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如何解决GraphQL中的命名冲突?

graphql-tools graphql javascript

left click · 技术社区 · 6 年前

import { makeExecutableSchema } from 'graphql-tools';

const fish = { length:50 },
      rope = { length:100 };

const typeDefs = `
    type Query {
        rope: Rope!
        fish: Fish!
    }

    type Mutation {
        increase_fish_length: Fish!
        increase_rope_length: Rope!
    }

    type Rope {
        length: Int!
    }

    type Fish {
        length: Int!
    }
`;

const resolvers = {
    Mutation: {
        increase_fish_length: (root, args, context) => {
            fish.length++;
            return fish;
        },
        increase_rope_length: (root, args, context) => {
            rope.length++;
            return rope;
        }
    }
};

export const schema = makeExecutableSchema({ typeDefs, resolvers });

增加长度 而不是和 增加绳索长度 .

鱼/增加长度 和 绳索/增加长度

GraphQl是否支持命名空间的任何解决方案?

2 回复 | 直到 6 年前

yussenn 6 年前

Graphql不支持类似名称空间的东西

Dan Crews 6 年前

我一直在玩弄关于名称空间的一些想法。如果您的类型定义看起来像这样:

type Mutation {
    increase_length: IncreaseLengthMutation!
}

type IncreaseLengthMutation {
    fish: Fish!
    rope: Rope!
}

你的分解器是这样的:

const resolvers = {
    Mutation: {
        increase_Length: () => {
            return {}
        }
    },
    IncreaseLengthMutation {
        fish: (root, args, context) => {
            fish.length++;
            return fish;
        },
        rope: (root, args, context) => {
            rope.length++;
            return rope;
        }
    }
};