode__(self):
        return self.caption

我已经包含了以下代码，以防它可能有用：

from designs.models import *
from django.contrib import admin
class DesignImageInline(admin.TabularInline):
    model = DesignImage
    extra = 3 # number of extra inline form fields to display
class DesignAdmin(admin.ModelAdmin): 
    inlines = [ DesignImageInline, ]
admin.site.register(Design, DesignAdmin)
admin.site.register(DesignImage)

Answer 1

如果ImageField上传到的位置取决于实例(就像这里所做的那样(，则upload_to参数可以是可调用的，而不是字符串，在字符串中它将传递当前实例和文件名。你需要这样的东西：

import os 
def upload_to(instance, filename):
    return os.path.join('designs', instance.design.machine_name, 'images', filename)
...
    image = models.ImageField(upload_to=upload_to)

Answer 2

问题出在这一行：

image=模型。ImageField(upload_to='designs/'+design.get_machine_name+'/images'(

字段设计不是在"编译时"定义的，它是模型中的另一个字段。

您需要在运行时将其应用于您正在使用的每个表单中进行编辑。