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我想組裝一個文章的URL並在保存實例後將它發送到某個地方。訪問ManyToMany對象用Django創建url
一切都正常,除非我無法獲取URL的一部分,'category',它應該是ManyToMany對象的第一項。
我想用這樣的事情結束了: http://www.example.com/category/article.html
代碼:
class Categories(models.Model):
...
name = models.CharField(max_length=150, blank=False)
slug = models.SlugField()
class Texts(models.Model):
...
slug = models.SlugField()
title = models.CharField(max_length=150, blank=False)
subtitle = models.TextField(blank=True)
cetegory = models.ManyToManyField(to=Categories, blank=True)
def get_absolute_url(self):
#firstpart = self.category.all() #returns empty list
#firstpart = self.category #returns empty list
#firstpart = Categories.objects.all().filter(texts__slug = self.slug) #returns empty list
lastpart = self.Slug #this is ok
return firstpart[0] + "/" + lastpart + ".html"
models.signals.post_save.connect(post_to_twitter, sender=Texts)
...是的,我知道的有關反向(),我會使用它,但首先我想要知道參數
感謝
你有什麼試過?您的網址來自您的'urls.py'及其相應的'views.py' - 並且您只發布了您的模型。 –