在此javascript倒计时脚本中添加前导零？

时间：2012-12-12 11:01:52

标签： javascript countdown

我正在使用以下倒计时脚本，效果很好，但我无法弄清楚如何在数字中添加前导零（例如，它显示09而不是9）。有人可以帮我吗？这是当前的脚本：

function countDown(id, end, cur){
        this.container = document.getElementById(id);
        this.endDate = new Date(end);
        this.curDate = new Date(cur);


var context = this;

var formatResults = function(day, hour, minute, second){
    var displayString = [
                '<div class="stat statBig">',day,'</div>',
                '<div class="stat statBig">',hour,'</div>',
                '<div class="stat statBig">',minute,'</div>',
                '<div class="stat statBig">',second,'</div>'
    ];
    return displayString.join("");
}

var update = function(){
    context.curDate.setSeconds(context.curDate.getSeconds()+1);

    var timediff = (context.endDate-context.curDate)/1000; 

    // Check if timer expired:
    if (timediff<0){ 
        return context.container.innerHTML = formatResults(0,0,0,0);
    }

    var oneMinute=60; //minute unit in seconds
    var oneHour=60*60; //hour unit in seconds
    var oneDay=60*60*24; //day unit in seconds

    var dayfield=Math.floor(timediff/oneDay);
    var hourfield=Math.floor((timediff-dayfield*oneDay)/oneHour);
    var minutefield=Math.floor((timediff-dayfield*oneDay-hourfield*oneHour)/oneMinute);
    var secondfield=Math.floor((timediff-dayfield*oneDay-hourfield*oneHour-minutefield*oneMinute));

    context.container.innerHTML = formatResults(dayfield, hourfield, minutefield, secondfield);

    // Call recursively
    setTimeout(update, 1000);
};

// Call the recursive loop
update();
}

6 个答案:

答案 0 :(得分：3)

您只需要检查变量是否小于10并将它们添加到前导零。 请尝试以下方法：

function countDown(id, end, cur){
        this.container = document.getElementById(id);
        this.endDate = new Date(end);
        this.curDate = new Date(cur);


var context = this;

var formatResults = function(day, hour, minute, second){
    day = (day < 10) ? "0"+day : day;
    hour = (hour < 10) ? "0"+hour : hour;
    minute = (minute < 10) ? "0"+minute : minute;
    second = (second < 10) ? "0"+second: second; 
    var displayString = [
                '<div class="stat statBig">',day,'</div>',
                '<div class="stat statBig">',hour,'</div>',
                '<div class="stat statBig">',minute,'</div>',
                '<div class="stat statBig">',second,'</div>'
    ];
    return displayString.join("");
}

var update = function(){
    context.curDate.setSeconds(context.curDate.getSeconds()+1);

    var timediff = (context.endDate-context.curDate)/1000; 

    // Check if timer expired:
    if (timediff<0){ 
        return context.container.innerHTML = formatResults(0,0,0,0);
    }

    var oneMinute=60; //minute unit in seconds
    var oneHour=60*60; //hour unit in seconds
    var oneDay=60*60*24; //day unit in seconds

    var dayfield=Math.floor(timediff/oneDay);
    var hourfield=Math.floor((timediff-dayfield*oneDay)/oneHour);
    var minutefield=Math.floor((timediff-dayfield*oneDay-hourfield*oneHour)/oneMinute);
    var secondfield=Math.floor((timediff-dayfield*oneDay-hourfield*oneHour-minutefield*oneMinute));

    context.container.innerHTML = formatResults(dayfield, hourfield, minutefield, secondfield);

    // Call recursively
    setTimeout(update, 1000);
};

// Call the recursive loop
update();
}

<强>更新

你也可以使用@Alnitak解决方案并用函数包装它，效果是一样的，你将集中你的逻辑：

function countDown(id, end, cur){
        this.container = document.getElementById(id);
        this.endDate = new Date(end);
        this.curDate = new Date(cur);


var context = this;

var addLeadingZeros = function(number){
    return (number < 10) ? "0"+number : number;
}

var formatResults = function(day, hour, minute, second){
    day = addLeadingZeros(day);
    hour = addLeadingZeros(hour);
    minute = addLeadingZeros(minute);
    second = addLeadingZeros(second); 
    var displayString = [
                '<div class="stat statBig">',day,'</div>',
                '<div class="stat statBig">',hour,'</div>',
                '<div class="stat statBig">',minute,'</div>',
                '<div class="stat statBig">',second,'</div>'
    ];
    return displayString.join("");
}

var update = function(){
    context.curDate.setSeconds(context.curDate.getSeconds()+1);

    var timediff = (context.endDate-context.curDate)/1000; 

