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Python 从另一面获取多对多的一面_Python_Django_Django Models - Fatal编程技术网

Python 从另一面获取多对多的一面

Python 从另一面获取多对多的一面,python,django,django-models,Python,Django,Django Models,我将以下三个模型与多对多关系联系起来: class TeamMember(models.Model): member_name = models.CharField(max_length=140) title = models.CharField(max_length=140, null=True, blank=True) email = models.EmailField() report = models.ManyToManyField(Report, thr

我将以下三个模型与多对多关系联系起来:

class TeamMember(models.Model):
    member_name = models.CharField(max_length=140)
    title = models.CharField(max_length=140, null=True, blank=True)
    email = models.EmailField()
    report = models.ManyToManyField(Report, through='Owner')

class Report(models.Model):
    report_name = models.CharField(max_length=140)
    report_description = models.TextField(null=True, blank=True)
    last_updated = models.DateTimeField(null=True, blank=True)
    metric = models.ManyToManyField(Metric, through='MetricAssociation')

class Owner(models.Model):
    teammember = models.ForeignKey(TeamMember, on_delete=models.CASCADE) 
    report = models.ForeignKey(Report, on_delete=models.CASCADE)
我想建立一个团队成员的对象列表,这些成员属于某个特定的报告(由所有者指定),但遇到了问题。目前我有报告的ID。我试过这个,但似乎不对:

owners = TeamMember.objects.filter(report__id=id).distinct().values_list('name', flat=True)

TeamMember
模型中没有字段
name
。您有
成员名称

owners = TeamMember.objects.filter(report_id=id).distinct().values_list('member_name', flat=True)

另一方面,您可以使用
report\u id
而不是需要联接操作的
report\u id

TeamMember
模型中没有字段
name
。您有
成员名称

owners = TeamMember.objects.filter(report_id=id).distinct().values_list('member_name', flat=True)
另一方面,您可以使用
report\u id
而不是需要联接操作的
report\u id