I have an HTML form in a JSP file in my WebContent/jsps
folder. I have a servlet class servlet.java
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="/servlet.java">
<form action="/src/servlet.java">
<form action="../servlet.java">
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?
Best Answer
Introduction
This can have a lot of causes which are broken down in following sections:
package
url-pattern
@WebServlet
works only on Servlet 3.0 or newerjavax.servlet.*
doesn't work anymore in Servlet 5.0 or newer*.class
file is present in built WARPut 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):In case of a Maven project, the class needs to be placed in its package structure inside
main/java
and thus notmain/resources
, this is for non-class files and absolutely also notmain/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: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.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 newerIn order to use
@WebServlet
, you only need to make sure that yourweb.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).Or, in case you're not on Servlet 3.0+ yet (e.g. Tomcat 6 or older), then remove the
@WebServlet
annotation.And register the servlet instead in
web.xml
like this: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 newerSince 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 tojakarta.*
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 fromto
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.HttpServletMake sure compiled
*.class
file is present in built WARIn 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 ofpackage 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:And find in the server log a
java.lang.ClassNotFoundException: com.example.YourServlet
, followed by ajava.lang.NoClassDefFoundError: com.example.YourServlet
, in turn followed byjavax.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. In case you're using Eclipse, you can find instructions in How do I import the javax.servlet / jakarta.servlet API in my Eclipse project?
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 athttp://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 anydoGet()
or if it incorrectly callssuper.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 ;) See also Design Patterns web based applications.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 thehttp://
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 byfile://...
), and you need to submit to a servlet located inhttp://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.
The leading slash
/
makes the URL relative to the domain, thus the form will submit toBut this will likely result in a 404 as it's in the wrong context.
Form action submits to an URL without a leading slash.
This makes the URL relative to the current folder of the current URL, thus the form will submit to
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.
This will go one folder up (exactly like as in local disk file system paths!), thus the form will submit to
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.
This will generate
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="..."
oraction='...'
and thus not curly quotes likeaction=”...”
oraction=’...’
. 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.See also:
Other cases of HTTP Status 404 error: