This question was already asked, however since then all answers (that I could found) are no longer valid.
Essentially I want to implement a website with Vaadin (V23), that communicates with a WebApp via POST requests that is running on another server (physically). To do it, I want to create separate Servlet that would handle the communication (receiving side) with another Server. Let's say, this is not implemneted version:
#WebServlet(urlPatterns = "/communication", name = "QuizServlet", asyncSupported = true)
public class QuizServlet extends HttpServlet {
#Override
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
response.sendError(400, "Not implemented");
}
#Override
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
response.sendError(400, "Not implemented");
}
}
The problem is however, that I always get redirected to default dispatcher Servlet, and it seems, regardless of what I do:
SpringVaadinServlet was deprecated and no longer exists, extending VaadinServlet does not work.
Changing mappings in properties (vaadin.url-mapping=) also does not work, I just get redirected to this new mapping in all cases.
Trying to do servlets on separate ports yields same redirection on all ports, even if explicitly registering my custom Servlet on the Connector, with separate Sevice (WebMvcConfigurer Tomcat configuration). Answer from this post, also too old.
Registering servlet directly also does not do anything (by implementing WebApplicationInitializer).
There for the question, how to make use of two different servlets with new Vaadin 23 and Spring Boot 2.7.1?
I have found some kind of a solution to my problem. Namely on startup of my BootAplication, I am also starting the second separate Tomcat server that uses my custom Servlet :
#Service
public class QuizServer {
private final Log log = LogFactory.getLog(QuizServer.class);
#PostConstruct
public void startServer() throws IOException, LifecycleException {
start();
}
private void start() throws IOException, LifecycleException {
Tomcat tomcat = new Tomcat();
String contextPath = "/";
String appBase = new File(".").getAbsolutePath();
Context ctx = tomcat.addContext(contextPath, appBase);
Tomcat.addServlet(ctx, "quizServlet", new QuizServlet());
ctx.addServletMappingDecoded("/*", "quizServlet");
tomcat.setPort(8085);
tomcat.start();
tomcat.getConnector();
log.info("Quiz server started");
}
}
#WebServlet(urlPatterns = "/*", name = "quizServlet", asyncSupported = true)
public class QuizServlet extends HttpServlet {
#Override
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
response.getWriter().println("Test");
}
#Override
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
response.sendError(400, "Not implemented");
}
}
It is a bit crude though, since ideally, it shouldn't require a separate server.
Related
I have two classes Server (with the main method, starting the server) and StartPageServlet with a Servlet.
The most important part of the code is:
public class Server {
public static void main(String[] args) throws Exception {
// some code
// I want to pass "anObject" to every Servlet.
Object anObject = new Object();
Server server = new Server(4000);
ServletContextHandler context =
new ServletContextHandler(ServletContextHandler.SESSIONS);
context.addServlet(StartPageServlet.class, "/");
// more code
}
And the StartPageServlet:
public class StartPageServlet extends HttpServlet {
#Override
public void doGet(HttpServletRequest request,
HttpServletResponse response)
throws ServletException, IOException
{
// Here I want to access "anObject"
}
How do I do this?
Embedded Jetty is so wonderful here.
You have a few common options:
Direct instantiation of the servlet, use constructors or setters, then hand it off to Jetty via the ServletHolder (can be any value or object type)
Add it to the ServletContext in your main, and then access it via the ServletContext in your application (can be any value or object type).
Examples:
package jetty;
import java.io.IOException;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import org.eclipse.jetty.server.Server;
import org.eclipse.jetty.servlet.ServletContextHandler;
import org.eclipse.jetty.servlet.ServletHolder;
public class ObjectPassingExample
{
public static void main(String args[]) throws Exception
{
Server server = new Server(8080);
ServletContextHandler context = new ServletContextHandler();
context.setContextPath("/");
// Option 1: Direct servlet instantiation and ServletHolder
HelloServlet hello = new HelloServlet("everyone");
ServletHolder helloHolder = new ServletHolder(hello);
context.addServlet(helloHolder, "/hello/*");
// Option 2: Using ServletContext attribute
context.setAttribute("my.greeting", "you");
context.addServlet(GreetingServlet.class, "/greetings/*");
server.setHandler(context);
server.start();
server.join();
}
public static class HelloServlet extends HttpServlet
{
private final String hello;
public HelloServlet(String greeting)
{
this.hello = greeting;
}
#Override
protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException
{
resp.setContentType("text/plain");
resp.getWriter().println("Hello " + this.hello);
}
}
public static class GreetingServlet extends HttpServlet
{
private String greeting;
#Override
public void init() throws ServletException
{
this.greeting = (String) getServletContext().getAttribute("my.greeting");
}
#Override
protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException
{
resp.setContentType("text/plain");
resp.getWriter().println("Greetings to " + this.greeting);
}
}
}
Singleton
You want to pass the same single instance to each servlet?
Use the Singleton pattern to create a single instance that is available globally.
The simplest fool-proof way to do that in Java is through an Enum. See Oracle Tutorial. Also see this article and the book Effective Java: Programming Language Guide, Second Edition (ISBN 978-0-321-35668-0, 2008) by Dr. Joshua Bloch.
So no need to pass an object. Each servlet can access the same single instance through the enum.
Per web app
If you want to do some work when your web app is first launching but before any servlet in that web app has handled any request, write a class that implements the ServletContextListener interface.
Mark your class with the #WebListener annotation to have your web container automatically instantiate and invoke.
