Question

我是JavaScript的新手（但不是编程），我很难搞清楚我在这个函数中犯了哪个错误，在这里找到：http://mikeryan.webatu.com/function.html [dead link] [猜测原文底部的代码]

该函数应采用 DDMMYY 时间戳并将其转换为人类可读的字符串。例如， 210710 会变成 July 21st, 2010 。

更新：代码可能与OP的死链接相似：

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

一个问题：你错过了一个括号。变化：

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

到

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

话虽如此，这是一种更直接的计算方法：

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

接下来，您的数组初始化不必要地冗长。而不是：

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

只是这样做：

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

您的日期后缀不正确。它将“nd”置于12和“4月12日”之后显然不正确。我只是使用逻辑来做这个而不是查找数组，其中大多数元素是“th”。

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);
}

最后，您的“timestamp”在其当前形式中是一个整数，几乎没有价值。 YYYYMMDD更典型的格式有两种：

数字排序符合日期排序;以及
这是毫不含糊的。北美人前一个月（即MMDDYY）。世界上的其他人都把第一天放在第一位（即DDMMYY）。没有人做YYDDMM。

Answer 2

我使用% - modulo：X modulo 100丢弃除最后2位数之外的任何内容。有用！

也使用floor而不是round

Answer 3

使用日期对象。它的速度要快得多。这是一个快速的例子，它假设年份将在2000年代，所以你必须做一些修改。输出并不完全是你所拥有的，但它非常接近，而且代码要短得多。

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获取小数的最低值，然后在你的函数中有一些拼写错误。这是可行的代码（注意：只是通过几个例子进行了测试），但应该足以让你开始：

          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);
     }

将DDMMYY号码转换成日期？

4 个答案: