Servlet returns “HTTP Status 404 The requested resource (/servlet) is not available” [Answered]

Sample problem:

I have an HTML form in a JSP file in my WebContent/jsps folder. I have a servlet class in my default package in src folder. In my web.xml it is mapped as /servlet.

I have tried several URLs in action attribute of the HTML form:

<form action="/servlet">
<form action="/">
<form action="/src/">
<form action="../">

But none of those work. They all keep returning a HTTP 404 error like below in Tomcat 6/7/8:

HTTP Status 404 — /servlet

Description: The requested resource (/servlet) is not available.

Or as below in Tomcat 8.5/9:

HTTP Status 404 — Not Found

Message: /servlet

Description: The origin server did not find a current representation for the target resource or is not willing to disclose that one exists

Why is it not working?

Answer #1:


This can have a lot of causes which are broken down in following sections:

  • Put servlet class in a package
  • Set servlet URL in url-pattern
  • @WebServlet works only on Servlet 3.0 or newer
  • javax.servlet.* doesn’t work anymore in Servlet 5.0 or newer
  • Make sure compiled *.class file is present in built WAR
  • Test the servlet individually without any JSP/HTML page
  • Use domain-relative URL to reference servlet from HTML
  • Use straight quotes in HTML attributes

Put servlet class in a package

First of all, put the servlet class in a Java package. You should always put publicly reuseable Java classes in a package, otherwise they are invisible to classes which are in a package, such as the server itself. This way you eliminate potential environment-specific problems. Packageless servlets work only in specific Tomcat+JDK combinations and this should never be relied upon.

In case of a “plain” IDE project, the class needs to be placed in its package structure inside the “Java Sources” folder, not inside “Web Content” folder, which is for web files such as JSP. Below is an example of the folder structure of a default Eclipse Dynamic Web Project as seen in Navigator view (the “Java Sources” folder is in such project by default represented by src folder):

 |-- src
 |    `-- com
 |         `-- example
 |              `--
 |-- WebContent
 |    |-- WEB-INF
 |    |    `-- web.xml
 |    `-- jsps
 |         `-- page.jsp

In case of a Maven project, the class needs to be placed in its package structure inside main/java and thus not main/resourcesthis is for non-class files and absolutely also not main/webapp, this is for web files. Below is an example of the folder structure of a default Maven webapp project as seen in Eclipse’s Navigator view:

 |-- src
 |    `-- main
 |         |-- java
 |         |    `-- com
 |         |         `-- example
 |         |              `--
 |         |-- resources
 |         `-- webapp
 |              |-- WEB-INF
 |              |    `-- web.xml
 |              `-- jsps
 |                   `-- page.jsp

Note that the /jsps subfolder is not strictly necessary. You can even do without it and put the JSP file directly in webcontent/webapp root, but I’m just taking over this from your question.

Set servlet URL in url-pattern

The servlet URL is specified as the “URL pattern” of the servlet mapping. It’s absolutely not per definition the classname/filename of the servlet class. The URL pattern is to be specified as value of @WebServlet annotation.

package com.example; // Use a package!

@WebServlet("/servlet") // This is the URL of the servlet.
public class YourServlet extends HttpServlet { // Must be public and extend HttpServlet.
    // ...

In case you want to support path parameters like /servlet/foo/bar, then use an URL pattern of /servlet/* instead. See also Servlet and path parameters like /xyz/{value}/test, how to map in web.xml?

@WebServlet works only on Servlet 3.0 or newer

In order to use @WebServlet, you only need to make sure that your web.xml file, if any (it’s optional since Servlet 3.0), is declared conform Servlet 3.0+ version and thus not conform e.g. 2.5 version or lower. Below is a Servlet 4.0 compatible one (which matches Tomcat 9+, WildFly 11+, Payara 5+, etc).

<?xml version="1.0" encoding="UTF-8"?>
    <!-- Config here. -->

Or, in case you’re not on Servlet 3.0+ yet (e.g. Tomcat 6 or older), then remove the @WebServlet annotation.

package com.example;

public class YourServlet extends HttpServlet {
    // ...

And register the servlet instead in web.xml like this:

    <url-pattern>/servlet</url-pattern>  <!-- This is the URL of the servlet. -->

Note thus that you should not use both ways. Use either annotation based configuarion or XML based configuration. When you have both, then XML based configuration will override annotation based configuration.

javax.servlet.* doesn’t work anymore in Servlet 5.0 or newer

Since Jakarta EE 9 / Servlet 5.0 (Tomcat 10, TomEE 9, WildFly 22 Preview, GlassFish 6, Payara 6, Liberty 22, etc), the javax.* package has been renamed to jakarta.* package.

In other words, please make absolutely sure that you don’t randomly put JAR files of a different server in your WAR project such as tomcat-servlet-api-9.x.x.jar merely in order to get the javax.* package to compile. This will only cause trouble. Remove it altogether and edit the imports of your servlet class from

import javax.servlet.*;
import javax.servlet.http.*;


import jakarta.servlet.*;
import jakarta.servlet.http.*;

In case you’re using Maven, you can find examples of proper pom.xml declarations for Tomcat 10+, Tomcat 9-, JEE 9+ and JEE 8- in this answer: Tomcat casting servlets to javax.servlet.Servlet instead of jakarta.servlet.http.HttpServlet

Make sure compiled *.class file is present in built WAR

In case you’re using a build tool such as Eclipse and/or Maven, then you need to make absolutely sure that the compiled servlet class file resides in its package structure in /WEB-INF/classes folder of the produced WAR file. In case of package com.example; public class YourServlet, it must be located in /WEB-INF/classes/com/example/YourServlet.class. Otherwise you will face in case of @WebServlet also a 404 error, or in case of <servlet> a HTTP 500 error like below:

HTTP Status 500

Error instantiating servlet class com.example.YourServlet

And find in the server log a java.lang.ClassNotFoundException: com.example.YourServlet, followed by a java.lang.NoClassDefFoundError: com.example.YourServlet, in turn followed by javax.servlet.ServletException: Error instantiating servlet class com.example.YourServlet.

An easy way to verify if the servlet is correctly compiled and placed in classpath is to let the build tool produce a WAR file (e.g. rightclick project, Export > WAR file in Eclipse) and then inspect its contents with a ZIP tool. If the servlet class is missing in /WEB-INF/classes, or if the export causes an error, then the project is badly configured or some IDE/project configuration defaults have been mistakenly reverted (e.g. Project > Build Automatically has been disabled in Eclipse).

You also need to make sure that the project icon has no red cross indicating a build error. You can find the exact error in Problems view (Window > Show View > Other…). Usually the error message is fine Googlable. In case you have no clue, best is to restart from scratch and do not touch any IDE/project configuration defaults.

Test the servlet individually without any JSP/HTML page

Provided that the server runs on localhost:8080, and that the WAR is successfully deployed on a context path of /contextname (which defaults to the IDE project name, case sensitive!), and the servlet hasn’t failed its initialization (read server logs for any deploy/servlet success/fail messages and the actual context path and servlet mapping), then a servlet with URL pattern of /servlet is available at http://localhost:8080/contextname/servlet.

You can just enter it straight in browser’s address bar to test it invidivually. If its doGet() is properly overriden and implemented, then you will see its output in browser. Or if you don’t have any doGet() or if it incorrectly calls super.doGet(), then a “HTTP 405: HTTP method GET is not supported by this URL” error will be shown (which is still better than a 404 as a 405 is evidence that the servlet itself is actually found).

Overriding service() is a bad practice, unless you’re reinventing a MVC framework — which is very unlikely if you’re just starting out with servlets and are clueless as to the problem described in the current question 😉

Regardless, if the servlet already returns 404 when tested invidivually, then it’s entirely pointless to try with a HTML form instead. Logically, it’s therefore also entirely pointless to include any HTML form in questions about 404 errors from a servlet.

Use domain-relative URL to reference servlet from HTML

Once you’ve verified that the servlet works fine when invoked individually, then you can advance to HTML. As to your concrete problem with the HTML form, the <form action> value needs to be a valid URL. The same applies to <a href><img src><script src>, etc. You need to understand how absolute/relative URLs work. You know, an URL is a web address as you can enter/see in the webbrowser’s address bar. If you’re specifying a relative URL as form action, i.e. without the http:// scheme, then it becomes relative to the current URL as you see in your webbrowser’s address bar. It’s thus absolutely not relative to the JSP/HTML file location in server’s WAR folder structure as many starters seem to think.

