在 GraphQL 中使用特定字符串搜索特定结果

Question

我想使用我的查询 getFoodType 进行搜索，以根据特定餐厅/外卖店的 foodType 是否为 "Chicken","Pizza" 等返回结果

点赞foodType: "Chicken"

我尝试过使用参数和 mongoDB 过滤器（它是一个 MongoDB 服务器），但没有成功。

Schema

const EaterySchema = new Schema({
  name: {
    type: String,
    required: true
  },
  address: {
    type: String,
    required: true
  },
  foodType: {
    type: String,
    required: true
  }
});

我的架构类型

  type Eatery {
    id: String!
    name: String!
    address: String!
    foodType: String!
  }
  type Query {
    eatery(id: String!): Eatery
    eateries: [Eatery]
    getFoodType(foodType: String): [Eatery]
  }

我的Resolver

    getFoodType: () => {
      return new Promise((resolve, reject) => {
        Eatery.find({})
          .populate()
          .exec((err, res) => {
            err ? reject(err) : resolve(res);
          });
      });
    },

Apollo Playground 中的当前查询

{
   getFoodType (foodType: "Chicken") {
     id
     name
     address
     foodType
   }
 }

我基本上想以“鸡”作为foodType 返回所有结果。类似foodType: "Chicken"。

Answer 1

首先需要在Resolver中获取要查询的foodType的值

const resolvers = {
  Query: {
    getFoodType: (_, args) => {
      const { foodType } = args
      ...
    },
  },
}

然后查询时使用foodType

Eatery.find({ foodType })

最后需要返回结果

new Promise((resolve, reject) => {
  return Eatery.find({ foodType })
    .populate()
    .exec((err, res) => {
      err ? reject(err) : resolve(res)
    })
})

完整示例

const resolvers = {
  Query: {
    getFoodType: (_, args) => {
      const { foodType } = args
      return new Promise((resolve, reject) => {
        return Eatery.find({ foodType })
          .populate()
          .exec((err, res) => {
            err ? reject(err) : resolve(res)
          })
      })
    },
  },
}

使用async/await

const resolvers = {
  Query: {
    getFoodType: async (_, { foodType }) => {
      try {
        const eaterys = await Eatery.find({ foodType }).populate()
        return eaterys
      } catch (e) {
        // Handling errors
      }
    },
  },
}