计算从时间差 - 总工作时间

我需要计算时差24小时时间。有些事情似乎是错误的代码计算从时间差 - 总工作时间

if (shiftToMin >= shiftFromMin){ 
     shiftTotalMin = shiftToMin - shiftFromMin; 
     if(shiftFromHr < shiftToHr){ 
      shiftTotalHr = shiftToHr - shiftFromHr; 
     }else{ 
      shiftTotalHr = (24 - shiftFromHr) + shiftToHr; 
     } 
    }else{ 
     shiftTotalMin = 60 - (shiftFromMin - shiftToMin); 
     if(shiftFromHr < shiftToHr){ 
      shiftTotalHr = ShiftToHr - shiftFromHr; 
     }else{ 
      shiftTotalHr = (24 - shiftFromHr) + shiftToHr; 
     } 
     shiftTotalHr = shiftTotalHr - 1; 
    }

来源

2013-12-12 epynic

精品！输出是什么？期望的输出是什么？ #ThingsWeNeedToKnow –

时间差喜欢，总工作时间 – epynic

制定并工作

// Time Difference Cal 
     if (shiftFromHr <= shiftToHr){ 
      shiftTotalHr = shiftToHr - shiftFromHr; 
     }else{ 
      shiftTotalHr = (24 - shiftFromHr) + shiftToHr; 
     } 

     if (shiftFromMin <= shiftToMin){ 
      shiftTotalMin = shiftToMin - shiftFromMin; 
     }else{ 
      shiftTotalMin = 60 - (shiftFromMin - shiftToMin); 
      shiftTotalHr = shiftTotalHr - 1; 
     } 
// Total time difference shiftTotalHr : shiftTotalMin

Thanks All :)

来源

2013-12-12 06:28:19 epynic

向我们展示你的full code了解，如果你想calculate从two dates的time difference的试试这个，

function time_diff(start, end) { 
    sdate = new Date(start);edate = new Date(end); 
    var sd = [];var ed = []; 
    sd[0] = sdate.getFullYear(); 
    sd[1] = sdate.getMonth(); 
    sd[2] = sdate.getDate(); 
    sd[3] = sdate.getHours(); 
    sd[4] = sdate.getMinutes(); 
    ed[0] = edate.getFullYear(); 
    ed[1] = edate.getMonth(); 
    ed[2] = edate.getDate(); 
    ed[3] = edate.getHours(); 
    ed[4] = edate.getMinutes(); 
    var startDate = new Date(sd[0], sd[1], sd[2], sd[3], sd[4], 0); 
    var endDate = new Date(ed[0], ed[1], ed[2], ed[3], ed[4], 0); 
    var diff = endDate.getTime() - startDate.getTime(); 
    var hours = Math.floor(diff/1000/60/60); 
    diff -= parseInt(hours) * 1000 * 60 * 60; 
    var minutes = Math.floor(diff/1000/60); 
    if (hours < 0) { 
     return 'error'; 
    } 
    return (hours <= 9 ? "0" : "") + hours + ":" + (minutes <= 9 ? "0" : "") + minutes; 
} 
alert(time_diff('12/11/2013 10:25:00', '12/12/2013 11:25:00')); 
alert(time_diff('12/12/2013 10:25:00', '12/12/2013 11:50:00'));

Demo

来源

2013-12-12 05:58:53

嘿，伙计，感谢您的照顾和时间，Ur真棒，我将需要此代码有一天 – epynic

使用date构造函数的getTime方法。它提供了通过毫秒的总数（自1970年以来，我认为）

var shiftStart = new Date(0,0,0,shiftFromHour,shiftFromMin); 
var shiftEnd = new Date(0,0,0,shiftToHour,shiftToMin);  
var msDiff = shiftEnd.getTime()-shiftStart.getTime(); 
var hours = Math.floor(msDiff/(1000*60*60));

hours将包含工时数shiftStart和shiftEnd之间（均为Date实例）

来源

2013-12-12 06:49:08 tewathia

忘记提及，shiftToHr，shiftToMin ...是从文本框val（） – epynic

固定（假设shiftTo ....等都是整数） – tewathia

计算从时间差 - 总工作时间

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