Question

所以我有2分贝，气候有1分贝，食谱有1分贝。当我添加食谱时，我提供了与其他数据库等效的气候的ID。但在答案中，我需要实际值。我想知道最佳做法是什么？

案例1：应要求，我在另一个数据库上获取数据

情况2：我使用给定ID的实际值存储数据

情况3：另一种解决方案？

模式

type ClimateZone {
    _id: ID!
    name: String!
  }
  
  input ClimateZoneInput {
    name: String!
  }
  
  
  type Mutation {
    addClimateZone(input: ClimateZoneInput!): ClimateZone
    addRecipe(input: RecipeInput!): Recipe
   }
  
  type Query {
    allClimateZone: [ClimateZone!]!
    climateZone(id: ID!): ClimateZone!
    allRecipe: [Recipe!]!
    recipe(id: String!): Recipe!
  }
  
  type Recipe {
    _id: ID!
    variety: String!
    climate: ClimateZone!
  }
  
  input RecipeInput {
    variety: String!
    climate: ID!
  }

解析器：


const Query={
        allClimateZone:async ()=>await client.db('test').collection('climateZones').find().toArray(),
        climateZone:async (parent:any,input:any)=>await client.db('test').collection('climateZones').findOne({"_id":new ObjectID(input.id)}),
        allRecipe:async ()=>await client.db('test').collection('recipes').find().toArray(),
        recipe:async (parent:any,input:any)=> {
            return await client.db('test').collection('recipes').findOne({"_id":new ObjectID(input.id)})
        }
    }

    const Mutation={
        addClimateZone:async (parent:any,args:ClimateZoneInput)=> {
            const data=JSON.parse(JSON.stringify(args)).input
            const r = await client.db('test').collection('climateZones').insertOne(data);
            return {_id:r.insertedId,...data};
        },
        addRecipe: async (parent:any,args:RecipeInput)=> {
            const data:RecipeInput=JSON.parse(JSON.stringify(args)).input
            const r= await client.db('test').collection('recipes').insertOne(data)
            // here is my probleme
            return {_id:r.insertedId,...data};
        }
    }




    const resolvers={
     Query,
     Mutation
    }

如何使用graphql从另一个mongo数据库中获取元素

0 个答案: