python - Django，外键的子集

Question

我需要一种方法来选择外键的子集，但仅限于一个子集。这是旧 PHP 数据库中使用的原始查询；

SELECT a.name FROM character_trait b
LEFT JOIN trait a ON b.id_trait = a.id
WHERE b.id_character = 1
AND a.id_traittype = 10

所有的表、character、trait、traittype 和character_trait 都可用，但我不知道如何在Djando 中实现。我的想法是这样的；

traits = CharacterTrait.objects.filter( id_character = character, id_trait.id_traittype = 10 )

但这只会给出一个错误“关键字不能是一个表达式”。我可以用一种非常丑陋的方式来做到这一点，只需使用像这样的 for 循环遍历生成的“特征”；

traits = CharacterTrait.objects.filter( id_character = character )  

for t in traits:
    print t.id_trait.id_traittype.id

    if t.id_trait.id_traittype.id == 10:
        print "Got One"

编辑，模块定义；

class Trait(models.Model):
    id = models.IntegerField(primary_key=True)  # AutoField? 
    id_traittype = models.ForeignKey(Traittype, null=True, db_column = 'id_traittype')

    name = models.CharField(max_length=32)

    class Meta:
        managed = False
        db_table = 'trait'

    def __unicode__(self):
                return self.name

class Traittype(models.Model):
    id = models.IntegerField(primary_key=True)  # AutoField?
    name = models.CharField(max_length=64)

    class Meta:
        managed = False
        db_table = 'traittype'

    def __unicode__(self):
                return self.name    

class Trait(models.Model):
    id = models.IntegerField(primary_key=True)  # AutoField? 
    id_traittype = models.ForeignKey(Traittype, null=True, db_column = 'id_traittype')

    name = models.CharField(max_length=32)

    class Meta:
        managed = False
        db_table = 'trait'

    def __unicode__(self):
                return self.name

class CharacterTrait(models.Model):
    id = models.IntegerField(primary_key=True)  # AutoField?
    ##id_character = models.IntegerField()
    ## id_trait = models.IntegerField()
    id_character = models.ForeignKey(Werewolfcharacter, null=True, db_column = 'id_character')
    id_trait     = models.ForeignKey(Trait,             null=True, db_column = 'id_trait')

    class Meta:
        managed = False
        db_table = 'character_trait'

    def __unicode__(self):
                return self.id_trait.name  

Answer 1

我打算建议你需要链接过滤器，比如

traits = CharacterTrait.objects.filter( id_character = character).filter(id_trait__id_traittype = 10 )

但是回顾一些代码，

traits = CharacterTrait.objects.filter( id_character = character, id_trait__id_traittype = 10 )

应该足够了，关键是替换 .按照 dm03514 的建议使用 __