將DDMMYY號碼轉換爲日期嗎？

Question

我是JavaScript新手（但不是編程），我很難弄清楚我在這個函數中犯了什麼錯誤，在這裏找到：http://mikeryan.webatu.com/function.html[死鏈接] [底部原始代碼的猜測]

該函數應採用DDMMYY時間戳，並將其轉換爲人類可讀字符串。例如，210710將變成July 21st, 2010。

UPDATE：代碼這可能是類似的OP的死鏈接：

function timestamp(d){ 
    var year = (d-(Math.round(d/100)*100); 

    var day = Math.floor(d/10000); 
    var dayfix = (day - (Math.floor(day/10)*10)); 

    // var month = ((d-year)-(day*100000)/100); 

    var a = (d - year); 
    var b = ((day * 100000)/10); 

    var month = (a - b)/100; 

    var months = new Array(); 
    months[1] = "January"; 
    months[2] = "February"; 
    months[3] = "March"; 
    months[4] = "April"; 
    months[5] = "May"; 
    months[6] = "June"; 
    months[7] = "July"; 
    months[8] = "August"; 
    months[9] = "September"; 
    months[10] = "October"; 
    months[11] = "November"; 
    months[12] = "December"; 

    var daysuffix = new Array(); 
    daysuffix[0] = "th"; 
    daysuffix[1] = "st"; 
    daysuffix[2] = "nd"; 
    daysuffix[3] = "rd"; 
    daysuffix[4] = "th"; 
    daysuffix[5] = "th"; 
    daysuffix[6] = "th"; 
    daysuffix[7] = "th"; 
    daysuffix[8] = "th"; 
    daysuffix[9] = "th"; 

    if(year>20){ 
     year = '19' + year; 
    } 
    else{ 
     year = '20' + year; 
    } 
    return (months[month] + ' ' + day + daysuffix[dayfix] + ', ' + year); 
} 

Answer 1

一個問題：你缺少一個括號。改變：

var year = (d-(Math.round(d/100)*100); 

到

var year = (d-(Math.round(d/100)*100)); 

如此說來，這是一個更簡單的計算方法：

var year = d % 100; 
var month = Math.floor(d/100) % 100; 
var day = Math.floor(d/10000) % 100; 

接着，您的數組初始化是不必要的冗長。相反的：

var arr = new Array(); 
arr[0] = "foo"; 
arr[1] = "bar"; 

只是做：

var arr = ["foo", "bar"]; 

你一天後綴不正確。它在12和12月4日後顯示「nd」顯然不正確。我只是用邏輯來做這件事，而不是使用大多數元素都是「th」的查找數組。

所以：

function timestamp(d){ 
    var year = d % 100; 
    var month = Math.floor(d/100) % 100; 
    var day = Math.floor(d/10000) % 100; 
    var months = ["January", "February", "March", "April", "May", "June", 
    "July", "August", "September", "October", "November", "December"]; 
    if (year>20) { 
    year = '19' + year; 
    } else { 
    year = '20' + year; 
    } 
    if (day == 1 || day == 21 || day == 31) { 
    var suffix = "st"; 
    } else if (day == 2 || day == 22) { 
    var suffix = "nd"; 
    } else { 
    var suffix = "th"; 
    } 
    return (months[month-1] + ' ' + day + suffix + ', ' + year); 
} 

最後是在你的「時間戳」是目前形式的整數價值不大。這種類型的東西更典型的格式是YYYYMMDD，原因有兩個：

1. 數字排序匹配日期排序;和

2. 這是毫不含糊的。北美人前一個​​月（即MMDDYY）。世界上其他人都首先把日子放在首位（即DDMMYY）。沒有人會做YYDDMM。

Answer 2

我想用% - modulo：X modulo 100丟棄除最後2位數以外的任何東西。有用！

也使用floor不round

Answer 3

使用日期對象。速度快了很多。這是一個假設年份將在2000年代的快速示例，因此您需要進行一些修改。輸出結果並不完全符合您的要求，但它非常接近，代碼縮短了很多。

function date(e){ 
    var d = new Date(); 
    d.setYear(2000+e.substring(4)/1,e.substring(2,4)-1,e.substring(0,2)-1); 
    alert(d.toDateString()); 
} 

Answer 4

首先使用Math.floor來獲取小數點的底值，然後在函數中出現一些拼寫錯誤。這裏是工作的代碼（注意：只是一對夫婦的實例進行測試），但應該足以讓你開始：

  function timestamp(d){ 
     var year = (d-(Math.floor(d/100)*100)); 

     var day = Math.floor(d/10000); 
     var dayfix = (day - (Math.floor(day/10)*10)); 

     // var month = ((d-year)-(day*100000)/100); 

     var a = (d - year); 
     var b = ((day * 100000)/10); 

     var month = (a - b)/100; 

     var months = new Array(); 
     months[1] = "January"; 
     months[2] = "February"; 
     months[3] = "March"; 
     months[4] = "April"; 
     months[5] = "May"; 
     months[6] = "June"; 
     months[7] = "July"; 
     months[8] = "August"; 
     months[9] = "September"; 
     months[10] = "October"; 
     months[11] = "November"; 
     months[12] = "December"; 

     var daysuffix = new Array(); 
     daysuffix[0] = "th"; 
     daysuffix[1] = "st"; 
     daysuffix[2] = "nd"; 
     daysuffix[3] = "rd"; 
     daysuffix[4] = "th"; 
     daysuffix[5] = "th"; 
     daysuffix[6] = "th"; 
     daysuffix[7] = "th"; 
     daysuffix[8] = "th"; 
     daysuffix[9] = "th"; 

     if(year>20){ 
      year = '19' + year; 
     } 
     else{ 
      year = '20' + year; 
     } 
     return (months[month] + ' ' + day + daysuffix[dayfix] + ', ' + year); 
    }