Javascript 在另一个文件上使用函数，无需 背景：

我有3个文件，

parent.js

，

child1.js

，和

child2.js

let child1 = require("./child1.js")
let child2 = require("./child2.js")
let key = "*****"

child1.start(key)
child2.start();

let key = false;

module.exports = {
    action: async() => {
        return someApi.get(key);
    },
    start: async(_key) => {
        key = key;
    }
}

module.exports = {
    action: async() => {
        let res = await child1.action()
        ...
    },
    start: async() => {
        // startup actions
    }
}

parent.js

let child1 = require("./child1.js")
let child2 = require("./child2.js")
let key = "*****"

child1.start(key)
child2.start();

let key = false;

module.exports = {
    action: async() => {
        return someApi.get(key);
    },
    start: async(_key) => {
        key = key;
    }
}

module.exports = {
    action: async() => {
        let res = await child1.action()
        ...
    },
    start: async() => {
        // startup actions
    }
}

child1.js

let child1 = require("./child1.js")
let child2 = require("./child2.js")
let key = "*****"

child1.start(key)
child2.start();

let key = false;

module.exports = {
    action: async() => {
        return someApi.get(key);
    },
    start: async(_key) => {
        key = key;
    }
}

module.exports = {
    action: async() => {
        let res = await child1.action()
        ...
    },
    start: async() => {
        // startup actions
    }
}

child2.js

let child1 = require("./child1.js")
let child2 = require("./child2.js")
let key = "*****"

child1.start(key)
child2.start();

let key = false;

module.exports = {
    action: async() => {
        return someApi.get(key);
    },
    start: async(_key) => {
        key = key;
    }
}

module.exports = {
    action: async() => {
        let res = await child1.action()
        ...
    },
    start: async() => {
        // startup actions
    }
}

问题 我需要在

child2

内部运行

child1

中的函数，但我不能使用

require

，因为只能有一个

child1

---

有人知道这个问题的解决方案吗？谢谢

我想你问错问题了：）
如果您需要child1成为独一无二的，您应该使用单音，并在任何需要的地方使用它。

//child1
let instance

module.export = () => {
  if(!instance) {
     instance = {
    action: async() => {
        return someApi.get(key);
    },
    start: async(_key) => {
        key = key;
    }
   }
   return instance
}

我个人不喜欢单音法，但它很方便

您可以尝试的另一种方法是将child1服务实例注入child2“构造函数”
您只需导出函数而不是对象
parent.js

let child1 = require("./child1.js")()
let child2 = require("./child2.js")(child1)
let key = "*****"

child1.start(key)
child2.start();

child2

module.exports = (child1) => {
    
    const action =  async() => {
        let res = await child1.action()
        ...
    }
    const start =  async() => {
        // startup actions
    }
    return {action, start}
}