Question

i的值应以5秒为间隔增加到500。但我需要将前一天的总数添加到第二天。为简单起见，我只需要在i的每日递增开始之前添加最大值。但是我正在初始化为0以进行递增，并且无论是否添加最大值都将重置。

最好的方法是什么？到目前为止，这是我的代码。

 <script type='text/javascript'>
    var div = document.getElementById('count');
    var i = 0; 
    var count = 0;
    var max = 500;
    var d = new Date();
    // set d to midnight
    d.setHours(0,0,0,0);
    d.setMonth(0);
    d.setDate(1);
    d.setFullYear(2017);
     var one_day=1000*60*60*24;
     var today = new Date().getTime();  
     var numberofdays = Math.round((today - d.getTime()) / one_day);


   function isPastMidnight() {
      var currentTime = new Date().getTime();
      if ((currentTime - d.getTime()) > 0) {
        return true;
      } else {
        return false;
      } 
    }
    function updateCount() {
      var currentTime = new Date().getTime();
      if (!isPastMidnight()) {
       document.getElementById('count').innerHTML = "Check back January 1, 2017.";
      }


     if (isPastMidnight())   {
      if (numberofdays > 0 && today == d.getHours)   {      
        addmax();

        }
      }
     if (isPastMidnight() && i < max) {
        i = Math.floor((currentTime - d.getTime() ) / 5000);
        if (i > max) i = max; 
        document.getElementById('count').innerHTML = i;
      }
        if (i >= max) {
        // Stop executing myself ("interval" is declared in global scope, so accessible from here)
        clearInterval(interval);
      }
    }
    updateCount();
    var interval = setInterval(updateCount, 1000);
</script>

Answer 1

您需要计算从1月1日开始的数量并乘以500，然后将5秒（或其他）的数量增加到500并将其添加到计数中。关于如何计算日期之间的天数以及对当前时间值进行数学计算，存在许多问题。

您的代码似乎过于复杂，如果您不想在2017年1月1日之前启动计数器，那么只需测试该日期是否在此之前。

＆＃13;

/* Return number of whole days between dates. Sets
** time to 00:00:00 so only measures whole days.
** @param {Date} d0 - start date
** @param {Date} d1 - end date
** @returns {number} whole days between dates
*/
function daysBetweenDates(startDate, endDate) {
  var d0 = new Date(+startDate);
  var d1 = new Date(+endDate);
  return Math.round((d1.setHours(0,0,0,0) - d0.setHours(0,0,0,0)) / 8.64e7);
}

/* Return a value that is 500 * the number of days since 1 Jan of the current year
** + 1 for each interval since 00:00:00 on the current day
** @param {Date} date
** @returns {number}
*/
function getCurrentCount(date) {
  // Don't run before 1 Jan 2017
  if (date < new Date(2017,0,1)) return 'Too early...';

  // Get count for whole days since 1 Jan 2017
  var count = 500 * daysBetweenDates(new Date(date.getFullYear(), 0), date);

  // Find milliseconds since midnight at start of day
  var dayPart = date - new Date(date).setHours(0,0,0,0)

  // Add increments
  var increment = 5000;
  dayPart = Math.floor(dayPart % 8.64e7 / increment);

  // Increment by dayPart if less than 500, or 500 otherwise
  count += dayPart > 500? 500 : dayPart;

  return count;
}

// Some tests
[new Date(),              // Now
 new Date(2017,0,1,0,30), // 2017-01-01 00:30:00
 new Date(2017,5,23,0,30) // 2017-06-23 00:30:00
].forEach(function(d) { 
  console.log('Count for ' + d.toString() + 'is: ' + getCurrentCount(d));
});

＆＃13;

Answer 2

乘以max变量的天数，并在数学语句中将其添加到i。所以第一天从0开始递增而不是500，我将它放入if语句中，只有当天数大于0时才允许它执行。再次感谢！

每天午夜添加最大值

2 个答案: