小编Mun*_*eeb的帖子

我正在使用Django休息招摇.在rest框架UI中添加了一个显示给序列化程序的图像.但是不要出现在Swagger UI中.

在序列化程序中添加了以下字段.upload_photo = serializers.ImageField(required = False)

但昂首阔步显示一个简单的文本字段.休息框架显示上传字段.任何人都可以帮助我在swagger ui中显示上传字段

Django Rest Swagger无法将Inner Serializer解析为对象数组,而只显示字符串列表

我的序列化器:

class InfluencerSerializer(serializers.Serializer):
     prices = PriceSerializer(many=True)
     first_name = serializer.CharField(max_length=100)

class PriceSerializer(serializers.Serializer):
     cost = serializers.IntegerField(default=0)

在Swagger UI上,它在Example中显示为json下面

{ 
  "first_name": "string", 
  "prices": ["string"],
}

虽然我期望Swagger UI显示{"first_name":"string","price":[{"cost":0}],}

我在用

Django==1.10.6
djangorestframework==3.6.1
django-rest-swagger==2.1.2

我已经用这个查询了 Django 模型

news = News.objects.filter(Q(likes__user__isnull=True)|Q(likes__user=user))
.extra(select={"is_liked":NewsLikes._meta.db_table+".user_id = %d" % user.id})

这给了我以下查询

SELECT (shows_newslikes.user_id = 143) AS `is_liked`, * FROM `shows_news` 
LEFT OUTER JOIN `shows_newslikes` ON ( `shows_news`.`id` = `shows_newslikes`.`news_id`)
WHERE (`shows_newslikes`.`user_id` IS NULL OR `shows_newslikes`.`user_id` = 143 )

我想要的是以下查询作为结果

SELECT (shows_newslikes.user_id = 143) AS `is_liked`, * 
FROM `shows_news` LEFT OUTER JOIN `shows_newslikes` ON ( `shows_news`.`id` = 
`shows_newslikes`.`news_id` and `shows_newslikes`.`user_id` = 143 ) WHERE 
(`shows_newslikes`.`user_id` IS NULL  )

那么我在查询 Django 模型时必须做什么