Spring MVC Login application
In this
page, I have given spring MVC login application step by step.
Steps to create spring MVC web application:
1. Load the required jar file.
2. Create the request page to give
input.
3. Create the Controller class.
4. Create the web.xml file.
5. Create the spring configuration xml
file.
6. Create output page
7. Run the project.
Load the required jar files:
Note: load the jar files in lib folder also
·
commons-logging-1.1.1
·
org.springframework.asm-3.0.1
·
org.springframework.beans-3.0.1
·
org.springframework.context-3.0.1
·
org.springframework.core-3.0.1
·
org.springframework.expression-3.0.1
·
org.springframework.web-3.0.1
·
org.springframework.web.servlet-3.0.1
Create the request page to give
input: This file
contains input fields.
index.jsp
<html>
<head>
<meta http-equiv="Content-Type"
content="text/html; charset=ISO-8859-1">
<title>Insert title here</title>
</head>
<body>
<h1>Login Page</h1>
<form action="./loginform.html"
method="post">
<p>username : narendar
and password : admin123</p>
Name : <input type="text"
name="name"/><br>
Password : <input type="password"
name="password"/><br>
<input type="submit"
value="submit"/>
</form>
</body>
</html>
Create the Controller class: in this file we get the details from
input page and send the result to success page.
LoginController.java
package com.javathub;
import
javax.servlet.http.HttpServletRequest;
import
javax.servlet.http.HttpServletResponse;
import
org.springframework.web.servlet.ModelAndView;
import
org.springframework.web.servlet.mvc.Controller;
public class LoginController implements Controller {
@Override
public ModelAndView
handleRequest(HttpServletRequest req,
HttpServletResponse res) throws Exception {
String name=req.getParameter("name");
String password=req.getParameter("password");
if(password.equals("admin123")&&name.equals("narendar"))
{
String msg="Hello...welcome..."+name;
return new ModelAndView("homepage","message",msg);
}
else{
return new ModelAndView("errorpage","message","incorrect
username and password, please try again");
}
}
}
Create the web.xml file: in this file we are specifying the
DispatcherServlet, it is act as front controller. This file forwards the input
requests to DispatcherServlet.
<?xml version="1.0"
encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://java.sun.com/xml/ns/javaee" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee
http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" id="WebApp_ID" version="2.5">
<display-name>springMVC_login</display-name>
<welcome-file-list>
<welcome-file>index.jsp</welcome-file>
</welcome-file-list>
<servlet>
<servlet-name>login</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>login</servlet-name>
<url-pattern>*.html</url-pattern>
</servlet-mapping>
</web-app>
Create the spring configuration xml
file: this file name
should be (what we specified in web.xml)servletname-servlet.xml, here we define
the handler mapping class, Controller class and configure the view resolver.
login-servlet.xml
<?xml version="1.0"
encoding="UTF-8"?>
<!DOCTYPE beans
PUBLIC "-//SPRING//DTD
BEAN 2.0//EN"
"http://www.springframework.org/dtd/spring-beans-2.0.dtd">
<beans>
<!--default
handler mapping -->
<bean class="org.springframework.web.servlet.handler.BeanNameUrlHandlerMapping">
</bean>
<!-- define
controller name -->
<bean name="/loginform.html"
class="com.javathub.LoginController"></bean>
<!-- view
resolver -->
<bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name="prefix"
value="/"></property>
<property name="suffix"
value=".jsp"></property>
</bean>
</beans>
Output page:
homepage.jsp
${message}
errorpage.jsp
${message}
<jsp:include page="index.jsp"></jsp:include>
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