Servlet Is not showing any output

Question

I am new to servlets so very simple answer will be much appreciated. My problem is I have written a code of servlet extracting the values from HTML form. But when I run my HTML form like

http://localhost:8080/secondTry/form.html

html form appears okay. When I enter firstname and lastname, and press the submit button it is not showing any output.. I don't know why..I have searched it on internet but all in vain. Moreover, is it has something to do with my tomcat?

Below is my file form.html:

<html>
<body>
<form action="/formServlet" method="get">
First name: <input type="text" name="fname"><br>
Last name: <input type="text" name="lname"><br>
<input type="button" value="submit">
</form>
</body>
</html>

secTry.java:

import java.io.*;
import javax.servlet.*;
import javax.servlet.http.*;
public class secTry extends HttpServlet{
    public void doGet(HttpServletRequest req, HttpServletResponse res) throws IOException,ServletException{
        String fname = req.getParameter("fname");
        String lname = req.getParameter("lname");
        PrintWriter out = res.getWriter();
        out.println("Hello "+ fname + " " +lname + "...!!!");
    }
}

web.xml file:

<web-app>
  <servlet>
    <servlet-name>secTry</servlet-name>
    <servlet-class>secTry</servlet-class>
  </servlet>
  <servlet-mapping>
    <servlet-name>secTry</servlet-name>
    <url-pattern>/formServlet</url-pattern>
  </servlet-mapping>
</web-app>

PS: I'm compiling my java file on command line using

javac -cp .;C:/apache-tomcat-8.5.23/lib/servlet-api.jar secTry.java

Please help me and tell me where I am doing wrong?

1. Don't, ever, create classes in the default package. 2. `action="/formServlet"` should be `action="/theContextPathOfYourApplication/formServlet"` 3. Respect the Java naming conventions. 4. Use the WebServlet annotation rather than xml: we're not in 2005 anymore. — JB Nizet, Oct 16 '17 at 16:06
you mean context path like "http://localhost:8084/secondTry/formServlet" — Mahnoor Nadir, Oct 16 '17 at 16:10
If /secondTry is your context path (i.e. the name of your war file, by default), then yes. Note that URLs are case-sensitive, too. — JB Nizet, Oct 16 '17 at 16:11
I have tried this but browser is showing an error that file cannot be run on port number 8084.. — Mahnoor Nadir, Oct 16 '17 at 16:13
And when try it like "localhost:8080/secondTry/formServlet" then it is showing.. Hello null null...!!! — Mahnoor Nadir, Oct 16 '17 at 16:14

score 1 · Answer 1 · answered Oct 16 '17 at 19:06

1

When I enter firstname and lastname, and press the submit button it is not showing any output.

Based on the code you have provided you are not submitting the form at all:

<input type="button" value="submit">

This is not a submit element. Form will be submitted when element type is submit:

<input type="submit" value="submit">

answered Oct 16 '17 at 19:06

Aleh Maksimovich

2,622
8
19

I figured it out early. Thanks. :) – Mahnoor Nadir Oct 17 '17 at 21:45

Servlet Is not showing any output

1 Answers1