这个JSP程序怎么错了?

来源:百度知道 编辑:UC知道 时间:2024/06/08 23:51:50
我初学JSP,MyEclipse的报错为:
String firstL=request.getParameter("1th");
String secondL=request.getParameter("2th");
String thirdL=request.getParameter("3th");
以下为代码:
<%@ page contentType="text/html;charset=GB2312"%>
<%@ import="java.lang.Math %>
<%
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">
<

我帮你改了一下, triangle类加个带参数构造方法
triangle(String firstL,String secondL,String thirdL)
{
a=Double.parseDouble(firstL);
b=Double.parseDouble(secondL);
c=Double.parseDouble(thirdL);
}
然后把
String firstL=request.getParameter("1th");
String secondL=request.getParameter("2th");
String thirdL=request.getParameter("3th"); 这三句放到triangle tt=new triangle();
上面,并将triangle tt=new triangle(); 改为triangle tt=new triangle(firstL,secondL,thirdL); 就行了

你把错误提示写出来吧

把<%!这个叹号去掉试试。不过我建议另写一个java类(创建一个Servlet),将action的地址指向这个servlet.