我不认为有一个真正简单和优雅的方法来做到这一点。我会从两个字符串中提取日期值，将它们转换为秒或任何浮动的值，然后减去并测量minTime，如下所示：

var lastYearSeconds = parseInt(lastExecution.substr(0,4))*31536000; //this is in seconds
var lastMonthSeconds = parseInt(lastExecution.substr(4,2))*2592000;

然后对所有其他值执行相同的操作，将它们相加，然后从currTime（）解析的值中减去。
等等，忘了JavaScript中的Date（）类。Yay for builtins！
基本上，只解析日期字符串中的int值，然后将它们提供给构造函数like so。然后，可以用dateObject.getTime() - otherDateObject.getTime()减去这两个日期值

试试这个，它使用你想要的格式：

function strToDate(str) {
  var year = parseInt(str.substring(0,4));
  var month = parseInt(str.substring(4,6));
  var day = parseInt(str.substring(6,8));
  var hour = parseInt(str.substring(8,10));
  var minute = parseInt(str.substring(10,12));
  var second = parseInt(str.substring(12,14));

  var date = new Date(year, month, day , hour, minute, second, 0);
  return date.getTime();
}

function timeStampToDate(timestamp) {
  var d = new Date(timestamp);

    var diffStr = ""+(d.getFullYear()-1970)+":"+d.getMonth()+":"+(d.getDate()-1)+":"+(d.getHours()-1)+":"+d.getMinutes()+":"+d.getSeconds();
    return diffStr;
}

var lastExecution = "20150116165100"; //--- 16:51:00 16/01/2015
var currentTime =   "20150116170120"; //--- 17:01:20 16/01/2015
var lastExecutionDate = strToDate(lastExecution);
var currentTimeDate = strToDate(currentTime);

var difference = (currentTimeDate - lastExecutionDate);
var diffInTime = timeStampToDate(difference);
alert("Time difference: "+diffInTime);

链接：http://jsfiddle.net/nxjk1L8s/

使用Java 1。8、我的剧本对我有用。试试这个...

LocalDateTime now3 = LocalDateTime.now();       
System.out.println(now3);
Thread.sleep(70000);
LocalDateTime now4 = LocalDateTime.now(); 
System.out.println(now4);   
Duration duration2 = Duration.between(now3, now4);
System.out.println("duration is "+duration2);

o/p：

2023-04-27T19:14:11.968
2023-04-27T19:15:21.974
duration is PT1M10.006S