xmlhttp.status一直为404 怎么回事
HTML代码
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<script type="text/javascript">
function showCustomer(str) {
var xmlhttp;
if (str == "") {
document.getElementById("txtHint").innerHTML = "";
return;
}
if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
}
else {// code for IE6, IE5
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function () {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
document.getElementById("txtHint").innerHTML = xmlhttp.responseText;
}
}
xmlhttp.open("GET", "~/ajax/getcustomer.asp?q=" + str, true);
xmlhttp.send(null);
}
</script>
</head>
<body>
<form action="" style="margin-top:15px;">
<label>请选择一位客户:
<select name="customers" onchange="showCustomer(this.value)" style="font-family:Verdana, Arial, Helvetica, sans-serif;">
<option value="ANATR">Apple Computer, Inc.</option>
<option value="ANTON ">BAIDU, Inc</option>
<option value="Canon">Canon USA, Inc.</option>
<option value="Google">Google, Inc.</option>
<option value="Nokia">Nokia Corporation</option>
<option value="SONY">Sony Corporation of America</option>
</select>
</label>
</form>
<br />
<div id="txtHint">客户信息将在此处列出 ...</div>
</body>
</html>
ASP代码:
<html>
<body>
<%
response.expires=-1
sql="SELECT * FROM CUSTOMERS WHERE CUSTOMERID="
sql=sql & "'" & request.querystring("q") & "'"
Dim sConnection, objConn , rs
sConnection = "DRIVER=={SQL Server}; SERVER=localhost; DATABASE=clients; UID=sa;PWD=123456"
Set objConn = Server.CreateObject("ADODB.Connection")
objConn.Open(sConnection)
Set rs = objConn.Execute(sql)
response.write("<table>")
do until rs.EOF
for each x in rs.Fields
response.write("<tr><td><b>" & x.name & "</b></td>")
response.write("<td>" & x.value & "</td></tr>")
next
rs.MoveNext
loop
response.write("</table>")
rs.close
conn.close
%>
</body>
</html>
------解决方案--------------------
xmlhttp.open("GET", "~/ajax/getcustomer.asp?q=" + str, true);
这里的ajax处理程序的路径是否正确
------解决方案--------------------输出信息看看
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