使用jQuery将数（天）转换为天，月和年

Question

我有一个计算功能，部分显示了达到目标所需的天数。

而不是仅仅显示天数，我想根据天数将其计算为天数或月数或天数，月数和年数。 我有一个if语句用于拆分，但是似乎无法计算出从132天到x天x个月的数学公式。有什么建议吗？

// GOAL
var timeToGoal = Math.round(goal / costPerDay);         

// if more than a year
if ( timeToGoal >= 365 ) {
    alert('days + months + years');

    // if more than a month but less than a year
} else if ( timeToGoal >= 30 && timeToGoal <=365 ) {
    alert('Days + months');
} else {
    alert('days');
    $('#savings-goal span').text(timeToGoal+' days');
}

Answer 1

尝试这样的事情；

function humanise (diff) {
  // The string we're working with to create the representation
  var str = '';
  // Map lengths of `diff` to different time periods
  var values = [[' year', 365], [' month', 30], [' day', 1]];

  // Iterate over the values...
  for (var i=0;i<values.length;i++) {
    var amount = Math.floor(diff / values[i][1]);

    // ... and find the largest time value that fits into the diff
    if (amount >= 1) {
       // If we match, add to the string ('s' is for pluralization)
       str += amount + values[i][0] + (amount > 1 ? 's' : '') + ' ';

       // and subtract from the diff
       diff -= amount * values[i][1];
    }
  }

  return str;
}

预计参数是您要表示的天数之差。 假设一个月为30天，一年为365。

您应该像这样使用它；

$('#savings-goal span').text(humanise(timeToGoal));

http://jsfiddle.net/0zgr5gfj/

Answer 2

来自我的尝试（考虑到leap年并基于当前日期）

function humanise(total_days)
{
    //var total_days = 1001;
    var date_current = new Date();
    var utime_target = date_current.getTime() + total_days*86400*1000;
    var date_target = new Date(utime_target);

    var diff_year  = parseInt(date_target.getUTCFullYear() - date_current.getUTCFullYear());
    var diff_month = parseInt(date_target.getUTCMonth() - date_current.getUTCMonth());
    var diff_day   = parseInt(date_target.getUTCDate() - date_current.getUTCDate());

    var days_in_month = [31, (date_target.getUTCFullYear()%4?29:28), 31, 30, 31, 30, 31, 31, 30, 31, 30, 31];
    var date_string = "";
    while(true)
    {
        date_string = "";
        date_string += (diff_year>0?diff_year + "Y":"");

        if(diff_month<0){diff_year -= 1; diff_month += 12; continue;}
        date_string += (diff_month>0?diff_month + "M":"");

        if(diff_day<0){diff_month -= 1; diff_day += days_in_month[((11+date_target.getUTCMonth())%12)]; continue;}
        date_string += (diff_day>0?diff_day + "D":"");
        break;
    }
    console.log(date_string);
    return date_string;
}

var timeToGoal = 1001;
$('#savings-goal span').text(humanise(timeToGoal));

Answer 3

这是一个没有for bucle的简单解决方案：

function jarh(x) {
var y = 365;
var y2 = 31;
var remainder = x % y;
var casio = remainder % y2;
year = (x - remainder) / y;
month = (remainder - casio) / y2;

var result ="--- Year ---" + year + "--- Month ---" + month + "--- Day ---" + casio;

return result;
}

var call = jarh(6781);

http://jsfiddle.net/yHAcY/1/