为了澄清起见，我觉得我需要通过Placement模型来获得我需要的结果，因为我需要该模型的订单字段。

我正在尝试获取问题版本中属于部分的所有文章对象。即使在查看documentation django 2.x

之后，我也感到难过

我可以获得所有版本的所有版本，但是我无法弄清楚如何获得与每个版本的版块相关的文章。

对于上下文，这是我在 django视图中的工作查询而没有获取文章：

def issue_edit_form(request, iid=0):
    latest = utils.get_latest()
    issue = Issue.objects.get(pk=iid)
    editions = issue.edition_set.all()
    sections = []
    for i in editions:
        sections.append(i.section_set.all().order_by("order"))
    return render(request, 'issue_editor.html', {'latest': latest, 'issue': issue, 'editions': editions,
                                                 'sections': sections, 'article_form':
                                                     ArticleForm})

这是我的models.py

class Issue(models.Model):
        volume = models.SmallIntegerField(default=0)
        issue = models.SmallIntegerField(default=0)
        issue_date = models.DateField()
        published = models.BooleanField(default=False)

        class Meta:
            verbose_name = "   Issue"

        def __str__(self):
            return "Vol. " + str(self.volume) + " Issue: " + str(self.issue) + ": " + str(self.issue_date)


class Edition(models.Model):
        edition = models.CharField(max_length=100)
        issue = models.ForeignKey(Issue, on_delete=models.CASCADE, null=True)

        class Meta:
            verbose_name = "  Edition"

        @property
        def issue__name(self):
            return self.issue.issue_date


        def __str__(self):
            return self.edition


class Section(models.Model):
        section_name = models.CharField(max_length=100)
        section_image = models.ImageField(blank=True, null=True)
        order = models.SmallIntegerField(default=1)
        section_hidden = models.BooleanField(default=False);
        edition = models.ForeignKey(Edition, on_delete=models.CASCADE, null=True)

        class Meta:
            verbose_name = " Section"

        def __str__(self):
            return self.section_name


class Article(models.Model):
        title = models.CharField(max_length=255)
        article_url = models.URLField()
        admin_description = models.CharField(max_length=255, blank=True)
        # article_order = models.SmallIntegerField(default=1)
        sections = models.ManyToManyField(Section, through="Placement")

        class Meta:
            verbose_name = "Article"

        def __str__(self):
            return self.title


class Placement(models.Model):
        article = models.ForeignKey(Article, on_delete=models.DO_NOTHING, null=True)
        section = models.ForeignKey(Section, on_delete=models.DO_NOTHING, null=True)
        order = models.SmallIntegerField(default=1)

Answer 1

因此，如果您想要每个部分的文章列表，您可以执行此操作

edition_sections = edition.section_set.all()
for section in edition_sections:
    articles = section.articles_set.all()

或者，如果您不需要单独处理文章集，那么这也应该有效

edition_sections = edition.section_set.all()
articles = Article.objects.filter(sections__in=edition_sections)

这部分文档中有一些很好的例子 https://docs.djangoproject.com/en/2.0/topics/db/models/#intermediary-manytomany

在django中获取具有多对多关系的复杂查询集

为了澄清起见，我觉得我需要通过Placement模型来获得我需要的结果，因为我需要该模型的订单字段。

1 个答案: