function load_ajax(str)
{
var xmlHttp;
if(window.XMLHttpRequest){
xmlHttp = new XMLHttpRequest();
} else{
xmlHttp = new ActiveXobject("Microsoft.XMLHTTP");
}
var url = "control/act_ajax.php?target="+str+"&sid=" + Math.random();
xmlHttp.open("get",url,true);
xmlHttp.setRequestHeader("Content-type","application/x-www-form-urlencoded; charset=utf-8");
xmlHttp.send(null);
xmlHttp.onreadystatechange =function () {
if (xmlHttp.readyState == 4) {
if (xmlHttp.status == 200) {
var return_str = xmlHttp.responseText;
alert( return_str);
return return_str;
}
}
}}
function get_blackpic()
{
var bg_img = load_ajax("blackpic");
alert(bg_img);
}
get_blackpic();为什么弹出来 的信息是 undefind
=>
xmlhttp.ResponseText; 试试
xmlHttp.setRequestHeader("Content-type","application/x-www-form-urlencoded; charset=utf-8");//用get请求这句可以不要。另外你php端是怎么写的呢?是可以得到数据的!只是调取函数就得不到数据了,是不是这样传递ajax数据方式不对?