<!DOCTYPE>
<html>
<head>
<meta http-equiv="Content-Type" content="text/htmlcharset=UTF-8">
<title>RunJS</title>
</head>
<body>
<input value="秒杀" type="button" id="btnTest"/>
<script type="text/javascript">
var btnTest=document.getElementById("btnTest")
var timerId,temp,timerNum
if(!!btnTest)
{
temp=0,timerNum=0
btnTest.onclick=function(e)
{
temp++
if(!timerId)
{
timerId=setInterval(function(){timerNum++btnTest.value=timerNum+"秒内点击"+temp+"次"},1000)
setTimeout(function(){if(temp<10){reset()}},6000)
}
else if(temp>9 &&timerNum<=5)
{
btnTest.disabled=true
reset()
alert("您点击的太猛了，会吧电脑累坏的，休息3秒吧！")
setTimeout(function(){btnTest.disabled=false},3000)
return false
}
else if(temp<9 &&timerNum==5)
{
reset()
}
}
}
var reset=function(){
clearInterval(timerId)
timerId=null
temp=0
timerNum=0
btnTest.value="秒杀"
}
</script>
</body>
</html>
用JQ的，ajax 无刷新就能做到了<img src="" id="abc" value="addnum">
<script>
$("#abc").click(function(){
var value = $(this).attr('value')
$.ajax({
type: "POST",
url: "add.php",
data: "type="+value,
success: function(msg){
// alert( "Data Saved: " + msg )
}
})
return false
})
</script>
<?php
/*
*add.php
*参数：type 判断是否是那个图片的点击
*talbe_name 为表名，替换成自己的表名
*/
$type= $_POST['type']
if($type=='addnum'){
$query = mysql_query("upadae talbe_name set num = num +1 where 条件") //执行sql语句
return $query
}
?>
var count = 0document.onclick = function(){
count++
alert(count)
}
