感谢您阅读我的问题。
我正在使用这段代码,效果很好,但我正在尝试将按钮自定义为链接
原始码
<?xml version="1.0" encoding="utf-8"?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
<head>
<title>An XHTML 1.0 Strict standard template</title>
<meta http-equiv="content-type" content="text/html; charset=utf-8" />
<meta http-equiv="Content-Style-Type" content="text/css" />
<style type="text/css">
</style>
<script type="text/javascript">
function ajaxFunction(id, url){
var xmlHttp;
try {// Firefox, Opera 8.0+, Safari
xmlHttp = new XMLHttpRequest();
} catch (e) {// Internet Explorer
try {
xmlHttp = new ActiveXObject("Msxml2.XMLHTTP");
} catch (e) {
try {
xmlHttp = new ActiveXObject("Microsoft.XMLHTTP");
} catch (e) {
alert("Your browser does not support AJAX!");
return false;
}
}
}
xmlHttp.onreadystatechange = function(){
if (xmlHttp.readyState == 4) {
//Get the response from the server and extract the section that comes in the body section of the second html page avoid inserting the header part of the second page in your first page's element
var respText = xmlHttp.responseText.split('<body>');
elem.innerHTML = respText[1].split('</body>')[0];
}
}
var elem = document.getElementById(id);
if (!elem) {
alert('The element with the passed ID doesn\'t exists in your page');
return;
}
xmlHttp.open("GET", url, true);
xmlHttp.send(null);
}
</script>
</head>
<body>
<div id="test"></div>
<form>
<input type="button" value="Make Ajax Call" id="ajax" onclick="ajaxFunction('test','one.htm');"/>
</form>
</body>
</html>
我正在尝试转换
<input type="button" value="Make Ajax Call" id="ajax" onclick="ajaxFunction('test','one.htm');"/>
部分放入链接而不是按钮来完成相同的工作。
我该怎么做?
我试过以下
<a href="test3.php" value="Make Ajax Call" id="ajax" onclick="ajaxFunction('test','one.htm');">xxx</a>
我也希望 div 中的内容发生变化
<div id="test"></div>
我的意思是如果内容是“示例”,在我单击链接后它会更改为其他内容
使用下面的回复它不会给我“找不到错误”它只是不向我显示文件中的内容