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javabean问题
摘要: javabean问题
这是 UserBean.java
package user;
public class UserBean
{
private String username="";
private int age=0;
public UserBean()
{
//username="";
//age=0;
}
public void setUserName(String username)
{
this.username=username;
}
public String getUserName()
{
return username;
}
public void setAge(int age)
{
this.age=age;
}
public int getAge()
{
return age;
}
}
这是 userbeanform.jsp
<%@ page contentType="text/html;charset=GB2312" %>
<head>
</head>
<body>
<form action="userbeanform1.jsp" method="post">
用户名: <input type="text" name="username" size="20"/>
年 龄:<input type="text" name="age" size="10"/>
<input type="submit" value="[ submit ]">
</form>
</body>
这是 userbeanform1.jsp
<%@ page contentType="text/html;charset=GB2312"%>
<head>
</head>
<body>
<%
//newuser.setUserName(request.getParameter("username"));
out.println("姓名: "+newuser.getUserName()+", 年龄: "+newuser.getAge());
%>
</body>
UserBean.java 已编译.UserBean放在 WEB-INF 的classes的userxia
运行userbeanform.jsp -> 输入用户名: yoiu 年龄 33 -> submit 显示 姓名:,年龄33
为何 不能给username赋值,高手指点,
如果使用 newuser.setUserName(request.getParameter("username"));可以给 username赋值
如果将userbeanform1.jsp 中的
换成
提示如下错误
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: Cannot find any information on property 'username' in a bean of type 'user.UserBean'
org.apache.jasper.runtime.JspRuntimeLibrary.internalIntrospecthelper(JspRuntimeLibrary.java:363)
org.apache.jasper.runtime.JspRuntimeLibrary.introspecthelper(JspRuntimeLibrary.java:306)
org.apache.jsp.userbeanform1_jsp._jspService(userbeanform1_jsp.java:61)
org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:94)
javax.servlet.http.HttpServlet.service(HttpServlet.java:802)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:324)
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/5.0.30 logs.
--------------------------------------------------------------------------------
Apache Tomcat/5.0.30
newuser.setUserName(request.getParameter("username"));
就这样不就行了吗??
"
"
NAME怎么都一样??
我是初学者哦,不知道该怎么写,为何这能输入 age 的值呢??
INT
STRING.....
?
反正使用
就报错:-(
解决此问题的.我奉为jsp牛人
我不想当你眼中的牛..........
(一个月后你自然就懂...)
解决了
把UserBean.java的变量username改为 userName 编译,同时改 userbeanform1.jsp的
为
就好了
沾沾自喜:-)
tomcat的弱智成就了一个天才
或者将UserBean.java中的
public class setUserName(String username)改为
public class setUsername(String username)
也是可以的
public void setUsername(String username)
{
this.username=username;
}
public String getUsername()
{
return username;
}
这样就对了