function subit(){
ajax= new ActiveXObject("Microsoft.XMLHTTP");
f=document.form1;
username=f.username.value;
userqq=f.userqq.value;
title=f.title.value;
contents=f.contents.value;
var poststr="username="+username+"&userqq="+userqq+"&title="+title+"&contents="+contents;
var url="handle.php";
ajax.open("GET",url,true);
ajax.setRequestHeader("Content-Type","application/x-www-form-urlencoded");
ajax.onreadystatechange=function()
{
if(ajax.readystate==4&&ajax.status==200){
f.username.value="";
f.userqq.value="";
f.title.value="";
f.contents.value="";
showguest();
}
}
ajax.send("username="+username+"&userqq="+userqq+"&title="+title+"&contents="+contents);
}
ajax= new ActiveXObject("Microsoft.XMLHTTP");
f=document.form1;
username=f.username.value;
userqq=f.userqq.value;
title=f.title.value;
contents=f.contents.value;
var poststr="username="+username+"&userqq="+userqq+"&title="+title+"&contents="+contents;
var url="handle.php";
ajax.open("GET",url,true);
ajax.setRequestHeader("Content-Type","application/x-www-form-urlencoded");
ajax.onreadystatechange=function()
{
if(ajax.readystate==4&&ajax.status==200){
f.username.value="";
f.userqq.value="";
f.title.value="";
f.contents.value="";
showguest();
}
}
ajax.send("username="+username+"&userqq="+userqq+"&title="+title+"&contents="+contents);
}
}
后使用$_GET['username'],得到的结果是空。
或者去掉这行
是发送post数据发送get数据要
ajax.open("GET",url+'?'+poststr,true);
当然需要严格判定数据来源的场合例外
你说的
发送get数据要
ajax.open("GET",url+'?'+poststr,true);
我的代码中已经有了啊
见“ajax.open("GET",url,true);”