注册 登录
编程论坛 J2EE论坛

[求助]谁能看看这个程序那个地方错了

zplove 发布于 2006-07-30 12:32, 593 次点击
<%@ page contentType="text/html;charset=gb2312"%>
<html>
<body>
<form name="form1" action="forwardExample1.jsp" method="post" >
程序示例链接:
<select name="goaddress" onchange="javascript:form1.submit()">
<option value="novalue"></option>
<option value="1">实例13</option>
<option value="2">实例14</option>
<option value="3">实例15</option>
</select>
</form>
<%// forward应用示例Java程序片
String s=null;
s=request.getParameter("goaddress");
if(s!=null)
{ switch(s.charAt(0))
{ case '1':
%>
<jsp:forward page="xhApp1.jsp">
</jsp:forward>
<%
break;
case '2':
%>
<jsp:forward page="simpleCounterApp1.jsp">
</jsp:forward>
<%
break;
case '3':
%>
<jsp:forward page="includeSample1.jsp">
</jsp:forward>
<%
break;
default:
out.println("您没有选择。");
}
}
else
out.println("您没有选择。");
%>
</body>
</html>
7 回复
#2
ziyehanbin2006-07-30 16:44
if(s!=null && s.equals("")){

.........................

}
#3
zplove2006-07-30 17:03

你给改的怎么还运行不出来啊
#4
zplove2006-07-30 17:06

这个是运行时的错误
HTTP Status 500 -

--------------------------------------------------------------------------------

type Exception report

message

description The server encountered an internal error () that prevented it from fulfilling this request.

exception

org.apache.jasper.JasperException: /forwardExample1.jsp(22,7) Expecting "jsp:param" standard action with "name" and "value" attributes
org.apache.jasper.compiler.DefaultErrorHandler.jspError(DefaultErrorHandler.java:39)
org.apache.jasper.compiler.ErrorDispatcher.dispatch(ErrorDispatcher.java:409)
org.apache.jasper.compiler.ErrorDispatcher.jspError(ErrorDispatcher.java:90)
org.apache.jasper.compiler.Parser.parseParam(Parser.java:852)
org.apache.jasper.compiler.Parser.parseBody(Parser.java:1800)
org.apache.jasper.compiler.Parser.parseOptionalBody(Parser.java:1060)
org.apache.jasper.compiler.Parser.parseForward(Parser.java:902)
org.apache.jasper.compiler.Parser.parseStandardAction(Parser.java:1213)
org.apache.jasper.compiler.Parser.parseElements(Parser.java:1559)
org.apache.jasper.compiler.Parser.parse(Parser.java:126)
org.apache.jasper.compiler.ParserController.doParse(ParserController.java:220)
org.apache.jasper.compiler.ParserController.parse(ParserController.java:101)
org.apache.jasper.compiler.Compiler.generateJava(Compiler.java:203)
org.apache.jasper.compiler.Compiler.compile(Compiler.java:470)
org.apache.jasper.compiler.Compiler.compile(Compiler.java:451)
org.apache.jasper.compiler.Compiler.compile(Compiler.java:439)
org.apache.jasper.JspCompilationContext.compile(JspCompilationContext.java:511)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:295)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:292)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:236)
javax.servlet.http.HttpServlet.service(HttpServlet.java:802)


note The full stack trace of the root cause is available in the Apache Tomcat/5.0.28 logs.

#5
神vLinux飘飘2006-07-30 17:39
很遗憾,我运行一点错误都没有
#6
zplove2006-07-30 18:18
不会吧
你把你的的全代码发上来
让我看看
#7
ziyehanbin2006-07-31 07:34
昨天没注意看,呵呵!sorry!!

----------------------------------------------------------

<%// forward应用示例Java程序片
String s=null;
s=request.getParameter("goaddress");
if(s!=null && !s.equals(""))
{ switch(s.charAt(0))
{ case '1':
%>
<jsp:forward page="xhApp1.jsp"/>
<%
break;
case '2':
%>
<jsp:forward page="simpleCounterApp1.jsp"/>
<%
break;
case '3':
%>
<jsp:forward page="includeSample1.jsp"/>
<%
break;
default:
out.println("您没有选择。");
}
}
else
out.println("您没有选择。");
%>

------------------------------------------------------------------
都是初学者,共同学习!!
#8
zplove2006-07-31 08:32
谢谢你啊
运行出来了
1