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
<!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><meta http-equiv="Content-Type" content="text/html; charset=utf-8" /><title>无标题文档</title><script type="text/javascript" language="javascript" src="worinima.js"></script></head><body><a href="#" onclick="upsdowns('l','f','fsdf','sdf')">dsds</a></body></html>这是js：
var xmlHttp;function upsdowns(ac,id,ud,mk)//,did){//获取xmlHttpObject对象，如果为空，提示浏览器不支持ajax xmlHttp=GetXmlHttpObject();var url; url="ajax.php"+"?ac="+escape(ac)+"&url="+url+"&id="+escape(id)+"&ud="+escape(ud)+"&mk="+escape(mk)+"&sid="+Math.random(); //回调函数，执行动作xmlHttp.onreadystatechange=stateChanged; //openxmlHttp.open("GET",url,true);xmlHttp.setRequestHeader("Content-Type","application/x-www-form-urlencoded");xmlHttp.send(null);} 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; }}}//获取xml对象function GetXmlHttpObject(){var xmlHttp=null;try{// Firefox, Opera 8.0+, SafarixmlHttp=new XMLHttpRequest();}catch (e){// Internet Explorertry { xmlHttp=new ActiveXObject("Msxml2.XMLHTTP"); }catch (e) { xmlHttp=new ActiveXObject("Microsoft.XMLHTTP"); }}return xmlHttp;}这是ajax.php
header("Content-type: text/html;charset=utf-8");header('Vary: Accept-Language'); $w=$_GET['ac'];$a=$_GET['id'];$r=$_GET['ud'];;$t=$_GET['mk'];if($w!==''&&$a!=''&&$r!==''&&$t!==''){echo"good";}
回复讨论(解决方案)
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
XML/HTML code?12345678910111213141516nbsp;html PUBLIC "…… 里面那个是if(str=="good")我复制错了~~其实打得是对的
用火狐的firebug看看JS提示什么错误
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()
{
var str=xmlHttp.responseText;
if(str=="dasda")
alert(str);
//document.getElementById("txtHint").innerHTML=str;
}
}
}
同意楼上，应该是语法错误 其它的没问题
xmlHttp.setRequestHeader("Content-Type","application/x-www-form-urlencoded");
你的是get方式提交，这句也不需要了。
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