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!无法获取XMLHttpRequest.send方法传

发布时间: 2012-01-05 22:36:54 作者: rapoo

求助!无法获取XMLHttpRequest.send方法传的值。
如题~求助
<SCRIPT CODE>
<script language= "javascript " type= "text/javascript ">
var xmlHttp;

function createXMLHttpRequest() {
xmlHttp = new ActiveXObject( "Microsoft.XMLHTTP ");
}

function startRequest() {
createXMLHttpRequest();
xmlHttp.onreadystatechange = handleStateChange;

xmlHttp.open( "GET ",createQueryString(), true);
xmlHttp.send();
}

function createQueryString() {
var QueryString;
if(document.getElementById( "City ").value != " ")
{
QueryString = "DealData.aspx?ID= " + document.getElementById( "City ").value;
}
return QueryString;
}
</SCRIPT CODE>

<C# CODE>
try
{
string ff = Request.Url.ToString();
City = Page.Request.QueryString[ "CityID "].ToString();
if(City.Equals( " "))
{
City = "aa ";
}
}
catch(Exception err)
{
City = "bb ";
}
</C# CODE>

但是。。catch不到。。而且Page.Request.QueryString[ "CityID "].ToString();没有任何值

[解决办法]
DealData.aspx?ID= "
这个传过去的是ID,不是CityID

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