用以下代码在JavaScript中实现工厂模式的正确方法

Question

下面的代码应该为涉及Arduino的项目抽象一个温度传感器。

我有表示硬件传感器的传感器函数，用于创建传感器实例的sensorFactory函数，以及存储/从数组中返回传感器对象的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：

为什么在运行行

1. 时不能看到温度变量的值？

const SensorFactory =需要量(“./senor.js”)；var sensorFactory = SensorFactory()；sensorFactory.createSensor (1，"DS18B20"，“低温储罐”)；var sensor1 = sensorFactory.getSensor(1)；senor1.readTemperature( 200 )；senor1.getTemperature(200)；//prints 200 sensor1.readTemperature( 100 )；//print 100 sensor1.getTemperature()；sensorFactory.createSensor 1.sensor1.getTemperature( 10 )；//打印10个温度传感器(10)；

预期：

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

实际产出：

{"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，因此不应该从对象外部访问它，但在运行console.log(JSON.stringify(sensor1))时不应该看到温度变量(B)吗？

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;
};

1. 我对JS很生疏。你能帮我弄清楚我在概念上哪里错了吗？如何修改代码以使其按预期工作？

Answer 1

@Bergi的第二个评论是我的问题的解决方案。我还遵循了他的建议，并消除了返回sensorFactory对象的sensorFactory函数。代码现在很清楚了。

Answer 2

这里有一些重构到现代Javascript (回答第二个问题，并作为一个副作用，也是第一个;)

​

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))

​