Python Django Rest框架将模型嵌套为列表而不是dict

有没有办法将嵌套模型作为列表而不是Dict？
我正试图用ListField实现它，但遇到了困难。
下面是一个例子，以更好地解释我正在尝试做什么

示例模型：

class Album(models.Model):
    name = models.CharField(max_length=250)

class Track(models.Model):
    title = models.CharField(max_length=250)
    number = models.IntegerField()
    album = models.ForeignKey(Album, related_name="tracks")

Class TrackSerializer(serializers.ModelSerializer):
    class Meta:
    model = Track
    fields = ['number', 'title']

class AlbumSerializer(serializers.ModelSerializer):
    tracks = TrackSerializer(many=True)
    model = Album
    fields = ['name', 'tracks']

{
    "name": "ALBUM NAME",
    "tracks": [
        {
            "number": 1,
            "title": "TRACK TITLE"
        },
        {
            "number": 2,
            "title": "OTHER TRACK TITLE"
        }
    ]
}

{
    "name": "ALBUM NAME",
    "tracks": [
        [1, "TRACK TITLE"],
        [2, "OTHER TRACK TITLE"]
    ]
}

示例序列化程序：

class Album(models.Model):
    name = models.CharField(max_length=250)

class Track(models.Model):
    title = models.CharField(max_length=250)
    number = models.IntegerField()
    album = models.ForeignKey(Album, related_name="tracks")

Class TrackSerializer(serializers.ModelSerializer):
    class Meta:
    model = Track
    fields = ['number', 'title']

class AlbumSerializer(serializers.ModelSerializer):
    tracks = TrackSerializer(many=True)
    model = Album
    fields = ['name', 'tracks']

{
    "name": "ALBUM NAME",
    "tracks": [
        {
            "number": 1,
            "title": "TRACK TITLE"
        },
        {
            "number": 2,
            "title": "OTHER TRACK TITLE"
        }
    ]
}

{
    "name": "ALBUM NAME",
    "tracks": [
        [1, "TRACK TITLE"],
        [2, "OTHER TRACK TITLE"]
    ]
}

上述代码导致输出错误：

class Album(models.Model):
    name = models.CharField(max_length=250)

class Track(models.Model):
    title = models.CharField(max_length=250)
    number = models.IntegerField()
    album = models.ForeignKey(Album, related_name="tracks")

Class TrackSerializer(serializers.ModelSerializer):
    class Meta:
    model = Track
    fields = ['number', 'title']

class AlbumSerializer(serializers.ModelSerializer):
    tracks = TrackSerializer(many=True)
    model = Album
    fields = ['name', 'tracks']

{
    "name": "ALBUM NAME",
    "tracks": [
        {
            "number": 1,
            "title": "TRACK TITLE"
        },
        {
            "number": 2,
            "title": "OTHER TRACK TITLE"
        }
    ]
}

{
    "name": "ALBUM NAME",
    "tracks": [
        [1, "TRACK TITLE"],
        [2, "OTHER TRACK TITLE"]
    ]
}

所需输出：

class Album(models.Model):
    name = models.CharField(max_length=250)

class Track(models.Model):
    title = models.CharField(max_length=250)
    number = models.IntegerField()
    album = models.ForeignKey(Album, related_name="tracks")

Class TrackSerializer(serializers.ModelSerializer):
    class Meta:
    model = Track
    fields = ['number', 'title']

class AlbumSerializer(serializers.ModelSerializer):
    tracks = TrackSerializer(many=True)
    model = Album
    fields = ['name', 'tracks']

{
    "name": "ALBUM NAME",
    "tracks": [
        {
            "number": 1,
            "title": "TRACK TITLE"
        },
        {
            "number": 2,
            "title": "OTHER TRACK TITLE"
        }
    ]
}

{
    "name": "ALBUM NAME",
    "tracks": [
        [1, "TRACK TITLE"],
        [2, "OTHER TRACK TITLE"]
    ]
}

---

解决方案：带有一个字段的相册序列化程序，该字段是方法的结果，可以是任何内容（SerializerMethodField）

序列化程序.py

class AlbumSerializer(serializers.ModelSerializer):
    track_list = serializers.SerializerMethodField()
    class Meta:
        model = Album
        fields = ['name', 'track_list']
    class get_track_list(self, obj):
        output = []
        for i in Track.objects.filter(album = obj.id):
            output.append([i.number, i.title])
        return output

这将返回一个带有属性“name”和“track_list”的JSON，这正是我需要的方式：

{
    "name": "ALBUM NAME",
    "track_list": [
        [1, "TRACK TITLE"],
        [2, "OTHER TRACK TITLE"]
    ]
}

---

您试图构建的格式很奇怪，但仍然可以使用

SerializerMethodField

实现。除非您也希望它是可写的，否则在这种情况下，您必须实现自己的序列化器子类。这确实很奇怪，但我能够通过

SerializerMethodField

实现它。谢谢！