RT,现在想试试不用ajaxpro.dll,代码如下:
function push()
{
document.getElementById("Result").value = "Transporting......";
var ID = document.getElementById("ID").value;
var URL = "/AJAX.ashx?ID=" + escape(ID);
request.open("GET", URL, true);
request.onreadystatechange = pop;
request.send(null);
}
function pop()
{
if (request.readyState == 4)
document.getElementById("Result").value = request.responseText;
}现在的问题是:AJAX.ashx应该怎么写才能返回一个值给JS啊??谢谢各位啦。
function push()
{
document.getElementById("Result").value = "Transporting......";
var ID = document.getElementById("ID").value;
var URL = "/AJAX.ashx?ID=" + escape(ID);
request.open("GET", URL, true);
request.onreadystatechange = pop;
request.send(null);
}
function pop()
{
if (request.readyState == 4)
document.getElementById("Result").value = request.responseText;
}现在的问题是:AJAX.ashx应该怎么写才能返回一个值给JS啊??谢谢各位啦。
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