第 6 期(2019-05-13)：获取过去n天的日期 #8

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1. liwenkang commented on May 13, 2019

   

   liwenkang

   on May 13, 2019

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   /**
    * 获取当前时间 n天前的毫秒数
    * @param n
    * @returns {number}
    */
   function getPastDay(n) {
       return Date.now() - (n * 86400000);
   }
   
   /**
    * 月/日不够两位的补 0
    * @param num
    * @returns {string|*}
    */
   function padStart(num) {
       if (num < 10) {
           return '0' + num;
       }
       return num;
   }
   
   /**
    * @param time 是要转换的时间(以毫秒为单位)
    * @returns {string}
    */
   function formatTime(time) {
       const date = new Date(time);
       const year = date.getFullYear();
       const month = padStart(date.getMonth() + 1);
       const day = padStart(date.getDate());
       return `${year}-${month}-${day}`;
   }
   
   /**
    * @param {number} n - 天数
    * @return {Array} 日期数组，格式：[yyyy-MM-dd, yyyy-MM-dd]
    */
   function getPastDays(n) {
       let result = [];
       while (n > 0) {
           result.push(formatTime(getPastDay(n--)));
       }
       return result;
   }

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2. Wxh16144 commented on May 14, 2019

   

   Wxh16144

   on May 14, 2019

   Last edited by Wxh16144

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   早上上班看到这个题目,想到了微信公众号前端早读课推送的一篇文章【第1608期】What's New in JavaScript（视频）

   参考MDN Intl.DateTimeFormat
   参考Can I use

   /**
    * @param {number} n - 天数
    * @return {array} 日期数组，格式：[yyyy-MM-dd, yyyy-MM-dd]
    */
   const getPastDays = n => {
     return [...Array(n || 7).keys()].map(days => {
       const date = new Date(Date.now() - 86400000 * ( days + 1 ))
       return new Intl.DateTimeFormat('zh', {
         year: 'numeric',
         month: '2-digit',
         day: '2-digit'
       }).format(date)
         .replace(/\//g, '-')
     })
   }
   console.log(getPastDays(7)); 
   // ["2019-05-13", "2019-05-12", "2019-05-11", "2019-05-10", "2019-05-09", "2019-05-08", "2019-05-07"]

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3. wingmeng commented on May 14, 2019

   

   wingmeng

   on May 14, 2019

   CollaboratorAuthor

   More actions

   早上上班看到这个题目,想到了微信公众号前端早读课推送的一篇文章【第1608期】What's New in JavaScript（视频）

   参考MDN Intl.DateTimeFormat
   参考Can I use

   /**
    * @param {number} n - 天数
    * @return {array} 日期数组，格式：[yyyy-MM-dd, yyyy-MM-dd]
    */
   const getPastDays = n => {
     return [...Array(n || 7).keys()].map(days => {
       const date = new Date(Date.now() - 86400000 * ( days + 1 ))
       return new Intl.DateTimeFormat('zh', {
         year: 'numeric',
         month: '2-digit',
         day: '2-digit'
       }).format(date)
         .replace(/\//g, '-')
     })
   }
   console.log(getPastDays(7)); 
   // ["2019-05-13", "2019-05-12", "2019-05-11", "2019-05-10", "2019-05-09", "2019-05-08", "2019-05-07"]

   答案非常精彩，然而返回的日期顺序应该是从远到近。