我目前正在使用以下代码,从 action.payload
中获取一些属性并分配给对象weather
。
我想知道在 ES6 中是否有更优雅、更简洁的方法来执行相同的操作。
let weather = {}
weather.name = action.payload.name
weather.country = action.payload.sys.country
weather.temperature = action.payload.main.temp
weather.temperatureMin = action.payload.main.temp_min
weather.temperatureMax = action.payload.main.temp_max
weather.weatherMain = action.payload.weather[0].main
weather.weatherDescription = action.payload.weather[0].description
weather.weatherIcon = action.payload.weather[0].icon
weather.updatedTime = new Date().toString()
weather.windDegree = action.payload.wind.deg
weather.windSpeed = action.payload.wind.speed
weather.visibility = action.payload.visibility
最佳答案
这里没有特定的语言级别的魔法。您正在从 action
对象的多个级别进行拉取,并动态重命名许多属性。如果名称保持不变,对象解构和结构化赋值可能会彻底解决问题。但就目前情况而言,您可以获得的最大清晰度是不要重复这么多名称。例如:
let payload = action.payload
let weather = {
name: payload.name,
country: payload.sys.country,
temperature: payload.main.temp,
temperatureMin: payload.main.temp_min,
temperatureMax: payload.main.temp_max,
weatherMain: payload.weather[0].main,
weatherDescription: payload.weather[0].description,
weatherIcon: payload.weather[0].icon,
updatedTime: new Date().toString(),
windDegree: payload.wind.deg,
windSpeed: payload.wind.speed,
visibility: payload.visibility
}
当然可以使用更多 ES2015 解构功能作为此作业的一部分。例如:
const { payload } = action
const { main, weather, wind, sys } = payload
let weather = {
name: payload.name,
country: sys.country,
temperature: main.temp,
temperatureMin: main.temp_min,
temperatureMax: main.temp_max,
weatherMain: weather[0].main,
weatherDescription: weather[0].description,
weatherIcon: weather[0].icon,
updatedTime: new Date().toString(),
windDegree: wind.deg,
windSpeed: wind.speed,
visibility: payload.visibility
}
我通常不建议采取进一步的步骤。在我看来,这并不一定能让事情变得更清楚。至少,需要考虑一些权衡。原始物体与其多层结构之间的自然对应关系变得更加隐藏;人们需要记住并解释诸如 wind
、weather
、main
和 sys
等子对象的起源。但从好的方面来说,它在 ES2015 的使用中更简洁、更激进。您必须判断这是更好还是更奇特。
关于javascript - ES6中如何简化对象映射?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/43708503/