    // Check if timer expired:
    if (timediff<0){ 
        return context.container.innerHTML = formatResults(0,0,0,0);
    }

    var oneMinute=60; //minute unit in seconds
    var oneHour=60*60; //hour unit in seconds
    var oneDay=60*60*24; //day unit in seconds

    var dayfield=Math.floor(timediff/oneDay);
    var hourfield=Math.floor((timediff-dayfield*oneDay)/oneHour);
    var minutefield=Math.floor((timediff-dayfield*oneDay-hourfield*oneHour)/oneMinute);
    var secondfield=Math.floor((timediff-dayfield*oneDay-hourfield*oneHour-minutefield*oneMinute));

    context.container.innerHTML = formatResults(dayfield, hourfield, minutefield, secondfield);

    // Call recursively
    setTimeout(update, 1000);
};

// Call the recursive loop
update();
}

答案 1 :(得分：2)

IMO，获得领先零的最简单方法是使用substr：

var n = 10;
console.log(('00' + n).substr(-2));//logs 10
n = 2;
console.log(('00' + n).substr(-2));//logs 02

易peasy。如果要将其倒入函数中，则返回一个字符串：

function addLeadingZeroes(n)
{
    return ('00' + n).substr(-2);
}

就是这样。

答案 2 :(得分：1)

最简单的方法是投入实用功能来填充你的数字：

function pad2(n) {
    return (n < 10) ? '0' + n : n;
}

答案 3 :(得分：0)

这应该有所帮助：

function countDown(id, end, cur){
        this.container = document.getElementById(id);
        this.endDate = new Date(end);
        this.curDate = new Date(cur);


var context = this;

var formatNum = function (num) {
    if (num.toString().length < 2) {
        return '0' + num;
    }    
    return num;
}

var formatResults = function(day, hour, minute, second){
    formatNum(day);
    formatNum(hour);
    formatNum(minute);
    formatNum(second);    
    var displayString = [
                '<div class="stat statBig">',day,'</div>',
                '<div class="stat statBig">',hour,'</div>',
                '<div class="stat statBig">',minute,'</div>',
                '<div class="stat statBig">',second,'</div>'
    ];
    return displayString.join("");
}

var update = function(){
    context.curDate.setSeconds(context.curDate.getSeconds()+1);

    var timediff = (context.endDate-context.curDate)/1000; 

    // Check if timer expired:
    if (timediff<0){ 
        return context.container.innerHTML = formatResults(0,0,0,0);
    }

    var oneMinute=60; //minute unit in seconds
    var oneHour=60*60; //hour unit in seconds
    var oneDay=60*60*24; //day unit in seconds

    var dayfield=Math.floor(timediff/oneDay);
    var hourfield=Math.floor((timediff-dayfield*oneDay)/oneHour);
    var minutefield=Math.floor((timediff-dayfield*oneDay-hourfield*oneHour)/oneMinute);
    var secondfield=Math.floor((timediff-dayfield*oneDay-hourfield*oneHour-minutefield*oneMinute));

    context.container.innerHTML = formatResults(dayfield, hourfield, minutefield, secondfield);

    // Call recursively
    setTimeout(update, 1000);
};

// Call the recursive loop
update();
}

唯一的区别是方法formatNum，它有四个调用：

formatNum(day);
formatNum(hour);
formatNum(minute);
formatNum(second);

答案 4 :(得分：0)

如果数字小于10，那么你应该添加前面的零，白天，小时，秒...... 

day = day<10?'0' + day : day;
hour = hour<10?'0' + hour : hour;
second = second<10?'0' + second : second;

答案 5 :(得分：0)

我是新手，但我做到了，而且奏效了！

if (days< 10) {
        day =  "0"+days;}

  if (hours < 10) {
    hours = "0"+hours;
}

  if (minutes < 10) {
    minutes = "0"+minutes;
  }

  if (seconds < 10) {
    seconds = "0"+seconds;
}

  // Output the result
  document.getElementById("demo").innerHTML = days + ":" + hours + ":"
  + minutes + ":" + seconds + "";
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