请检查出来并做需要 [英] please check it out and do the need
本文介绍了请检查出来并做需要的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>Digitl clock</title>
<style type="text/css">
#time{
font-size:50pt;
}
</style>
</head>
<body önload="timer()">
<script type="text/javascript" language=javascript>
//var digiclock = "00:00:00";
i = 0;
function timer()
{
var digiformat = "";
if(i>3599)
{
var H = Math.floor(i/3600);
}
else
{
var H = 0;
}
var M = i - (H*3600)
if(M>59)
{
M = Math.floor(M/60)
}
else
{
M = 0
}
var S = i - (M*60)
if(H<10)
{
H = "0"+H;
}
if(M<10)
{
M = "0"+M;
}
if(S<10)
{
S = "0"+S;
}
// document.getElementById('time').innerHTML = H+":"+M+":"+S;
document.getElementById('lblTimer').innerHTML = H+":"+M+":"+S;
setTimeout('timer()', 1000);
i++;
}
</script>
<%--<asp:HiddenField ID="hidenfld" runat="server" />--%>
<asp:Label ID="lblTimer" runat="server" Font-Size="30px" ></asp:Label>
<input type=button ID="btnrefresh" runat="server" value="refresh"/>
</body>
</html>
我尝试了所有方法，如何避免刷新计数器.
How can I avoid refresh of counter, I tried all the ways.
推荐答案
也许这样的方法可以为您提供帮助:
Maybe something like this will help you:
<body onload="timer()">
<form id="form1" runat="server">
<div>
<asp:hiddenfield runat="server" id="timerValue" value="0" xmlns:asp="#unknown" />
<script type="text/javascript" language="javascript">
function timer() {
var timerValue = document.getElementById("timerValue");
var i = timerValue.value;
var digiformat = "";
if (i > 3599) {
var H = Math.floor(i / 3600);
} else {
var H = 0;
}
var M = i - (H * 3600)
if (M > 59) {
M = Math.floor(M / 60)
} else {
M = 0
}
var S = i - (M * 60)
if (H < 10) {
H = "0" + H;
}
if (M < 10) {
M = "0" + M;
}
if (S < 10) {
S = "0" + S;
}
document.getElementById('lblTimer').innerHTML = H + ":" + M + ":" + S;
setTimeout('timer()', 1000);
i++;
timerValue.value = i;
}
</script>
<label id="lblTimer" style="font-size:30px"></label>
<asp:button runat="server" text="Refresh" xmlns:asp="#unknown" />
</div>
</form>
</body>
使用隐藏字段
1.从服务器(功能的第一行)中读取i
2.i++;之后，使用__doPostBack将值发布到服务器.
我认为每次刷新后i都不会闪烁.
Use hidden fields
1. Readifrom Server (First line of your function)
2. afteri++;, use __doPostBack to post the value to server.
I think after every refresh theiis not flashing.
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