<script type="text/javascript">
function GetResult()
{
var xmlhttp;
var strResult;
var prefixes = ["MSXML3.XMLHTTP","MSXML2.XMLHTTP","Microsoft.XMLHTTP","MSXML.XMLHTTP","Msxml2.XMLHTTP.5.0","Msxml2.XMLHTTP.4.0","Msxml2.XMLHTTP.3.0"];
if (window.ActiveXObject)
{
for(var i=0;i<prefixes.length;i++)
{
try
{
xmlhttp=new ActiveXObject(prefixes[i]);
break;
}
catch (e)
{
}
}
}
else if (window.XMLHttpRequest)
{
try
{
xmlhttp = new XMLHttpRequest();
}
catch (e)
{
alert("你的浏览器不支持XMLHttpRequest对象,请升级");
}
} xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4)
{
strResult=xmlhttp.responseText;
}
}; xmlhttp.open("GET","dataread.jsp",false);
xmlhttp.send(null); };
</script>
<form method=post action='datadisplay.jsp'>
<input type=hidden name=action value="sta">
<input type=submit name=suboper value="静态数据">
</form>上面的代码本身正确,但我想给dataread.jsp传递一个参数,比如:上面隐藏控件action的值“sta”,(这两段代码在同一页面),该怎么修改上面的程序,我用
xmlhttp.send("sta");却不正确,该怎么写请高手指教!
function GetResult()
{
var xmlhttp;
var strResult;
var prefixes = ["MSXML3.XMLHTTP","MSXML2.XMLHTTP","Microsoft.XMLHTTP","MSXML.XMLHTTP","Msxml2.XMLHTTP.5.0","Msxml2.XMLHTTP.4.0","Msxml2.XMLHTTP.3.0"];
if (window.ActiveXObject)
{
for(var i=0;i<prefixes.length;i++)
{
try
{
xmlhttp=new ActiveXObject(prefixes[i]);
break;
}
catch (e)
{
}
}
}
else if (window.XMLHttpRequest)
{
try
{
xmlhttp = new XMLHttpRequest();
}
catch (e)
{
alert("你的浏览器不支持XMLHttpRequest对象,请升级");
}
} xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4)
{
strResult=xmlhttp.responseText;
}
}; xmlhttp.open("GET","dataread.jsp",false);
xmlhttp.send(null); };
</script>
<form method=post action='datadisplay.jsp'>
<input type=hidden name=action value="sta">
<input type=submit name=suboper value="静态数据">
</form>上面的代码本身正确,但我想给dataread.jsp传递一个参数,比如:上面隐藏控件action的值“sta”,(这两段代码在同一页面),该怎么修改上面的程序,我用
xmlhttp.send("sta");却不正确,该怎么写请高手指教!
解决方案 »
免费领取超大流量手机卡,每月29元包185G流量+100分钟通话, 中国电信官方发货