I'm trying to deploy a simple REST example in JBOSS AS 7 but seems that I'm doing something wrong and I'm now clueless. I'm using Eclipse plugin to deploy.
The class I wrote is the following
#Path("/resources")
#Consumes({MediaType.APPLICATION_JSON})
#Produces({MediaType.APPLICATION_JSON})
public class ReceivedImagePersister {
#POST
#Path("/image")
public Response save(String entry) {
return Response.ok().build();
}
}
Then I create a web.xml
<web-app id="WebApp_ID" 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">
<display-name>API</display-name>
<servlet-mapping>
<servlet-name>javax.ws.rs.core.Application</servlet-name>
<url-pattern>/resources/*</url-pattern>
</servlet-mapping>
</web-app>
And a jboss-web.xml
<jboss-web>
<context-root>api</context-root>
</jboss-web>
When I deploy the application and try to send some data to the RESTful servlet the response is always:
(Could not find resource for relative : /image of full path: http://localhost:8080/api/resources/image)
Do I have to make some changes in configuration?
Thanks a lot in advance.
I think your web.xml your url pattern should /* i.e.
<url-pattern>/resources/*</url-pattern>
should be
<url-pattern>/*</url-pattern>
What i am getting at is you need let javax.ws.rs.core.Application parse your url starting from http://localhost/api/ only then will it able to detect that there is a resource named "resources" which has method "image"
Right now it is trying to lookup a resource named "image" which is under the context path /api/resources/
The following url may work for the way its been setup now
http://localhost:8080/api/resources/resources/image
I don't have any experience with JBoss but you should probably use a Servlet 3.0 web.xml instead of a 2.4 if you want the best compatibility with the Java EE 6 deployment model.
The correct url for accessing what you have declared is:
http://localhost:8080/api/resources/resources/image
The first resources is the servlet mapping, and the second is the name of the rest controller.
Related
This question already has answers here:
Servlet returns "HTTP Status 404 The requested resource (/servlet) is not available"
(19 answers)
Closed 6 years ago.
I have a servlet register in class p1. I have a JSP jsp1.jsp. I run JSP file and see it, but when I try to apply to the servlet, Tomcat shows an error:
HTTP Status 404
The requested resource (/omgtuk/Register) is not available.
Servlet:
#WebServlet("/register")
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_3_0.xsd" id="WebApp_ID" version="3.0">
<display-name>omgtuk</display-name>
<servlet>
<description></description>
<display-name>register</display-name>
<servlet-name>register</servlet-name>
<servlet-class>p1.register</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>register</servlet-name>
<url-pattern>/register</url-pattern>
</servlet-mapping>
<welcome-file-list>
<welcome-file>jsp1.jsp</welcome-file>
</welcome-file-list>
</web-app>
I'm using Eclipse.
The requested resource (/omgtuk/Register) is not available.
This simply means that the servlet isn't listening on an URL pattern of /Register. In other words, you don't have a #WebServlet("/Register").
In your particular case, you made a case mistake in the URL. URLs are case sensitive. You're calling /Register, but your servlet is listening on /register. Fix your form action accordingly.
So, it should not look like this:
<form action="Register">
But it should look like this:
<form action="register">
Or this, which is more robust in case you happen to move around JSPs when you're bored:
<form action="${pageContext.request.contextPath}/register">
Unrelated to the concrete problem, please note that you registered the servlet via both a #WebServlet annotation on the class and a <servlet> entry in web.xml. This is not right. You should use the one or the other. The #WebServlet is the new way of registering servlets since Servlet 3.0 (Java EE 6) and the <servlet> is the old way of registering servlets.
Just get rid of the whole <servlet> and <servlet-mapping> in web.xml. You don't need to specify both. Make sure that you're reading up to date books/tutorials. Servlet 3.0 exist since December 2009 already.
Another detail is that p1 is not a class, it's a package. I'd warmly recommend to invest a bit more time in learning basic Java before diving into Java EE.
See also:
Our servlets wiki page
I would like to set the welcome-file of my JSP/JavaBeans project. I have a servlet named 'Controller.java' with the following #WebServlet annotation:
#WebServlet(name="Controller", urlPatterns={"/login", "/show_dbs"})
and I hava a web.xml file with the following content:
<?xml version="1.0" encoding="ISO-8859-1"?>
<web-app xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee
http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" version="3.0">
<welcome-file-list>
<welcome-file>Controller</welcome-file>
</welcome-file-list>
</web-app>
Almost all things are going well, I can open http://localhost:8080/PROJECT/login and http://localhost:8080/PROJECT/show_dbs and I come to Controller.java. But when I open http://localhost:8080/PROJECT/ I get a 404 error.
I'm using Eclipse with a 'Dynamic Web Project', the Controller.java file is located under /src (default package) and the web.xml file is under /WebContent/WEB-INF.
I hope you have a tip for me.
Thank you for your help. Here comes my solution:
If you want to set your servlet as welcome file you have to do the following:
Define a standard html as welcome-file such as index.html in your web.xml:
<?xml version="1.0" encoding="ISO-8859-1"?>
<web-app xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee
http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" version="3.0">
<welcome-file-list>
<welcome-file>index.html</welcome-file>
</welcome-file-list>
</web-app>
Make sure this file (index.html) doesn't exist.
Define your urlPatterns in #WebServlet like this:
#WebServlet(name="Controller", urlPatterns={"/index.html", "/login", "/show_dbs"})
Now every request to http://.../PROJECT/ (root) will be redirected to http://.../PROJECT/index.html and this calls the servlet.
In the welcome file list you must specify the URIs. But you have specified the name of the servlet.
Quote from the Java™ Servlet Specification version 3.0 (emphasis mine):
10.10 Welcome Files
Web Application developers can define an
ordered list of partial URIs called welcome files in the Web
application deployment descriptor. The deployment descriptor syntax
for the list is described in the Web application deployment descriptor
schema.
The purpose of this mechanism is to allow the deployer to
specify an ordered list of partial URIs for the container to use for
appending to URIs when there is a request for a URI that corresponds
to a directory entry in the WAR not mapped to a Web component. This
kind of request is known as a valid partial request.
The use for this facility is made clear by the following common example: A welcome file
of 'index.html' can be defined so that a request to a URL like
host:port/webapp/directory/, where 'directory' is an entry in the WAR
that is not mapped to a servlet or JSP page, is returned to the client
as 'host:port/webapp/directory/index.html'.
If a Web container receives a valid partial request, the Web container must examine the
welcome file list defined in the deployment descriptor. The welcome
file list is an ordered list of partial URLs with no trailing or
leading /. The Web server must append each welcome file in the order
specified in the deployment descriptor to the partial request and
check whether a static resource in the WAR is mapped to that request
URI. If no match is found, the Web server MUST again append each
welcome file in the order specified in the deployment descriptor to
the partial request and check if a servlet is mapped to that request
URI. The Web container must send the request to the first resource in
the WAR that matches. The container may send the request to the
welcome resource with a forward, a redirect, or a container specific
mechanism that is indistinguishable from a direct request.
If no matching welcome file is found in the manner described, the container
may handle the request in a manner it finds suitable. For some configurations this may
mean returning a directory listing or for others returning a 404 response.
P.S.
Also see the examples in the specification in the chapter 10.10
This is what your web.xml should be. Create an index.jsp. Use the index.jsp as your welcome file. The controller class is your sevlet. So define a servlet in your web.xml as follows. This should cause all requests to be directed to the Controller class. And you should still be able to access localhost:8080/PROJECT/, in which case it will be directed to your welcome file.
If you don't want to create index.jsp, you can use your existing jsp file, may be your login.jsp file as your welcome file. In which case, just replace the index.jsp with login.jsp
<?xml version="1.0" encoding="ISO-8859-1"?>
<web-app xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee
http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" version="3.0">
<welcome-file-list>
<welcome-file>index.jsp</welcome-file>
</welcome-file-list>
<servlet>
<servlet-name>Controller</servlet-name>
<servlet-class>com.company.Controller</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Controller</servlet-name>
<url-pattern>*</url-pattern>
</servlet-mapping>
</web-app>
You can specify the url "/Controller" in urlPatterns in #WebServlet annotation. I think it will work.
The default welcome file is index.html, so just add the URL pattern.
#WebServlet(name="Controller", urlPatterns={"/index.html","/login", "/show_dbs"})
I'm new to Spring. I worked through the sections of this tutorial that cover web flow. My ultimate goal is to use Spring to implement new features in a legacy servlet webapp and then gradually replace the existing code making the servlet webapp into a Spring webabb.
So, to that end, I decided to go through the web flow part of the tutorial again, changing names to make my own first "hello world" screen with Spring within the development copy of the legacy servlet application.
My problem is that when I put the servlet mappings for Spring into my web.xml, I get 404s trying to get to my landing page.
I'm using WebLogic 9.2
I converted the directory tree of my webapp to mimic the layout that Spring webapps use
I copied spring.jar and spring-mvc.jar into my WEB-INF/lib
I made a simplified version of my web.xml, with just one legacy servlet in it ( for the landing page ) and Spring. It works with the Spring stuff commented out, but not otherwise. Here it is, my WEB-INF/web.xml for the "abc" webapp
<?xml version="1.0" encoding="UTF-8"?>
<web-app id="WebApp_ID" 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">
<servlet>
<servlet-name>LogIn</servlet-name>
<servlet-class>
com.utilities.LogIn
</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>LogIn</servlet-name>
<url-pattern>/home</url-pattern>
</servlet-mapping>
<servlet>
<servlet-name>abc</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>abc</servlet-name>
<url-pattern>*.htm</url-pattern>
</servlet-mapping>
<!-- Welcome File List -->
<welcome-file-list>
<welcome-file>index.jsp</welcome-file>
</welcome-file-list>
</web-app>
This is my WEB-INF/abc-servlet.xml
<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-2.5.xsd">
<!-- the application context definition for the NSD webapp DispatcherServlet -->
<beans name = "/hello.htm" class = "com.somecompany.web.HelloController"/>
</beans>
Here is the code for my elementary controller:
package com.somecompany.web;
import org.springframework.web.servlet.mvc.Controller;
import org.springframework.web.servlet.ModelAndView;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import java.io.IOException;
import org.apache.log4j.Logger;
public class HelloController implements Controller {
protected static final Logger logger = Logger.getLogger(HelloController.class);
public ModelAndView handleRequest(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
logger.info("Returning view for CBS Search ....");
return new ModelAndView("hello.jsp");
}
}// end class
Again, my problem isn't with the controller or the view (jsp)...yet. Right now, when I include the Spring servlet mappings in my web.xml, I can't get to my landing page, I get a 404. When I yank the Spring servlet mappings, that problem goes away.
I'm a raw beginner with Spring, so I am not sure where to look.
The issue is probably with the <url-pattern> in your web.xml. You cannot use ".jsp" as your mapped extension.
There is some pretty detailed discussion about the low-level reasons in this thread... but the long and short of it is that the "*.jsp" file extension is special to the underlying Java servlet specification. You're trying to make Spring hijack that. This is why the tutorial you linked to is using "*.htm" instead.
Trying changing your <url-pattern> in web.xml to something else (such as *.htm), and see if you can pull up your test URL with that extension.
What is your landing page? /hello.htm? your have just mapped everything with htm extension to a dispatcher, but you only have a controller which deals with hello.htm. You should get 404 if you are trying to access /index.htm. Change your mapping or create a controller for other urls.
However, if I were you, I would use Spring annotations in Spring 3.0 to configure the web application. It is much simpler.
As mentioned by others its better to use Spring 3.0 which is more easier to configure. Since you have started with this i am giving some of my thoughts. You need to use something like SimpleUrlHandlerMapping which takes care of resolving your url pattern to appropriate controllers. Take a look at this spring reference documentation
http://static.springsource.org/spring/docs/2.0.x/reference/mvc.html.
Hope this helps.
Try a forward slash on your ModelAndView view name:
return new ModelAndView("/hello.jsp");
I'm trying to deploy a web service onto WebSphere using a WAR file, which I have been told directly is completely possible and has been done many times before. WebSphere allows me to upload the file, specify the context root, and even start the application. However, when I try to access it by specifying my underlying URIs, WebSphere 404s on me. The relatively useless error message displayed is:
Error 404: SRVE0202E: Servlet [Jersey REST Service]: com.sun.jersey.spi.container.servlet.ServletContainer was found, but is corrupt: SRVE0227I: 1. Check that the class resides in the proper package directory. SRVE0228I: 2. Check that the classname has been defined in the server using the proper case and fully qualified package. SRVE0229I: 3. Check that the class was transferred to the filesystem using a binary transfer mode. SRVE0230I: 4. Check that the class was compiled using the proper case (as defined in the class definition). SRVE0231E: 5. Check that the class file was not renamed after it was compiled.
I have checked my naming conventions, modified my web.xml according to this blog post, attempted packaging it into an ear file (which threw out its own errors when I tried to upload it), and am trying to figure out what configurations I might have wrong. Any ideas of what I could change to make this work?
Edit
Here is the relevant part of my web.xml file:
<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.4"
id="WebAppId"
xmlns="http://java.sun.com/xml/ns/j2ee"
xsi="http://www.w3.org/2001/XMLSchema-instance"
schemalocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
<display-name>MYPROJECT'SDISPLAYNAME</display-name>
<servlet>
<servlet-name>Jersey REST Service</servlet-name>
<servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>com.sun.jersey.config.property.packages</param-name>
<param-value>MYPROJECTNAME</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Jersey REST Service</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
</web-app>
Another Edit
I'm using the newest release of Jersey- is that part of the problem?
Yet Another Edit
I'm pretty sure that's the entire problem. WebSphere 6.1 runs jdk1.5, and Jersey stopped supporting that after Jersey 1.2...
As you suspect your problem is lack for WebSphere support for Jersey (or rather JAX-RS).
I don't see JAX-RS in the list of supported APIs by WAS.
http://publib.boulder.ibm.com/infocenter/wasinfo/v6r1/index.jsp?topic=/com.ibm.help.ic.WS.doc/info_sching.html
WAS 6.1 runs on J2SE 1.5 (as seen in the URL above)
Specification or API Version 6.1
Java 2 Standard Edition (J2SE) specification J2SE 5
These probably are the reasons behind the errors that you get to see in your WAS 6.1
HTH
Manglu
In a JSP page(index.jsp):
${requestContext.requestURL} is the URL
just shows the expression itself. It used to be evaluated to something like "http://.../somerset/"
I created the Maven project with maven-archetype-webapp archetype in Eclipse. The Jetty version is jetty-6.1.14.
My web.xml is simple:
<web-app>
<display-name>Archetype Created Web Application</display-name>
<servlet>
<servlet-name>SomersetServlet</servlet-name>
<display-name>SomersetServlet</display-name>
<description></description>
<servlet-class>com.foo.somerset.SomersetServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>SomersetServlet</servlet-name>
<url-pattern>/som.do</url-pattern>
</servlet-mapping>
</web-app>
See Javascript String.replace(/\$/,str) works weirdly in jsp file for some possible reasons.
Your web.xml should contain reference to web-app_2_4.xsd schema, like
<web-app 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/web-app_2_4.xsd"
version="2.4">
This enables servlet 2.4 and jsp 2.0 processing, which includes EL.
Btw requestContext is not valid implicit object.
Incorrectly matched quotes can cause this behavior, where the expression just gets treated as a string. Your IDE would normally highlight this in a different color if this is the case.
Be sure you have directive, and other libraries you use included
<jsp:root .....
More info on definition here
http://java.sun.com/products/jsp/tags/12/syntaxref123.html