java - How does servlet container process this http request? -


i have created dynamic web project using eclipse first time work servlet , jsp.

below servlet code, 

package com.example.tutorial;  import java.io.ioexception; import java.io.printwriter;  import javax.servlet.servletexception; import javax.servlet.http.httpservlet; import javax.servlet.http.httpservletrequest; import javax.servlet.http.httpservletresponse;   public class servletexample extends httpservlet {     private static final long serialversionuid = 1l;      protected void service(httpservletrequest request, httpservletresponse response) throws servletexception, ioexception {         printwriter out = response.getwriter();         string firstname = request.getparameter("firstname");         string lastname = request.getparameter("lastname");          out.println(firstname + " " + lastname);     }  }

and corresponding web.xml,

<?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" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" 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>servletsjspexample</display-name>   <welcome-file-list>     <welcome-file>index.html</welcome-file>     <welcome-file>index.htm</welcome-file>     <welcome-file>index.jsp</welcome-file>     <welcome-file>default.html</welcome-file>     <welcome-file>default.htm</welcome-file>     <welcome-file>default.jsp</welcome-file>   </welcome-file-list>   <servlet>     <description></description>     <display-name>servletexample</display-name>     <servlet-name>servletexample</servlet-name>     <servlet-class>com.example.tutorial.servletexample</servlet-class>   </servlet>   <servlet-mapping>     <servlet-name>servletexample</servlet-name>     <url-pattern>/servletexample</url-pattern>   </servlet-mapping> </web-app>

i have written index.jsp has below form:

<?xml version="1.0" encoding="iso-8859-1" ?> <%@ page language="java" contenttype="text/html; charset=iso-8859-1"     pageencoding="iso-8859-1"%> <!doctype html public "-//w3c//dtd xhtml 1.0 transitional//en" "http://www.w3.org/tr/xhtml1/dtd/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="content-type" content="text/html; charset=iso-8859-1" /> <title>insert title here</title> </head> <body>     <form action="servletexample" method="post" >        <table border="0">          <tr>             <td>first name:</td> <td><input type="text" name="firstname" /></td>          </tr>          <tr>             <td>last name:</td> <td><input type="text" name="lastname"  /></td>          </tr>          <tr>             <td colspan="2"> <input type="submit" value="submit" /></td>          </tr>        </table>     </form> </body> </html>

in first case, within eclipse, if select run -> run on server, browser within eclipse shows presentation code of index.jsp uri: http://localhost:8081/servletsjspexample/

i later add below line of code in servlet,

this.getservletcontext().getrequestdispatcher("/index.jsp").forward(request, response);

as shown below,

public class servletexample extends httpservlet {     private static final long serialversionuid = 1l;      protected void service(httpservletrequest request, httpservletresponse response) throws servletexception, ioexception {         printwriter out = response.getwriter();          this.getservletcontext().getrequestdispatcher("/index.jsp").forward(request, response);          string firstname = request.getparameter("firstname");         string lastname = request.getparameter("lastname");          out.println(firstname + " " + lastname);     }  }

in second case, within eclipse, if select run -> run on server, browser within eclipse shows presentation code of index.jsp uri: http://localhost:8081/servletsjspexample/servletexample

so,

wrt these 2 cases, how uri getting changed? how control flowing servlet container application servletsjspexample before , after adding line of code this.getservletcontext().getrequestdispatcher("/index.jsp").forward(request, response);

please me understand this!!!

note: servletsjspexample "dynamic web project" name in eclipse

first, forget eclipse , how launch program within it. servlets, url-mappings, , urls hit.

in first sample, hit 

http://localhost:8081/servletsjspexample/

assuming servletsjspexample context root, you're reaching root of app. since have welcome-file registered (which jsp)

<welcome-file>index.jsp</welcome-file>

the servlet container serve directly.

in second case, send request to

http://localhost:8081/servletsjspexample/servletexample

which has path of /servletexample. have servletexample servlet mapped to

<url-pattern>/servletexample</url-pattern>

so servlet container chooses servletexample servlet handle request. handling request means invoking service method, goes on invoke 

this.getservletcontext().getrequestdispatcher("/index.jsp").forward(request, response);

with call getrequestdispatcher container

returns requestdispatcher object acts wrapper resource located @ given path.

the resource located @ given path, in case, jsp (another servlet). call forward, are, down line, invoking service on jsp servlet. renders content of jsp , sends response body.


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