我的django模型如下所示:
class Article(model.Model):
slug = models.SlugField(db_index=True, max_length=255, unique=True)
title = models.CharField(db_index=True, max_length=255)
body = models.TextField()
tags = models.ManyToManyField(
'articles.Tag', related_name='articles'
)
def __str__(self):
return self.title
class Tag(model.Model):
tag = models.CharField(max_length=255)
slug = models.SlugField(db_index=True, unique=True)
def __str__(self):
return self.tag
还有我的schema.py:
class ArticleType(DjangoObjectType):
class Meta:
model = Article
class Query(ObjectType):
article = graphene.Field(ArticleType, slug=graphene.String())
def resolve_article(self, info, slug):
article = Article.objects.get(slug=slug)
return article
使用以下方法查询此模型:
query {
article(slug: "my_slug") {
id
title
body
slug
tagList: tags {
tag
}
}
}
产生:
{
"data": {
"article": {
"id": "1",
"title": "How to train your dragon 1",
"slug": "how-to-train-your-dragon-y41h1x",
"tagList": [
{
"tag": "dragon",
"tag": "flies"
}
]
}
}
}
**问题:**如何自定义返回的json输出?特别地,tagList是“tag”键是多余的对象的列表。相反,我想返回一个字符串列表,以使输出变为:
{
"data": {
"article": {
"id": "1",
"title": "How to train your dragon 1",
"slug": "how-to-train-your-dragon-y41h1x",
"tagList": ["dragon","flies"]
}
}
}
我怎样才能做到这一点??
最佳答案
使用可返回字符串列表的解析器,将自定义tag_list字段添加到ArticleType
中。就像是:
class ArticleType(DjangoObjectType):
tag_list = graphene.List(graphene.String)
class Meta:
model = Article
def resolve_tag_list(self, info):
return [tag.tag for tag in self.tags.all()]
关于django - 获取查询以返回值列表,而不是graphene-django中的对象,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/51762817/