__in 过滤器只返回一个值，通过中间显示查询 table

Question

编程新手，需要帮助。我试图通过过滤模型 Spots 来渲染视图 article。我有一个中间 table ArticleSpots 到 link 2 tables Spots 和 Articles。在视图 article 中，我只想显示 link 编辑到该特定文章的位置。我的问题是 Spots.objects.filter(id__in=articleSpots) 只显示第一个值，而不是 linked 的所有点。我在这里做错了什么？

views.py

def article(request, slug):
    articles = get_object_or_404(Articles, slug=slug)
    article_id = articles.id
    articleSpots = ArticleSpots.objects.filter(article__id=article_id)    
    spots = Spots.objects.filter(id__in=articleSpots)
    
    context = {"spots": spots, "articles": articles}
    template_name = "articletemplate.html"
    return render(request, template_name, context)

models.py

class ArticleSpots(models.Model):
    article = models.ForeignKey('Articles', models.DO_NOTHING)
    spot = models.ForeignKey('Spots', models.DO_NOTHING)

    class Meta:
        managed = True
        db_table = 'article_spots'
        verbose_name_plural = 'ArticleSpots'
        
    def __str__(self):
        return str(self.article) + ": " + str(self.spot)

class Articles(models.Model):
    title = models.CharField(max_length=155)
    metatitle = models.CharField(max_length=155)
    slug = models.SlugField(unique=True, max_length=155)
    summary = models.TextField(blank=True, null=True)
    field_created = models.DateTimeField(db_column='_created', blank=True, null=True)  
    field_updated = models.DateTimeField(db_column='_updated', blank=True, null=True)  
    cover = models.ImageField(upload_to="cover", blank=True, default='logo-00-06.png')

    class Meta:
        managed = True
        db_table = 'articles'
        verbose_name_plural = 'Articles'
        
    def __str__(self):
        return str(self.id) + ": " + str(self.title)

class Spots(models.Model):
    title = models.CharField(max_length=155)
    metatitle = models.CharField(max_length=155)
    slug = models.SlugField(unique=True, max_length=155)
    author = models.ForeignKey(Authors, models.DO_NOTHING)
    field_created = models.DateTimeField(db_column='_created', blank=True, null=True)  
    field_updated = models.DateTimeField(db_column='_updated', blank=True, null=True)  
    cover = models.ImageField(upload_to="cover", blank=True, default='logo-00-06.png')
    summary = models.TextField(blank=True, null=True)
    content1 = models.TextField(blank=True, null=True)
    content2 = models.TextField(blank=True, null=True)
    

    class Meta:
        managed = True
        db_table = 'spots'
        verbose_name_plural = 'Spots'
        
    def __str__(self):
        return str(self.id) + ": " + str(self.title)

html

<!-- START MAIN -->
    <main class="page"></main>
    <p>
      {{ spots.title }} <br />
      {{ spots.content1 }} <br />
      {{ articles.title }}
    </p>
    {% for spots in spots %} {{ spots.title}} {% endfor %}
<!-- END MAIN -->

Answer 1

您当前检索的 Spots 与 ArticleSpots 对象具有相同的主键，但这没有多大意义：可能是这种情况，但即使那样碰巧，返回的 Spots 本身并没有链接到给定文章的相关 ArticleSpots。

您可以通过以下方式检索相关的 Spots：

def article(request, slug):
    article = get_object_or_404(Articles, slug=slug)
    spots = Spots.objects.filter(<strong>articlespots__article=article</strong>)
    context = {'spots': spots, 'article': article}
    return render(request, 'articletemplate.html', context)

我强烈建议您将对象命名为 Article 对象 article 因为它是一个单个 Article，而不是 Article秒。 spots另一方面是斑点的集合。

呈现 {{ spots.content1 }} 和 {{ spots.title }} 没有任何意义，因为 spots 是 Spots 的集合，可以包含零个、一个或多个项目。

因此模板应如下所示：

<p>
    {{ <strong>article.title</strong> }}
</p>
{% for <strong>spot in spots</strong> %} {{ <strong>spot</strong>.title}} {% endfor %}

Note: normally a Django model is given a singular name, so Articles instead of ~~Article~~.

__in 过滤器只返回一个值，通过中间显示查询 table

__in filter only returning one value, show query through intermediate table

python

django

django-models

django-queryset

django-views