Human readable duration format

Description:
Your task in order to complete this Kata is to write a function which formats a duration, given as a number of seconds, in a human-friendly way.
The function must accept a non-negative integer. If it is zero, it just returns  "now" . Otherwise, the duration is expressed as a combination of  years ,  days ,  hours ,  minutes  and  seconds .
It is much easier to understand with an example:

formatDuration(62) // returns "1 minute and 2 seconds"
formatDuration(3662) // returns "1 hour, 1 minute and 2 seconds"

For the purpose of this Kata, a year is 365 days and a day is 24 hours.
Note that spaces are important.
Detailed rules
The resulting expression is made of components like  4 seconds ,  1 year , etc. In general, a positive integer and one of the valid units of time, separated by a space. The unit of time is used in plural if the integer is greater than 1.
The components are separated by a comma and a space ( ", " ). Except the last component, which is separated by  " and " , just like it would be written in English.
A more significant units of time will occur before than a least significant one. Therefore,  1 second and 1 year  is not correct, but  1 year and 1 second  is.
Different components have different unit of times. So there is not repeated units like in  5 seconds and 1 second .
A component will not appear at all if its value happens to be zero. Hence,  1 minute and 0 seconds  is not valid, but it should be just  1 minute .
A unit of time must be used "as much as possible". It means that the function should not return  61 seconds , but  1 minute and 1 second  instead. Formally, the duration specified by of a component must not be greater than any valid more significant unit of time.
My solution:

function formatDuration (seconds) {
  const years = (seconds- seconds % 31536000) / 31536000;
  const days = parseInt((seconds % 31536000) / 86400);
  const hours = parseInt((seconds % 86400) / 3600);
  const minutes = parseInt((seconds % 3600) / 60);
  const second = parseInt(seconds % 60);
  let object = {a: years, b: days, c: hours, d: minutes, e: second};
  for(item of Object.keys(object)) {
    if(item == "a") {
      object[item] += " year";
    } else if(item == "b") {
      object[item] += " day";
    } else if(item == "c") {
      object[item] += " hour";
    } else if(item == "d") {
      object[item] += " minute";
    } else if(item == "e") {
      object[item] += " second";
    }

    if(object[item].match(/(\d+)/)[0] >1) {
      object[item] += "s";
    }

    if(object[item].match(/(\d+)/)[0] == 0) {
      delete object[item];
    }
  }
  let array = Object.values(object);
  for(i=0; i<array.length; i++) {
    if(array.length>1 && i==array.length-2) {
      array[i]+=" and ";
    } else if(i == array.length-1) {
      continue;
    } else {
      array[i]+=", ";
    }
  }
  const newstring = array.join('');
  if(newstring == ""){
    return 'now';
  }

  return newstring;
}

Best solutions:

function formatDuration (seconds) {
  var time = { year: 31536000, day: 86400, hour: 3600, minute: 60, second: 1 },
      res = [];

  if (seconds === 0) return 'now';

  for (var key in time) {
    if (seconds >= time[key]) {
      var val = Math.floor(seconds/time[key]);
      res.push(val += val > 1 ? ' ' + key + 's' : ' ' + key);
      seconds = seconds % time[key];
    }
  }

  return res.length > 1 ? res.join(', ').replace(/,([^,]*)$/,' and'+'$1') : res[0]
}

オブジェクト上でfor...文.
実行するとkeyが順番に返されます.

正規表現/,([^,]*)$/了解

[^,]

[a,b,c] : a || b || c||

[^a-z]: not a~z

*は、前のパリティを満たす文字列を0以上返します.

$文末

→先頭の、含まない文の末尾にある文字列

正規表現()と$1
正規表現()で()の文字列を取得して記憶します.$1でアクセスできます.3つの()がある場合は、$1、$2、$3の順に文字列を返すことができます.