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我打算干掉我下次重建数据库相关的名字......我现在用的模型更加的试验模型。因此,有了Class Creator和作者,cover_artist等,我将如何去展示Creator创建的问题(一旦我摆脱了相关的名称,除非有办法绕过它)?显示具有不同名称的多对多子集?在Django模板
class Creator(models.Model):
name = models.CharField(max_length=256)
desc = models.TextField("description", blank=True, null=True)
#type = writer, artist, editor, letterer
slug = models.SlugField(blank=True, null=True)
def __unicode__(self):
return self.name
class Meta:
ordering = ['name']
def get_absolute_url(self):
return "/comics2/creators/%s" % self.slug
class Issue(models.Model):
....
writer = models.ManyToManyField(Creator, related_name="written by", help_text="Use cmd/ctrl + click to select multiple names. The same applies to the rest of the creators below.", blank=True, null=True)
artist = models.ManyToManyField(Creator, related_name="drawn by", blank=True, null=True)
cover_artist = models.ManyToManyField(Creator, related_name="cover drawn by", blank=True, null=True)
colorist = models.ManyToManyField(Creator, related_name="colored by", blank=True, null=True)
inker = models.ManyToManyField(Creator, related_name="inked by", blank=True, null=True)
penciller = models.ManyToManyField(Creator, related_name="pencilled by", blank=True, null=True)
letterer = models.ManyToManyField(Creator, related_name="lettered by", blank=True, null=True)
editor = models.ManyToManyField(Creator, related_name="edited by", blank=True, null=True)
arc = models.ManyToManyField(Arc, related_name="related story arc", blank=True, null=True)
...
def __unicode__(self):
return u'%s %s' % (self.title, self.number)
def get_absolute_url(self):
return "/comics2/issues/%s" % self.slug
class Meta:
ordering = ['title', 'number']
def get_stars(self):
star_rating = self.rating * 16
return star_rating
....
{% for issue in creator.____?__.all %}
<ul>
<li>{{ issue }}</li>
</ul>
{% endfor %}
是行不通的。
谢谢。
这似乎有点混乱,因为我不知道如何作用将看在Django的示例模型后作家,艺术家等区分,在模板中。这是我现在为我准备的:http://amlaplant.webfactional.com/comics2/issues/astonishing-x-men-1/ – AAA 2011-04-19 08:30:15
谢谢。这可以让我显示创作者通过模板参与的问题吗?我会怎么做呢? – AAA 2011-04-19 09:24:21
是的。问题相同。 – DTing 2011-04-19 11:17:07