Get servlet path in java

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How do I get the location of my web application context in the file system?

Web Tutorials :: Servlets :: 18. How do I get the location of my web application context in the file system?

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Description: This tutorial describes how to use the ServletContext to get the location of the web application context directory in the file system.

Tutorial created using: Windows XP || JDK 1.5.0_09 || Eclipse Web Tools Platform 2.0 (Eclipse 3.3.0) || Tomcat 5.5.20

A web application’s context path is the directory that contains the web application’s WEB-INF directory. It can be thought of as the ‘home’ of the web app. Often, when writing web applications, it can be important to get the actual location of this directory in the file system, since this allows you to do things such as read from files or write to files.

This location can be obtained via the ServletContext object’s getRealPath() method. This method can be passed a String parameter set to File.separator to get the path using the operating system’s file separator («/» for UNIX, «\» for Windows).

The TestServlet class below demonstrates getting and displaying this path in a servletW.

TestServlet.java

package test; import java.io.File; import java.io.IOException; import java.io.PrintWriter; import javax.servlet.ServletContext; import javax.servlet.http.HttpServlet; import javax.servlet.http.HttpServletRequest; import javax.servlet.http.HttpServletResponse; public class TestServlet extends HttpServlet < private static final long serialVersionUID = 1L; protected void doGet(HttpServletRequest request, HttpServletResponse response) throws IOException < ServletContext servletContext = getServletContext(); String contextPath = servletContext.getRealPath(File.separator); PrintWriter out = response.getWriter(); out.println("
File system context path (in TestServlet): " + contextPath); > >

The TestFilter class demonstrates getting and displaying the path in a filter.

TestFilter.java

package test; import java.io.File; import java.io.IOException; import java.io.PrintWriter; import javax.servlet.Filter; import javax.servlet.FilterChain; import javax.servlet.FilterConfig; import javax.servlet.ServletContext; import javax.servlet.ServletException; import javax.servlet.ServletRequest; import javax.servlet.ServletResponse; import javax.servlet.http.HttpServletRequest; import javax.servlet.http.HttpServletResponse; public class TestFilter implements Filter < FilterConfig filterConfig = null; public void init(FilterConfig filterConfig) throws ServletException < this.filterConfig = filterConfig; > public void destroy() < >public void doFilter(ServletRequest servletRequest, ServletResponse servletResponse, FilterChain filterChain) throws IOException, ServletException < HttpServletRequest request = (HttpServletRequest) servletRequest; HttpServletResponse response = (HttpServletResponse) servletResponse; ServletContext servletContext = filterConfig.getServletContext(); String contextPath = servletContext.getRealPath(File.separator); PrintWriter out = response.getWriter(); out.println("File system context path (in TestFilter): " + contextPath); filterChain.doFilter(request, response); > >

If I fire up my web application and hit http://localhost:8080/tomcat-demo/test in a web browser, I see that both the filter and the servlet display the path to my context directory in the file system.

In case it is useful, the project above utilizes the following structure:

Here is my web.xmlW file. Both the servlet and the filter are mapped to «/test».

web.xml

 version="1.0" encoding="UTF-8"?>  id="tomcat-demo" version="2.4" xmlns="http://java.sun.com/xml/ns/j2ee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">  TestFilter test.TestFilter   TestFilter /test   TestServlet test.TestServlet   TestServlet /test   

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Web Tutorials :: Servlets :: 18. How do I get the location of my web application context in the file system?

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Context Path vs. Servlet Path in Spring

As always, the writeup is super practical and based on a simple application that can work with documents with a mix of encrypted and unencrypted fields.

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Get started with Spring 5 and Spring Boot 2, through the Learn Spring course:

1. Introduction

DispatcherServlet plays a significant role in Spring applications and provides a single entry point for the application. Whereas the context path defines the URL that the end-user will access the application.

In this tutorial, we’re going to learn about the differences between context path and servlet path.

2. Context Path

Simply put, the context path is a name with which a web application is accessed. It is the root of the application. By default, Spring Boot serves the content on the root context path (“/”).

