servlet

Get all Request Parameters in Servlet

With this tutorial we shall show you how to get all requests parameters in a Java Servlet. This the most basic step you have to consider when developing a Servelt application because HTTP is based mostly on parameters exchange.

Basically in order to get all Request Parameters in Servlet, one should take the following steps:

  • Create a handleRequest method so you can use it both in doGet and doPost methods.
  • Use HttpServletRequest.getParameterNames to get an Enumeration of parameter names.
  • Use HttpServletRequest.getParameterValues(paramName) to get the parameters values.


 
Let’s see the simple code snippets that follow:
package com.javacodegeeks.snippets.enterprise;

import java.io.IOException;
import java.io.PrintWriter;
import java.util.Enumeration;

import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

public class GetAllRequestParametersInServlet extends HttpServlet {

	private static final long serialVersionUID = -2128122335811219481L;

	public void doGet(HttpServletRequest req, HttpServletResponse res) throws IOException {
		handleRequest(req, res);
	}

	public void doPost(HttpServletRequest req, HttpServletResponse res) throws IOException {
		handleRequest(req, res);
	}

	public void handleRequest(HttpServletRequest req, HttpServletResponse res) throws IOException {

		PrintWriter out = res.getWriter();
		res.setContentType("text/plain");

		Enumeration<String> parameterNames = req.getParameterNames();

		while (parameterNames.hasMoreElements()) {

			String paramName = parameterNames.nextElement();
			out.write(paramName);
			out.write("n");

			String[] paramValues = req.getParameterValues(paramName);
			for (int i = 0; i < paramValues.length; i++) {
				String paramValue = paramValues[i];
				out.write("t" + paramValue);
				out.write("n");
			}

		}

		out.close();

	}

}

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"
  version="2.5">

	<display-name>JCG Snippets Web Project</display-name>

	<servlet>
		<servlet-name>JCG Snippets Application</servlet-name>
		<servlet-class>com.javacodegeeks.snippets.enterprise.GetAllRequestParametersInServlet</servlet-class>
	</servlet>

	<servlet-mapping>
		<servlet-name>JCG Snippets Application</servlet-name>
		<url-pattern>/jcgservlet</url-pattern>
	</servlet-mapping>

</web-app>

URL:

http://myhost:8080/jcgsnippets/jcgservlet?param1=paramvalue1&param2=paramvalue2a&param2=paramvalue2b

Output:

param2
	paramvalue2a
	paramvalue2b
param1
	paramvalue1

 
This was an example on how to get all Request Parameters in Servlet.

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Byron Kiourtzoglou

Byron is a master software engineer working in the IT and Telecom domains. He is an applications developer in a wide variety of applications/services. He is currently acting as the team leader and technical architect for a proprietary service creation and integration platform for both the IT and Telecom industries in addition to a in-house big data real-time analytics solution. He is always fascinated by SOA, middleware services and mobile development. Byron is co-founder and Executive Editor at Java Code Geeks.
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3 Comments
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alnebras
alnebras
5 years ago

I’am copy and past this code but when insert manual parameters Nothing happen

Vivek Bhadiyadra
Vivek Bhadiyadra
1 year ago

I’am copy and past this code but when insert manual parameters Nothing happen

Joe Dunn
Joe Dunn
1 year ago

I’am copy and past this code but when insert manual parameters Nothing happen

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