我的Jsp页面使用了下列JavaBean
<jsp:useBean id="User" scope="page" class="com.LunchMoney.UserInfo" />package com.LunchMoney;
import java.sql.*;/**
* @author Administrator
*
* TODO 要更改此生成的类型注释的模板,请转至
* 窗口 - 首选项 - Java - 代码样式 - 代码模板
*/
public class UserInfo
{
private String kName;
public UserInfo()
{
this.kName = "kkk";
}
public void setkName(String NameStr)
{
this.kName = NameStr;
}
public String getkName()
{
return (this.kName);
}
}在执行Jsp页面时出现下列提示:
type Exception reportmessage description The server encountered an internal error () that prevented it from fulfilling this request.exception org.apache.jasper.JasperException: /Index.jsp(25,0) The value for the useBean class attribute com.LunchMoney.UserInfo is invalid.
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:150)
org.apache.jasper.compiler.Generator$GenerateVisitor.visit(Generator.java:1227)
org.apache.jasper.compiler.Node$UseBean.accept(Node.java:1116)
org.apache.jasper.compiler.Node$Nodes.visit(Node.java:2163)
org.apache.jasper.compiler.Node$Visitor.visitBody(Node.java:2213)
org.apache.jasper.compiler.Node$Visitor.visit(Node.java:2219)
org.apache.jasper.compiler.Node$Root.accept(Node.java:456)
org.apache.jasper.compiler.Node$Nodes.visit(Node.java:2163)
org.apache.jasper.compiler.Generator.generate(Generator.java:3272)
org.apache.jasper.compiler.Compiler.generateJava(Compiler.java:244)
org.apache.jasper.compiler.Compiler.compile(Compiler.java:470)
org.apache.jasper.compiler.Compiler.compile(Compiler.java:451)
org.apache.jasper.compiler.Compiler.compile(Compiler.java:439)
org.apache.jasper.JspCompilationContext.compile(JspCompilationContext.java:511)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:295)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:292)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:236)
javax.servlet.http.HttpServlet.service(HttpServlet.java:802)
note The full stack trace of the root cause is available in the Apache Tomcat/5.0.28 logs.请各位高手赐教!!!小弟不胜感激!!
<jsp:useBean id="User" scope="page" class="com.LunchMoney.UserInfo" />package com.LunchMoney;
import java.sql.*;/**
* @author Administrator
*
* TODO 要更改此生成的类型注释的模板,请转至
* 窗口 - 首选项 - Java - 代码样式 - 代码模板
*/
public class UserInfo
{
private String kName;
public UserInfo()
{
this.kName = "kkk";
}
public void setkName(String NameStr)
{
this.kName = NameStr;
}
public String getkName()
{
return (this.kName);
}
}在执行Jsp页面时出现下列提示:
type Exception reportmessage description The server encountered an internal error () that prevented it from fulfilling this request.exception org.apache.jasper.JasperException: /Index.jsp(25,0) The value for the useBean class attribute com.LunchMoney.UserInfo is invalid.
org.apache.jasper.compiler.DefaultErrorHandler.jspError(DefaultErrorHandler.java:39)
org.apache.jasper.compiler.ErrorDispatcher.dispatch(ErrorDispatcher.java:409)
org.apache.jasper.compiler.ErrorDispatcher.jspError(ErrorDispatcher.java:150)
org.apache.jasper.compiler.Generator$GenerateVisitor.visit(Generator.java:1227)
org.apache.jasper.compiler.Node$UseBean.accept(Node.java:1116)
org.apache.jasper.compiler.Node$Nodes.visit(Node.java:2163)
org.apache.jasper.compiler.Node$Visitor.visitBody(Node.java:2213)
org.apache.jasper.compiler.Node$Visitor.visit(Node.java:2219)
org.apache.jasper.compiler.Node$Root.accept(Node.java:456)
org.apache.jasper.compiler.Node$Nodes.visit(Node.java:2163)
org.apache.jasper.compiler.Generator.generate(Generator.java:3272)
org.apache.jasper.compiler.Compiler.generateJava(Compiler.java:244)
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.请各位高手赐教!!!小弟不胜感激!!
package com.LunchMoney;
import java.sql.*;/**
* @author Administrator
*
* TODO 要更改此生成的类型注释的模板,请转至
* 窗口 - 首选项 - Java - 代码样式 - 代码模板
*/
public class UserInfo{
private String Name;
public void setName(String NameStr)
{
this.Name = NameStr;
}
public String getName()
{
return Name;
}
public ResultSet getAllUserInfo()
{
String SqlStr = "Select PersonID,Name From T_PersonID Where IsUsed = '0'";
ConnectDBMS Db = new ConnectDBMS();
ResultSet Rs = Db.executeQuerySql(SqlStr);
Db.closeConn();
return Rs;
}
}
我在Jsp页面中只要加入<jsp:useBean id="User" scope="page" class="com.LunchMoney.UserInfo" />
访问页面时就出现上面的情况了,我还没有使用它里面的属性呢?
将使用JavaBean的这句去掉就可以了,到底是什么原因啊?
因为你的信息不全.所以我举例出几个地方你看看是不是合适.
1.重要的一点,在你放了JAVABEAN到classes里,有没有将服务器重新启动.
2.你BEAN的路径是不是/classes/com/LunchMoney/UserInfo.class.
3.假如不是这个问题.你就试试
<%@ import="com.LunchMoney.*"%>
<jsp:useBean id="User" scope="page" class="UserInfo"/>
Db.closeConn();
return Rs;这样可以吗?
{
String SqlStr = "Select PersonID,Name From T_PersonID Where IsUsed = '0'";
ConnectDBMS Db = new ConnectDBMS();
ResultSet Rs = Db.executeQuerySql(SqlStr);
Db.closeConn();
return Rs;
}
这样的用法没见过,也许JAVABEAN没有这种用法
2.你BEAN的路径是不是/classes/com/LunchMoney/UserInfo.class.
3.假如不是这个问题.你就试试
<%@ page import="com.LunchMoney.*"%>
<jsp:useBean id="User" scope="page" class="UserInfo"/>