So, any Boot application with default configuration can be accessed as:

However, in some cases, we may wish to change the context of our application. There are multiple ways to configure the context path, and application.properties is one of them. This file resides under the src/main/resources folder.

Let’s configure it using the application.properties file:

server.servlet.context-path=/demo

As a result, the application main page will be:

When we deploy this application to an external server, this modification helps us to avoid accessibility issues.

3. Servlet Path

The servlet path represents the path of the main DispatcherServlet. The DispatcherServlet is an actual Servlet, and it inherits from HttpSerlvet base class. The default value is similar to the context path, i.e. (“/”):

In the earlier versions of Boot, the property was in the ServerProperties class and known as server.servlet-path =/.

From 2.1.x, this property is moved to the WebMvcProperties class and renamed as spring.mvc.servlet.path =/.

Let’s modify the servlet path:

spring.mvc.servlet.path=/baeldung

Because a servlet belongs to a servlet context, changing the context path will affect the servlet path too. So, after modifications, the application servlet path will become http://localhost:8080/demo/baeldung.

In other words, if a style sheet was being served as http://localhost:8080/demo/style.css, now will serve as http://localhost:8080/demo/baeldung/style.css.

Usually, we don’t configure the DispatcherServlet by ourselves. But, if we really need to do it, we have to provide the path of our custom DispatcherServlet.

4. Conclusion

In this quick article, we looked at the semantics of context path and servlet path. We also saw what these terms represent and how they work together in a Spring application.

Slow MySQL query performance is all too common. Of course it is. A good way to go is, naturally, a dedicated profiler that actually understands the ins and outs of MySQL.

The Jet Profiler was built for MySQL only, so it can do things like real-time query performance, focus on most used tables or most frequent queries, quickly identify performance issues and basically help you optimize your queries.

Critically, it has very minimal impact on your server’s performance, with most of the profiling work done separately — so it needs no server changes, agents or separate services.

Basically, you install the desktop application, connect to your MySQL server, hit the record button, and you’ll have results within minutes:

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How to get servlet path?

posted 2 years ago

I know that we get the servlet path using request.getServletPath(). But in my case, let’s say my request url is http://localhost:8080/feedback/questions/deletequestion/39 . I want to get the number 39 which is after the delete question. How can I get it using request.getServletPath().

Saloon Keeper

posted 2 years ago

• 1

getServletPath is likely not the right method to use (although that depends on your servlet mapping, about which we know nothing). Check out getRequestURI, possibly in conjunction with String.substring and String.lastIndexOf.

Sheriff

posted 2 years ago

• 1

The official way to get the part of the path after the servlet path is to use HttpServletRequest.getPathInfo(). It’s also mentioned on the article Tim linked to.

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Saloon Keeper

posted 2 years ago

• 1

In many cases, you can combine getServletPath() with getRequestURI(). Assuming you have a straightforward servlet URL mapping, the two strings will both have the same leading values, so you can just snip off the common part from the URI and the remainder should give you what you need.

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getServletPath() Method Example

This program uses getServletPath() of HttpServletRequest to retrieve the alias name (of ) tag. Client uses this alias name to call the Servlet.

Following is the method signature. Method is defined in HttpServletRequest interface.

public abstract java.lang.String getServletPath();

Example on getServletPath()

Client Program: ClientData.html

web.xml entry for ClientInformation servlet

 efgh ClientInformation  efgh /jasmine  

Let us go to the coding part.

Server Program: ClientInformation.java

import javax.servlet.*; import javax.servlet.http.*; import java.io.*; public class ClientInformation extends HttpServlet < public void service(HttpServletRequest req, HttpServletResponse res) throws ServletException, IOException < res.setContentType("text/html"); PrintWriter out = res.getWriter(); String str = req.getServletPath(); out.println("req.getServletPath() : " + str); out.close(); >>

Output screen of ClientData.html with text field filled up.

The output screen when submit button is clicked.

Observe, the getServletPath() prints just the alias name and not complete URL. To retrieve complete URL, use getRequestURL() method of HttpServletRequest.

Pass your comments and suggestions on this tutorial «getServletPath() Method Example». It improves the quality of site.

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