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问个简单的ajax有关问题，小弟我都急死了

WBOYOriginal: 2016-06-13 12:48:37776browse

问个简单的ajax问题，我都急死了~
刚学ajax，写了一段~~感觉js里面的那个if(str=="good")不起作用~~如果单独拿出来返回的话确实输出good
放进去判断就不行了不知道为什么啊求解~弄了我好长时间~~急死了新手求教！！！！

这是html

<br />
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"><br />
<html xmlns="http://www.w3.org/1999/xhtml"><br />
<head><br />
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" /><br />
<title>无标题文档</title><br />
<script type="text/javascript" language="javascript" src="worinima.js"></script><br />
</head><br />
<br />
<body><br />
<br />
<br />
<a href="#" onclick="upsdowns('l','f','fsdf','sdf')">dsds</a><br />
<br />
<br />
</body><br />
</html><br />

这是js：

<br />
<br />
var xmlHttp;<br />
function upsdowns(ac,id,ud,mk)//,did)<br />
{<br />
<br />
//获取xmlHttpObject对象，如果为空，提示浏览器不支持ajax<br />
  xmlHttp=GetXmlHttpObject();<br />
var url;<br />
 url="ajax.php"+"?ac="+escape(ac)+"&url="+url+"&id="+escape(id)+"&ud="+escape(ud)+"&mk="+escape(mk)+"&sid="+Math.random();<br />
 //回调函数，执行动作<br />
xmlHttp.onreadystatechange=stateChanged; <br />
 //open<br />
xmlHttp.open("GET",url,true);<br />
xmlHttp.setRequestHeader("Content-Type","application/x-www-form-urlencoded");<br />
xmlHttp.send(null);<br />
} <br />
<br />
function stateChanged() <br />
{<br />
	<br />
if (xmlHttp.readyState==4)<br />
{ <br />
  {if(xmlHttp.status==200)   // phparray=new Array()<br />
     <br />
	 var str=xmlHttp.responseText;<br />
	 if(str=="dasda")<br />
	 alert(str);<br />
	 //document.getElementById("txtHint").innerHTML=str;<br />
    <br />
   }<br />
}<br />
<br />
}<br />
//获取xml对象<br />
function GetXmlHttpObject()<br />
{<br />
var xmlHttp=null;<br />
try<br />
{<br />
// Firefox, Opera 8.0+, Safari<br />
xmlHttp=new XMLHttpRequest();<br />
}<br />
catch (e)<br />
{<br />
// Internet Explorer<br />
try<br />
 {<br />
 xmlHttp=new ActiveXObject("Msxml2.XMLHTTP");<br />
 }<br />
catch (e)<br />
 {<br />
 xmlHttp=new ActiveXObject("Microsoft.XMLHTTP");<br />
 }<br />
}<br />
return xmlHttp;<br />
}<br />
<br />

这是ajax.php

<br />
<br />
header("Content-type: text/html;charset=utf-8");<br />
header('Vary: Accept-Language'); <br />
$w=$_GET['ac'];<br />
$a=$_GET['id'];<br />
$r=$_GET['ud'];;<br />
$t=$_GET['mk'];<br />
if($w!==''&&$a!=''&&$r!==''&&$t!=='')<br />
{echo"good";<br />
}<br />
<br />

------解决方案--------------------
function stateChanged()
{

if (xmlHttp.readyState==4)
{
  {if(xmlHttp.status==200)   // phparray=new Array()

     var str=xmlHttp.responseText;
     if(str=="dasda")
     alert(str);
     //document.getElementById("txtHint").innerHTML=str;

   }
}

}
标红处大括号位置放错了吧，应该是
function stateChanged()
{

if (xmlHttp.readyState==4)
{
  if(xmlHttp.status==200)   // phparray=new Array()

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