鉴于此架构:
const typeDefs = gql`type Asset {
id: ID!
name: String!
ScanProfiles: [ScanProfile]
}
type ScanProfile {
id: ID!
name: String!
Assets: [Asset]
}
type Query {
Asset(id: ID!): Asset
ScanProfile(id: ID!): Asset
}`;
// Some generic resolvers...
const resolvers = {
Asset: (root, args, context) => context.get('asset', args),
ScanProfile: (root, args, context) => context.get('scanProfile, args),
Query: {
Asset: (root, args, context) => context.get('asset', args),
ScanProfile: (root, args, context) => context.get('scanProfile', args)
}
};
const schema = makeExecutableSchema({typeDefs, resolvers
const joinResolvers = {
Asset: {
ScanProfiles: (root, args, context) => {
console.log(root.ScanProfiles);
return root.ScanProfiles.map(id => context.get('scanProfile', {id});
}
},
ScanProfiles: {
Assets: (root, args, context) => {
console.log(root.id);
return context.filter('asset', {ScanProfiles: ({ScanProfiles}) => ScanProfiles.includes(root.id)});
}
}
};
addResolveFunctionsToSchema({ schema, resolvers: joinResolvers });
我希望context.get
给定的 id 返回这些形状:
// Asset
{
id: '1234',
name: 'abcd',
ScanProfiles: ['5678', '5679']
}
//ScanProfile
{
id: '5678',
name: 'efgh'
}
控制台正在返回[null]
。Asset.ScanProfiles
根解析器将字段作为字符串数组返回,但是当根随后传递给joinResolvers
这些文件时,这些文件为空。我假设是因为它们与定义的 typeDef 不匹配。
要解决此问题,我必须单独调用以重新获取父级,然后为子级提取存储的值。
我是否错误地定义了我的类型?我的解析器结构错了吗?