So, assuming that the JSP page with the HTML form is opened by http://localhost:8080/contextname/jsps/page.jsp (and thus not by file://...), and you need to submit to a servlet located in http://localhost:8080/contextname/servlet, here are several cases (note that you can here safely substitute <form action> with <a href><img src><script src>, etc):

  • Form action submits to an URL with a leading slash. <form action="/servlet"> The leading slash / makes the URL relative to the domain, thus the form will submit to http://localhost:8080/servlet But this will likely result in a 404 as it’s in the wrong context.
  • Form action submits to an URL without a leading slash. <form action="servlet"> This makes the URL relative to the current folder of the current URL, thus the form will submit to http://localhost:8080/contextname/jsps/servlet But this will likely result in a 404 as it’s in the wrong folder.
  • Form action submits to an URL which goes one folder up. <form action="../servlet"> This will go one folder up (exactly like as in local disk file system paths!), thus the form will submit to http://localhost:8080/contextname/servlet This one must work!
  • The canonical approach, however, is to make the URL domain-relative so that you don’t need to fix the URLs once again when you happen to move the JSP files around into another folder. <form action="${pageContext.request.contextPath}/servlet"> This will generate <form action="/contextname/servlet"> Which will thus always submit to the right URL.

Use straight quotes in HTML attributes

You need to make absolutely sure you’re using straight quotes in HTML attributes like action="..." or action='...' and thus not curly quotes like action=”...” or action=’...’. Curly quotes are not supported in HTML and they will simply become part of the value. Watch out when copy-pasting code snippets from blogs! Some blog engines, notably WordPress, are known to by default use so-called “smart quotes” which thus also corrupts the quotes in code snippets this way. On the other hand, instead of copy-pasting code, try simply typing over the code yourself. Additional advantage of actually getting the code through your brain and fingers is that it will make you to remember and understand the code much better in long term and also make you a better developer.

Answer #2:

Scenario #1: You accidentially re-deployed from the command line while tomcat was already running.

Short Answer: Stop Tomcat, delete target folder, mvn package, then re-deploy

Scenario #2: request.getRequestDispatcher(“MIS_SPELLED_FILE_NAME.jsp”)

Short Answer: Check file name spelling, make sure case is correct.

Scenario #3: Class Not Found Exceptions (Answer put here because: Question# 17982240 ) (java.lang.ClassNotFoundException for servlet in tomcat with eclipse ) (was marked as duplicate and directed me here )

Short Answer #3.1: web.xml has wrong package path in servlet-class tag.

Short Answer #3.2: java file has wrong import statement.

Below is further details for Scenario #1:

1: Stop Tomcat

  • Option 1: Via CTRL+C in terminal.
  • Option 2: (terminal closed while tomcat still running)
  • ———— 2.1: press:Windows+R –> type:”services.msc
  • ———— 2.2: Find “Apache Tomcat #.# Tomcat#” in Name column of list.
  • ———— 2.3: Right Click –> “stop

2: Delete the “target” folder. (mvn clean will not help you here)

3: mvn package


(Mine: java -jar target/dependency/webapp-runner.jar –port 5190 target/*.war )

Full Back Story:

Accidentially opened a new git-bash window and tried to deploy a .war file for my heroku project via:

java -jar target/dependency/webapp-runner.jar –port 5190 target/*.war

After a failure to deploy, I realized I had two git-bash windows open, and had not used CTLR+C to stop the previous deployment.

I was met with:

HTTP Status 404 – Not Found Type Status Report

Message /if-student-test.jsp

Description The origin server did not find a current representation for the target resource or is not willing to disclose that one exists.

Apache Tomcat/8.5.31

Below is further details for Scenario #3:

SCENARIO 3.1: The servlet-class package path is wrong in your web.xml file.

It should MATCH the package statement at top of your java servlet class.

File: my_stuff/

   package my_stuff;

File: PRJ_ROOT/src/main/webapp/WEB-INF/web.xml



You put the wrong “package” statement at top of your file.

For example:

File is in: “/my_stuff” folder

You mistakenly write:

package com.my_stuff

This is tricky because:

1: The maven build (mvn package) will not report any errors here.

2: servlet-class line in web.xml can have CORRECT package path. E.g:


Stack Used: Notepad++ + GitBash + Maven + Heroku Web App Runner + Tomcat9 + Windows10:

Answer #3:

Check if you have entered the correct URL Mapping as specified in the Web.xml

For example:

In the web.xml, your servlet declaration maybe:



What this snippet does is <url-pattern>/theController</url-pattern>will set the name that will be used to call the servlet from the front end (eg: form) through the URL. Therefore when you reference the servlet in the front end, in order to ensure that the request goes to the servlet “ControllerA”, it should refer the specified URL Pattern “theController” from the form.


<form action="theController" method="POST">

Hope you learned something from this post.

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