2009-08-18 77 views
1

我有以下2种型号:是否可以在Django注释中的相关项上过滤?

class Job(models.Model): 
    title = models.CharField(_('title'), max_length=50) 
    description = models.TextField(_('description')) 
    category = models.ForeignKey(JobCategory, related_name='jobs') 
    created_date = models.DateTimeField(auto_now_add=True) 

class JobCategory(models.Model): 
    title = models.CharField(_('title'), max_length=50) 
    slug = models.SlugField(_('slug')) 

这是我在哪里,在与查询迄今:

def job_categories(): 
    categories = JobCategory.objects.annotate(num_postings=Count('jobs')) 
    return {'categories': categories} 

的问题是,我只想计算在创建的工作过去30天。然而,我想要返回所有类别,不仅仅是那些具有合格工作的类别。

回答

1

我决定采取不同的方法,并选择不使用注释。我向作业模型中添加了一个管理器,该管理器仅返回活动(30天或更少的旧作业),并在JobCategory模型上创建一个属性,以查询实例的作业计数。我的模板标签只是返回所有类别。这是相关的代码。

class JobCategory(models.Model): 
    title = models.CharField(_('title'), max_length=50, help_text=_("Max 50 chars. Required.")) 
    slug = models.SlugField(_('slug'), help_text=_("Only letters, numbers, or hyphens. Required.")) 

    class Meta: 
     verbose_name = _('job category') 
     verbose_name_plural = _('job categories') 

    def __unicode__(self): 
     return self.title 

    def get_absolute_url(self): 
     return reverse('djobs_category_jobs', args=[self.slug]) 

    @property 
    def active_job_count(self): 
     return len(Job.active.filter(category=self)) 

class ActiveJobManager(models.Manager): 
    def get_query_set(self): 
     return super(ActiveJobManager, self).get_query_set().filter(created_date__gte=datetime.datetime.now() - datetime.timedelta(days=30)) 

class Job(models.Model): 
    title = models.CharField(_('title'), max_length=50, help_text=_("Max 50 chars. Required.")) 
    description = models.TextField(_('description'), help_text=_("Required.")) 
    category = models.ForeignKey(JobCategory, related_name='jobs') 
    employment_type = models.CharField(_('employment type'), max_length=5, choices=EMPLOYMENT_TYPE_CHOICES, help_text=_("Required.")) 
    employment_level = models.CharField(_('employment level'), max_length=5, choices=EMPLOYMENT_LEVEL_CHOICES, help_text=_("Required.")) 
    employer = models.ForeignKey(Employer) 
    location = models.ForeignKey(Location) 
    contact = models.ForeignKey(Contact) 
    allow_applications = models.BooleanField(_('allow applications')) 
    created_date = models.DateTimeField(auto_now_add=True) 

    objects = models.Manager() 
    active = ActiveJobManager() 

    class Meta: 
     verbose_name = _('job') 
     verbose_name_plural = _('jobs') 

    def __unicode__(self): 
     return '%s at %s' % (self.title, self.employer.name) 

和标签...

def job_categories(): 
    categories = JobCategory.objects.all() 
    return {'categories': categories} 
4

只是一个猜测......但会这样工作吗?

def job_categories(): 
    thritydaysago = datetime.datetime.now() - datetime.timedelta(days=30) 
    categories = JobCategory.objects.filter(job__created_date__gte=thritydaysago).annotate(num_postings=Count('jobs')) 
    return {'categories': categories} 

有关生成查询的更多详细信息,请参阅“lookups-that-span-relationships”。 嗯...可能需要在那里获得所有类别的另一个查询...

+1

这将返回只有那些在过去30天内创建工作的类别。我也试图避免返回多个查询集,因为这违反了使用注释的目的。这个函数实际上是作为模板标签存在的,我希望在模板层中包含每个类别的计数(即使计数为0)。谢谢你。 – gsiegman 2009-08-18 13:31:58

+0

是的,我认为你只需要执行两个查询。 – monkut 2009-08-19 00:47:30

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