Question

我正在尝试打印品牌和它的链接，我正在使用嵌套循环，但它没有显示正确的输出

这是我的代码

{% for brand_report in brand_reports %}
  <h1> REPORT for Brand: {{ brand_report.brand.name }}</h1>
  {% for link in brand_report.links.all %}
    <h3>link:</h3><a href="{{ link.url }}">{{link.url}}</a>
  {% endfor %}
{% endfor %}

所需的输出

REPORT for Brand: SizCom
link:
https://www.justdial.com/Kozhikode/SizCom-Near-to-Big-Bazaar-Calicut-City/0495PX495-X495-151217184614-Z4I2_BZDET
REPORT for Brand: SizCom
link:
https://www.justdial.com/Kozhikode/SizCom-Near-to-Big-Bazaar-Calicut-City/0495PX495-X495-151217184614-Z4I2_BZDET

但我得到这样的输出

REPORT for Brand: SizCom
REPORT for Brand: SizCom
link:
https://www.justdial.com/Kozhikode/SizCom-Near-to-Big-Bazaar-Calicut-City/0495PX495-X495-151217184614-Z4I2_BZDET
link:
https://www.justdial.com/Kozhikode/SizCom-Near-to-Big-Bazaar-Calicut-City/0495PX495-X495-151217184614-Z4I2_BZDET

不知道我做错了什么

模型：

class Brand(models.Model):
"""For the brand"""

name = models.CharField(max_length=500)
description = models.TextField(blank=True, null=True)
context = models.ManyToManyField(Context, blank=True)
variation = models.ManyToManyField('BrandVariation', blank=True, related_name='brand')
status = models.BooleanField(default=False)

def __str__(self):
    return self.name

class ResultLink(models.Model):
    """for results"""
    url = models.URLField(max_length=1000)
    country = models.ForeignKey(Country, blank=True, null=True)
    parent_site = models.ForeignKey(Domain, on_delete=models.CASCADE, blank=True, null=True)
    variation = models.ForeignKey(BrandVariation, on_delete=models.CASCADE, blank=True, null=True)
    brand = models.ForeignKey(Brand, on_delete=models.CASCADE, blank=True, null=True, related_name='link')
   date = models.DateField(auto_now_add=True)

  def __str__(self):
     return "%s" % self.url

class BrandReport(TimeStampedModel):
""""""
brand = models.ForeignKey(Brand, related_name='counterfeit_report')
links = models.ManyToManyField(ResultLink, blank=True)

class Meta:
    verbose_name = "Brand Report"

观点：

def view(request,brands):
  template_path = 'profile_brand_report.html'
  brand_reports = BrandReport.objects.filter(brand__id__in=brands)
  context = {'brand_reports': brand_reports}
  html = render_to_string(template_path, context)

Answer 1

我重新创建了你的项目。你唯一错过的是你的观点。所以我想这就是你的问题所在。以下是我的观点：

def demoview(request):
    objs = BrandReport.objects.all()
    c = {
        "brand_reports": objs,
    }
    return render(request, "blog/demoview.html", c)

这是我的模板：

{% for brand_report in brand_reports %}
  <h1> REPORT for Brand: {{ brand_report.brand.name }}</h1>
  {% for link in brand_report.links.all %}
    <h3>link:</h3><a href="{{ link.url }}">{{link.url}}</a>
  {% endfor %}
{% endfor %}

我的结果正是你所期望的。请检查您的观点。

Answer 2

你显然没有任何相关链接的第一个BrandReport和第二个带有两个相关链接的BrandReport - 检查你的数据库。

FWIW，我实际上并没有得到你的BrandReport模型，因为你已经拥有了ResultLink中的Brand的外键。您的BrandReport模型不仅是多余的（据我们所知），而且还很脆弱，因为它允许任何ReportLink与任何品牌相关联，而不考虑ReportLink.brand约束。

django模板中的嵌套循环未显示正确的输出

2 个答案: