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不要急，我与你一起学习JSP（一）――平台搭建，让你眼前一亮

2014-08-12 13:55 323 查看

<%@ page language="java" import="java.util.*" pageEncoding="UTF-8"%>
<%@page import="com.deng.testjs.model.Person"%>
<%
String path = request.getContextPath();
String basePath = request.getScheme()+"://"+request.getServerName()+":"+request.getServerPort()+path+"/";
%>

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<base href="<%=basePath%>">

<title>My JSP 'index.jsp' starting page</title>
<meta http-equiv="pragma" content="no-cache">
<meta http-equiv="cache-control" content="no-cache">
<meta http-equiv="expires" content="0">
<meta http-equiv="keywords" content="keyword1,keyword2,keyword3">
<meta http-equiv="description" content="This is my page">
<!--
<link rel="stylesheet" type="text/css" href="styles.css">
-->
<script type="text/javascript">

/**
利用递归实现:
1 1 2 3 5 8 13 21 34 根据规律(从第3项开始后项等于前面两项之和),请第30项的值
*/

function test(){
alert("非递归值为=" + cal(9));
alert("递归值为=" + calcForRecursive(9));

}

/**
不用递归实现
*/
function cal(number){
if(number <= 2){
return 1;
}

var a1 = 1;
var a2 = 1;
var result;
var temp;

for(var i = 2; i < number; i++){
result = a1 + a2;
temp = a2;
a2 = result;
a1 = temp;
}

return result;
}

function calcForRecursive(number){
if(number <= 2){
return 1;
}

return recursiveHandler(1, 1, 2,number);

}

function recursiveHandler(a1,a2,count,number){
var result = a1 + a2;
count++;
if(count == number){
return result;
}else{
return recursiveHandler(a2,result,count,number);
}
}

</script>

</head>

<body>
This is my JSP page. <br>
<input type="button" onclick="test()" value="test">
</body>
</html>

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