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controller cannot be mappped

 
Nilay Shankar
Greenhorn
Posts: 1
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I have built a small login application using spring mvc. but there is some error coming (HTTP Status 404 - /Myspringapp11/login).
I have used Multiactioncontroller with simpleurlhandlermapping.

Here goes the code:

web.xml file
<servlet>
<servlet-name>myServlet</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherSe rvlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/myServlet-servlet.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>

<servlet-mapping>
<servlet-name>myServlet</servlet-name>
<url-pattern>*.*</url-pattern>
</servlet-mapping>
----------------------------------------------------------------------------


myServlet-servlet.xml

<beans>

<bean id="commonResolver" class="org.springframework.web.servlet.mvc.multiac tion.ParameterMethodNameResolver">
<property name="paramName">
<value>method</value>
</property>
</bean>

<bean id="loginController" class="beans.LoginController">
<property name="methodNameResolver"><ref bean="commonResolver"/></property>
</bean>

<bean id="urlMapping" class="org.springframework.web.servlet.handler.Sim pleUrlHandlerMapping">
<property name="mappings">
<props>
<prop key="login">loginController</prop> <!-- there should be bean defined with the name of loginController -->
</props>
</property>
</bean>

<bean id="viewResolver" class="org.springframework.web.servlet.view.Intern alResourceViewResolver">
<property name="prefix"><value>/</value></property>
<property name="suffix"><value>.jsp</value></property>
</bean>

</beans>
--------------------------------------------------------------------------
login.jsp

<%@ page language="java" contentType="text/html; charset=ISO-8859-1"
pageEncoding="ISO-8859-1"%>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
<title>Insert title here</title>
<script language="javascript">

function submitForm(formName,action)

{

obj = eval('document.'+formName);

obj.action = action;
obj.submit();
}

</script>
</head>
<body>

<h3>
Please login to enter
</h3>
<form name="form1" method="post" action="">

Username:
<input type="text" name="user" />
<br><br>
Password:
<input type="password" name="pass" />
<br><br>

<input type="button" name="sub1" value="Login" onclick="submitForm('form1','login?method=loginhan dle')" />
<input type="button" name="sub2" value="Register" onclick="submitForm('form1','login?method=toNewUse r')" />

</form>

</body>
</html>
-----------------------------------------------------------------------


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

LoginController.java

package beans;
import javax.servlet.http.*;

import org.springframework.validation.BindException;
import org.springframework.web.servlet.ModelAndView;
import org.springframework.web.servlet.mvc.multiaction.Mu ltiActionController;

public class LoginController extends MultiActionController
{


protected ModelAndView loginhandle(HttpServletRequest arg0,
HttpServletResponse arg1)
{

Mydao myd=new Mydao();
Logininfo logininfo=new Logininfo();

if (myd.validate(logininfo.getUser(), logininfo.getPass()))
{
System.out.println("22222");
return new ModelAndView("home");
}

else
{
System.out.println("333333");
return new ModelAndView("failure");
}
}

protected ModelAndView toNewUser(HttpServletRequest arg0,
HttpServletResponse arg1)
{
return new ModelAndView("newuser");


}

}
--------------------------------------------------------------------------

Logininfo.java

package beans;
public class Logininfo {

private String user;
private String pass;

public String getUser() {
return user;
}
public void setUser(String user) {
this.user = user;
}
public String getPass() {
return pass;
}
public void setPass(String pass) {
this.pass = pass;
}



}
---------------------------------------------------------------------

mydao.java

package beans;

import java.sql.*;

public class Mydao {

public boolean validate(String user, String pass)
{
Boolean b1=false;
ResultSet rs;


try
{
Class.forName("com.mysql.jdbc.Driver");
Connection con=DriverManager.getConnection("jdbc:mysql://localhost:3306/mydb1","root","root");
PreparedStatement pst=con.prepareStatement("select * from loginuser where username=? and password=?");
pst.setString(1, user);
pst.setString(2, pass);
System.out.println("user:"+user+"pass"+pass);
rs=pst.executeQuery();


if(rs.next())
{
b1= true;
}

else
{
b1=false;
}

}
catch(Exception e)
{
System.out.println(e);
}
return b1;

}
}
--------------------------------------------------------------------------

Pleas help me. Its a very small and simple application. I am getting this error in the browser when i click on any of the 2 buttons.

HTTP Status 404 - /Myspringapp11/login

type Status report

message /Myspringapp11/login

description The requested resource (/Myspringapp11/login) is not available.

-->there is no error in the eclipse console. Why is my application not able to map the controller??? Please help....

Thanks in advance
Nilay
 
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