倒计时问题 - JavaScript论坛

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倒计时问题

程序执行到函数djs只执行一次，是一个动态显示倒计时的js，请高手相帮，不胜感激。
<body>
<span id="id1" style="display:none;">2009-10-01 00:00:00</span>
<script>
            var     dt = document.getElementById("id1").innerText;
            var     strArr  = dt.split(" ");
            var     strDay  = strArr[0].split("-");
            var     strTime = strArr[1].split(":");
            //document.write(strDay[0]+strDay[1]+strDay[2]+strTime[0]+strTime[1]+strTime[2]);
            //这个函数是倒计时
             function djs()
            {
                 minutes = 1000 * 60;
                 hours   = minutes  * 60;
                 days    = hours  * 24;
                //var years   = days * 365;
                 date1   = new Date();
                 date2   = new Date(Date.UTC(strDay[0], strDay[1]-1, strDay[2], strTime[0]-8, strTime[1], strTime[2]));
                 minudt  = date2 - date1;
                //var y       = Math.floor(minudt/years);
                 d       = Math.floor(minudt/days);
                 h       = Math.floor((minudt - d * days)/hours);
                 m       = Math.floor((minudt - d * days - h * hours)/minutes);
                 s       = Math.floor((minudt - d * days - h * hours - m * minutes)/1000);
                document.write(d + "天" + h +  "时" + m + "分" + s + "秒");
                setInterval("djs()",1000);
            }
            djs();
</script>
</body>

搜索更多相关主题的帖子: 倒计时　

<body> <span id="id1">2009-10-01 00:00:00</span> <script> var dt = $("id1").innerHTML; var strArr = dt.split(" "); var strDay = strArr[0].split("-"); var strTime = strArr[1].split(":"); //document.write(strDay[0]+strDay[1]+strDay[2]+strTime[0]+strTime[1]+strTime[2]); //这个函数是倒计时 //setInterval("djs()",1000); setInterval(showTime, 1000); function $(id) { return document.getElementById(id); } function showTime() { minutes = 1000 * 60; hours = minutes * 60; days = hours * 24; //var years = days * 365; date1 = new Date(); date2 = new Date(Date.UTC(strDay[0], strDay[1]-1, strDay[2], strTime[0]-8, strTime[1], strTime[2])); minudt = date2 - date1; //var y = Math.floor(minudt/years); d = Math.floor(minudt/days); h = Math.floor((minudt - d * days)/hours); m = Math.floor((minudt - d * days - h * hours)/minutes); s = Math.floor((minudt - d * days - h * hours - m * minutes)/1000); $('id1').innerHTML = d + "天" + h + "时" + m + "分" + s + "秒"; } </script> </body>