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问个简单的ajax有关问题，小弟我都急死了

WBOY
WBOYOriginal
2016-06-13 11:21:18922browse

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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