javascript - Uncaught ReferenceError : Ajax is not defined

Question

我的 mybb 论坛出现错误。我使用了一个插件来进行高级统计。但是我在控制台中收到关于“Uncaught ReferenceError: Ajax is not defined”的错误，这就是为什么我的网站没有从顶部加载最新帖子的原因。

网站:http://satfriends.net/index.php

这是我的代码

<!-- start: changuondyu_stats_script -->
	<script language="JavaScript" type="text/javascript">
		var fcmenu;
var listtab = new Array();
listtab[0] = "allforum";
listtab[1] = "inforum&listforumid=8";
listtab[2] = "inforum&listforumid=21,15,7";
listtab[3] = "inforum&listforumid=14";


function cswitch(taborder)
{
fcmenu = listtab[taborder];
document.getElementById('ct'+taborder).className = 'thead';
for (i = 0; i <= 3; i++)
{
 if (i != taborder)
   {
    document.getElementById('ct'+i).className = 'tcat';
   }
}
changstats_post();
}

function hshowpost(request)
{
  if (request.readyState == 4 && request.status == 200)
	{
	document.getElementById('chang_latestposts').innerHTML = request.responseText;
	document.getElementById('cprogress_post').style.display="none";
	}
}
function hshowtop_user(request)
{
  if (request.readyState == 4 && request.status == 200)
	{
	document.getElementById('chang_top_user').innerHTML = request.responseText;
	document.getElementById('cprogress_top').style.display="none";
	}
}
function hshowtop_forum(request)
{
  if (request.readyState == 4 && request.status == 200)
	{
	document.getElementById('chang_top_forum').innerHTML = request.responseText;
	document.getElementById('cprogress_top').style.display="none";
	}
}


function changstats_post()
{
	document.getElementById('cprogress_post').style.display="inline";
	fcresult = document.getmenu.choosekq.value;
	new Ajax.Request('xmlhttp.php?do='+fcmenu+'&result='+fcresult, {method: 'GET', postBody: null, onComplete: function(request) { hshowpost(request); }});
}
function changstats_top_user()
{
	document.getElementById('cprogress_top').style.display="inline";
	fcmenu_top_user = document.getmenu.choosetop_user.value;
	fcresult = document.getmenu.choosekq.value;
	new Ajax.Request('xmlhttp.php?do='+fcmenu_top_user+'&result='+fcresult, {method: 'GET', postBody: null, onComplete: function(request) { hshowtop_user(request); }});
}
function changstats_top_forum()
{
	document.getElementById('cprogress_top').style.display="inline";
	fcmenu_top_forum = document.getmenu.choosetop_forum.value;
	fcresult = document.getmenu.choosekq.value;
	new Ajax.Request('xmlhttp.php?do='+fcmenu_top_forum+'&result='+fcresult, {method: 'GET', postBody: null, onComplete: function(request) { hshowtop_forum(request); }});
}

cswitch(0);
changstats_top_user();
changstats_top_forum();


</script>
<!-- end: changuondyu_stats_script -->

请给我一个解决方案，我该如何解决。

谢谢

Answer 1

您必须包括对包含方法 Ajax.Request 的 api 的引用。在您的情况下，它似乎是 Prototype 库。

在您的 html 页面顶部包含库:

<script src="https://ajax.googleapis.com/ajax/libs/prototype/1.7.3.0/prototype.js"></script>