ajax无法显示返回的数据

Question

<script type="text/javascript">
var maxID = 0;
function showmsg()
var xhr = new XMLHttpRequest();
xhr.onreadystatechange = function(){
if(xhr.readyState==4){
eval("var jn_info="+xhr.responseText);
var s = "";
for(var i=0; i < jn_info.length; i++){
s += '<p>'+jn_info[i].jieshuo+" :"+jn_info[i].content+"</p>";
maxID = info[i].id;
}
var hm =document.getElementById('show_msg');
hm.innerHTML += s;
hm.scrollTop = hm.scrollHeight;
}
}
xhr.open('get','./jieshuo_data.php?maxID='+maxID);
xhr.send(null);
}

 window.onload = function(){
            setInterval("showmsg()",2000);
        }
</script>

//处理页面
$maxID = $_GET['maxID'];

$sql = "select * from jieshuo where id>".$maxID;
$rs=mysqli_query($conn,$sql);
$info = array();
while($row= mysqli_fetch_assoc($rs)){
$info[] = $row;
}

echo json_encode($info);

{

Answer 1

兄弟，你还是把代码格式化下，然后我在帮你看看，你这个看的头疼

<script type="text/javascript">
var maxID = 0;
function showmsg()
var xhr = new XMLHttpRequest();
xhr.onreadystatechange = function(){
if(xhr.readyState==4){
eval("var jn_info="+xhr.responseText);
var s = "";
for(var i=0; i < jn_info.length; i++){
s += '<p>'+jn_info[i].jieshuo+" :"+jn_info[i].content+"</p>";
maxID = info[i].id;    //没看懂这句有何意义，你也没有比较最大id啊，还有maxID = info[i].id;     这个你是不是调用错了？哪来的info，不是jn_info吗？
}
var hm =document.getElementById('show_msg');
hm.innerHTML += s;
hm.scrollTop = hm.scrollHeight;
}
}
xhr.open('get','./jieshuo_data.php?maxID='+maxID);
xhr.send(null);
}

 window.onload = function(){
            setInterval("showmsg()",2000);
        }
</script>