在以下代码中 JavaScript 中实现工厂模式的正确方法

Question

以下代码应该为涉及 Arduino 的项目抽象温度传感器。

我有代表硬件传感器的 Sensor 函数，一个创建传感器实例的 sensorFactory 函数，以及一个存储/returns 数组中的传感器对象的 saveSensor / findSensor 函数。

var sensors = [];
const Sensor = (sensorId, sensorType, sensorName) =>{
    var temperature;
    let sensor = {
        sensorId,
        sensorType,
        sensorName,
        temperature,
        readTemperature: (temp) => {
            temperature = temp //simulates reading from hardware
        },
        getTemperature: () => {
            console.log(temperature)
        }
    }
    return sensor; 
};
const SensorFactory = () =>({
    createSensor: (sensorId, sensorType, sensorName) => {
        var sensor = {};
        switch(sensorType){
            case "DS18B20":         
                sensor = Sensor(sensorId,sensorType, sensorName);
                saveSensor(sensor);
                return findSensor(sensorId);
                break;
        }
    },
    getSensor: (sensorId) =>{
    if (sensorId){
            return findSensor(sensorId);            
        }
        return sensors;
    }
});
const saveSensor = (sensor) => {
    sensors.push(sensor);
};
const findSensor = (sensorId) => {
    return sensors[sensorId -1];
};
module.exports = SensorFactory;

我有两个问题：

为什么我在运行行 console.log(JSON.stringify(sensor1)) 时看不到温度变量的值？

const SensorFactory = require ("./sensor.js");
var sensorFactory   = SensorFactory();
sensorFactory.createSensor (1,"DS18B20", "Cryogenic Tank");

var sensor1 = sensorFactory.getSensor(1);
sensor1.readTemperature(200);
sensor1.getTemperature(); //prints 200  
sensor1.readTemperature(100); //prints 100
sensor1.getTemperature();   
sensor1.readTemperature(10); //prints 10
sensor1.getTemperature();

console.log(JSON.stringify(sensor1))

预计：

{"sensorId":1,"sensorType":"DS18B20","sensorName":"Cryogenic Tank", "temperature": 10}

实际输出：

{"sensorId":1,"sensorType":"DS18B20","sensorName":"Cryogenic Tank"}

我认为以下块中的温度 (A) 是一个私有变量，因此不应从对象外部访问它，但是当我运行 console.log(JSON.stringify(sensor1))?

const Sensor = (sensorId, sensorType, sensorName) =>{
    var temperature; <----A----private. 
    let sensor = {
        sensorId,
        sensorType,
        sensorName,
        temperature, <----B----I should be able to see this, right?
        readTemperature: (temp) => {
            temperature = temp;
            console.log('readTemperature');             
        },
        getTemperature: () => {             
            console.log('getTemperature');
            console.log(temperature);           }
    }
    return sensor;
};

我对 JS 很生疏。你能帮我理解我在概念上错在哪里吗？您将如何修改代码以使其按预期工作？

Answer 1

这是对现代 Javascript 的一些重构（回答了第二个问题，并且，作为副作用，也回答了第一个问题；）

class Sensor {
    constructor(id, type, name) {
        this.id = id
        this.type = type
        this.name = name
        this.temperature = null
    }

    readTemperature(temp) {
        this.temperature = temp //simulates reading from hardware
    }
}


class SensorSet extends Map {
    add(id, type, name) {
        this.set(id, new Sensor(id, type, name))
        return this.get(id)
    }
}

let sensors = new SensorSet()

sensors.add(1, "DS18B20", "Cryogenic Tank")

let s1 = sensors.get(1)

s1.readTemperature(200)
console.log(s1.temperature)

s1.readTemperature(100)
console.log(s1.temperature)

s1.readTemperature(10)
console.log(s1.temperature)

console.log(JSON.stringify(s1))

Answer 2

@Bergi 的第二个评论是我问题的解决方案。我也听从了他的建议并删除了 returns sensorFactory 对象的 sensorFactory 函数。代码现在清楚多了。

在以下代码中 JavaScript 中实现工厂模式的正确方法

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