显示不了正确的文本，responseText
三个文件分别是
l.html
myAJAXlib.js
libtest.php
不能接收html传递的参数param,要怎么写呢？
- HTML code
<html><head><script language="javascript" src="myAJAXlib.js"></script><script language="javascript">function cback(text){ alert(text);}</script></head><body><form name="form1"><input type="button" value="test" onclick="doAjax('libtest.php','param=hello','cback','get','0')"></body></html>- JScript code
function createREQ(){ var req=false; try{ req=new XMLHttpRequest(); } catch(err1) { try{ req=new ActiveXObject("Msxml2.XMLHTTP"); } catch(err2) { try{ req=new ActiveXObject("Microsoft.XMLHTTP"); } catch(err3) { req=false; } } } return req;}function requestGET(url,query,req){ myRand=parseInt(Math.random()*99999); req.open("GET",url+'?'+query+'&rand='+myRand,true); req.send(null);}function requestPOST(url,query,req){ req.open("POST",url,true); req.setRequestHeader('Content-Type','application.x-www-form-urlencoded'); req.send(query);}function doCallback(callback,item){ eval(callback+'(item)');}function doAjax(url,query,callback,reqtype,getxml){ var myreq=createREQ(); myreq.onreadystatechange=function(){ if(myreq.readyState==4){ if(myreq.status==200){ var item=myreq.responseText; if(getxml==1){ item=myreq.responseXML; } doCallback(callback,item);}}}if(reqtype=='post'){requestPOST(url,query,myreq);}else{requestGET(url,query,myreq);}}- PHP code
<?phpecho "Parameter value was".$param;?>
[解决办法]
$_GET是获取GET提交的键值对，$_POST则是POST提交的
ajax提交的和普通form提交的或者直接在url后加参数【get】获取参数的方法一样