如何在javascript中将时间戳转换为天数？答案

【问题标题】：How to convert timestamp into days in javascript?如何在javascript中将时间戳转换为天数？
【发布时间】：2020-07-09 10:31:48
【问题描述】：

我得到 endtimestamp 作为 api 响应，我想从当前时间戳计算天数。我已经为此转换编写了一个函数，但给了我磨损的结果。

function timeDifference(current, previous) {

        var msPerMinute = 60 * 1000;
        var msPerHour = msPerMinute * 60;
        var msPerDay = msPerHour * 24;
        var msPerMonth = msPerDay * 30;
        var msPerYear = msPerDay * 365;
    
        var elapsed = current - previous;
    
        if (elapsed < msPerMinute) {
             return Math.round(elapsed/1000) + ' seconds ago';   
        }
    
        else if (elapsed < msPerHour) {
             return Math.round(elapsed/msPerMinute) + ' minutes ago';   
        }
    
        else if (elapsed < msPerDay ) {
             return Math.round(elapsed/msPerHour ) + ' hours ago';   
        }
    
        else if (elapsed < msPerMonth) {
            return '' + Math.round(elapsed/msPerDay) + ' days ago';   
        }
    
        else if (elapsed < msPerYear) {
            return '' + Math.round(elapsed/msPerMonth) + ' months ago';   
        }
    
        else {
            return '' + Math.round(elapsed/msPerYear ) + ' years ago';   
        }
    }

请建议我正确的解决方案，以便我可以得到正确的转换

【问题讨论】：

考虑附加示例输入和您期望的相应输出。

标签： javascript timestamp

【解决方案1】：

我建议使用一个库作为 moment.js https://momentjs.com/docs/#/displaying/difference/ 除了更容易编码之外，您不会遇到闰年、闰秒等问题。

【讨论】：

【解决方案2】：

function timeDifference(date) {
    var seconds = Math.floor((new Date() - date) / 1000);
    if(Math.round(seconds/(60*60*24*365.25)) >= 2) return Math.round(seconds/(60*60*24*365.25)) + " years ago";
    else if(Math.round(seconds/(60*60*24*365.25)) >= 1) return "1 year ago";
    else if(Math.round(seconds/(60*60*24*30.4)) >= 2) return Math.round(seconds/(60*60*24*30.4)) + " months ago";
    else if(Math.round(seconds/(60*60*24*30.4)) >= 1) return "1 month ago";
    else if(Math.round(seconds/(60*60*24*7)) >= 2) return Math.round(seconds/(60*60*24*7)) + " weeks ago";
    else if(Math.round(seconds/(60*60*24*7)) >= 1) return "1 week ago";
    else if(Math.round(seconds/(60*60*24)) >= 2) return Math.round(seconds/(60*60*24)) + " days ago";
    else if(Math.round(seconds/(60*60*24)) >= 1) return "1 day ago";
    else if(Math.round(seconds/(60*60)) >= 2) return Math.round(seconds/(60*60)) + " hours ago";
    else if(Math.round(seconds/(60*60)) >= 1) return "1 hour ago";
    else if(Math.round(seconds/60) >= 2) return Math.round(seconds/60) + " minutes ago";
    else if(Math.round(seconds/60) >= 1) return "1 minute ago";
    else if(seconds >= 2)return seconds + " seconds ago";
    else return seconds + "1 second ago";
}

你也可以使用这个插件javascript-time-ago

【讨论】：

【解决方案3】：

我喜欢减少东西，这是我的解决方案。

/**
 * @method msToTimeString : convert milliseconds to time string
 * @param {number} ms : milliseconds to convert
 */
const msToTimeString = ms => {
    const mods = [1000, 60, 60, 24, 365]; // time modulus
    const names = ["millisecond", "second", "minute", "hour", "day"]; // names
    return mods.reduce((acc, cur, index) => { // reduce modulus
        let unit = acc.val % cur; // calc current time scale
        acc.val = (acc.val - unit) / cur; // keep remaining
        acc[names[index]] = unit; // assign name to result
        return acc; // next
    }, {val: ms});
};

let past = new Date("2020-03-17T00:00:00").getTime(), // past date
    now = Date.now(); // now
    
let calc = msToTimeString(now - past);
console.log(calc.day, "days", calc.hour, "hours"); // also : calc.minute, calc.second, calc.millisecond

相同的代码，野蛮的一行

/**
 * @method msToTimeString : convert milliseconds to time string
 * @param {number} ms : milliseconds to convert
 */
const msToTimeString = ms => {
    const names = ["millisecond", "second", "minute", "hour", "day"]; // names
    return [1000, 60, 60, 24, 365].reduce((acc, cur, index) => ({...acc, [names[index]]: acc.val % cur, val: (acc.val - (acc.val % cur)) / cur}), {val: ms});
};

let past = new Date("2020-03-17T00:00:00").getTime(), // past date
    now = Date.now(); // now
    
console.log(msToTimeString(now - past)); // display !

【讨论】：