計算從時間差 - 總工作時間

我需要計算時差24小時時間。有些事情似乎是錯誤的代碼計算從時間差 - 總工作時間

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; 
    }

來源

2013-12-12 epynic

精品！輸出是什麼？期望的輸出是什麼？ #ThingsWeNeedToKnow –

時間差喜歡，總工作時間 – epynic

制定並工作

// 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 :)

來源

2013-12-12 06:28:19 epynic

向我們展示你的full code瞭解，如果你想calculate從two dates的time difference的試試這個，

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

來源

2013-12-12 05:58:53

嘿，夥計，感謝您的照顧和時間，Ur真棒，我將需要此代碼有一天 – epynic

使用date構造函數的getTime方法。它提供了通過毫秒的總數（自1970年以來，我認爲）

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將包含工時數shiftStart和shiftEnd之間（均爲Date實例）

來源

2013-12-12 06:49:08 tewathia

忘記提及，shiftToHr，shiftToMin ...是從文本框val（） – epynic

固定（假設shiftTo ....等都是整數） – tewathia

計算從時間差 - 總工作時間

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