Python Django Slug生成的URL返回：'；没有与给定查询匹配的基金；

我试图使用slug生成url，但是无论我在哪里访问url，我都会得到“Page not found at/funds/[slug\u generated\u value]”

Raised by:  apps.funds.views.details
No Fund matches the given query

这是my models.py，我使用fund.name生成fund_id以创建首字母缩略词，并添加DateDate的datetime值减去小时、分钟、秒值：

class Fund(models.Model):
    name = models.CharField(max_length=128, unique=True)
    description = models.TextField()
    duration = models.CharField(max_length=10)
    slug = models.SlugField(unique=True)

    def save(self, *args, **kwargs):
        self.slug = slugify(self.name)
        super(Fund, self).save(*args, **kwargs)

    def __str__(self):
        return self.name

class OfferedFunds(models.Model):
    fund_id = models.CharField(max_length=55, blank=True)
    fund_name = models.ForeignKey(Fund)
    deadline = models.DateTimeField('Deadline')
    slug = models.SlugField(unique=True)

    def save(self, *args, **kwargs):
        output = ''
        for i in self.fund_name.name.upper().split():
            output += i[0]
        deadline = str(self.deadline)[:-15] 
        deadline = deadline.replace('-', '')
        random_number = random.randint(1,9)
        self.fund_id = '%s%s%s' % (output, deadline, random_number)
        self.slug = slugify(self.fund_id)
        super(OfferedFunds, self).save(*args, **kwargs)

    class Meta:
        verbose_name_plural = "Offered funds"

    def __str__(self):
        return '%s' % (self.fund_id)

和我对应的views.py：

def details(request, slug):
    context_dict = {}

    fund = get_object_or_404(Fund, slug=slug)
    context_dict['fund'] = fund

    offered_funds = OfferedFunds.objects.filter(fund_name=fund, deadline__gte=timezone.now()).order_by('deadline')
    context_dict['offered_funds'] = offered_funds

    return render(request, 'funds/details.html', context_dict)

def of_details(request, slug):
    context_dict = {}

    of = get_object_or_404(OfferedFunds, slug=slug)
    context_dict['of'] = of

    return render(request, 'funds/offered-funds-details.html', context_dict)

最后是my URL.py：

url(r'^(?P<slug>[\w-]+)/$', views.details, name='details'),
url(r'^(?P<slug>[\w-]+)/$', views.of_details, name='of details'),

url（r'^（？P[\w-]+）/$，views.details，name='details'），
url（r'^（？P[\w-]+）/$，views.of_details，name='of details'），

---

My details（）视图工作正常，但_details（）视图不起作用。如果我取出details（）视图，则_details（）的作用是有效的。有人知道我如何解决这个错误吗？谢谢

好的，正如用户评论中指出的那样，我的URL太相似了，所以我改变了：

url(r'^(?P<slug>[\w-]+)/$', views.of_details, name='of details'),

url（r'^（？P[\w-]+）/$，views.of_details，name='of details'），

致：

url（r'^det/（？P[\w-]+）/$，views.of_details，name='of details'），

---

这是有效的。

Django如何区分相同的URL？您的两个正则表达式是相同的，因此当Django搜索适用的URL匹配项时，它将始终实现

视图。详细信息

函数就是这样。更改为

1r'^det/（？P[\w-]+）/$”

，效果非常好

url(r'^det/(?P<slug>[\w-]+)/$', views.of_details, name='of details'),