I had a similar situation but needed to share a singleton with a servlet deployed via war with hot (re)deploy in a Jetty container. The accepted answer wasn't quite what I needed in my case since the servlet has a lifecycle and context managed by a deployer.
I ended up with a brute-force approach, adding the object to the server context, which persists for the life of the container, and then fetching the object from within the servlet(s). This required loading the class of the object in a parent (system) classloader so that the war webapp doesn't load its own version of the class into its own classloader, which would cause a cast exception as explained here.
Embedded Jetty server code:
Server server = new Server(8090);
// Add all classes related to the object(s) you want to share here.
WebAppContext.addSystemClasses(server, "my.package.MyFineClass", ...);
// Handler config
ContextHandlerCollection contexts = new ContextHandlerCollection();
HandlerCollection handlers = new HandlerCollection();
handlers.setHandlers(new Handler[] { contexts });
server.setHandler(handlers);
// Deployer config (hot deploy)
DeploymentManager deployer = new DeploymentManager();
DebugListener debug = new DebugListener(System.err,true,true,true);
server.addBean(debug);
deployer.addLifeCycleBinding(new DebugListenerBinding(debug));
deployer.setContexts(contexts);
deployer.setContextAttribute(
"org.eclipse.jetty.server.webapp.ContainerIncludeJarPattern",
".*/[^/]*servlet-api-[^/]*\\.jar$|.*/javax.servlet.jsp.jstl-.*\\.jar$|.*/[^/]*taglibs.*\\.jar$");
WebAppProvider webapp_provider = new WebAppProvider();
webapp_provider.setMonitoredDirName("/.../webapps");
webapp_provider.setScanInterval(1);
webapp_provider.setExtractWars(true);
webapp_provider.setConfigurationManager(new PropertiesConfigurationManager());
deployer.addAppProvider(webapp_provider);
server.addBean(deployer);
// Other config...
// Tuck any objects/data you want into the root server object.
server.setAttribute("my.package.MyFineClass", myFineSingleton);
server.start();
server.join();
Example servlet:
public class MyFineServlet extends HttpServlet
{
MyFineClass myFineSingleton;
#Override
public void init() throws ServletException
{
// Sneak access to the root server object (non-portable).
// Not possible to cast this to `Server` because of classloader restrictions in Jetty.
Object server = request.getAttribute("org.eclipse.jetty.server.Server");
// Because we cannot cast to `Server`, use reflection to access the object we tucked away there.
try {
myFineSingleton = (MyFineClass) server.getClass().getMethod("getAttribute", String.class).invoke(server, "my.package.MyFineClass");
} catch (Exception ex) {
throw new ServletException("Unable to reflect MyFineClass instance via Jetty Server", ex);
}
}
#Override
protected void doGet( HttpServletRequest request,
HttpServletResponse response ) throws ServletException, IOException
{
response.setContentType("text/html");
response.setStatus(HttpServletResponse.SC_OK);
response.getWriter().println("<h1>Hello from MyFineServlet</h1>");
response.getWriter().println("Here's: " + myFineSingleton.toString());
}
}
My build file for the servlet (sbt) placed the my.package.MyFineClass dependency into the "provided" scope so it wouldn't get packaged into the war as it will already be loaded into the Jetty server.
I would recommend that you investigate Google's solution to this problem... namely: dependency injection with Guice. They have a special servlet package that deals with servlets specifically.
I have a problem with setting url-pattern for servlet. If I set it to something like "users/*", after forwart() or include() to jsp, I get a redirect looping.
Here is my code:
#WebServlet("/users/*")
public class UserServlet extends HttpServlet {
protected void doGet(HttpServletRequest req, HttpServletResponse res) throws ServletException, IOException {
String[] pathParts = request.getPathInfo().split("/");
int id = Integer.valueOf(pathParts[1]);
request.setAttribute("userId", id);
request.getRequestDispatcher("user.jsp").include(request, response);
}
}
Just add "/" to your jsp page
getServletContext().getRequestDispatcher("/user.jsp").forward(request, response);
So what made difference here ???
If you directly specify user.jsp Server check user.jsp in default directory for example if Tomcat is used as webserver then server checks for user.jsp in /webapp folder (Where all applications resides.).
so where is user.jsp located ??? Its in your application for example "JSPTurorial". If you want reference user.jsp in your application you should give the relative path "/user.jsp" so that your server will check here "http://localhost:8080/JSPTutorial/user.jsp" as server is executing the files in /JSPTutorial directory else it will check here "http://localhost:8080/user.jsp" which will not be available at that path.
Try to change your servlet as follow:
#WebServlet("/users")
public class UserServlet extends HttpServlet {
protected void doGet(HttpServletRequest req, HttpServletResponse res) throws ServletException, IOException {
String[] pathParts = request.getPathInfo().split("/");
int id = Integer.valueOf(pathParts[1]);
request.setAttribute("userId", id);
getServletContext().getRequestDispatcher("/user.jsp").forward(request, response);
}
}
I'm trying to integrate a 3rd party servlet into my Spring Boot application and when I try to submit a POST to the servlet, I see the following in the logs:
PageNotFound: Request method 'POST' not supported
I've made a simple test that show this. I started using an auto generated Spring Boot project. Then I created the following Servlet:
public class TestServlet extends HttpServlet {
private static final Logger log = LoggerFactory.getLogger(TestServlet.class);
#Override
protected void doPost(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
super.doPost(req, resp); //To change body of generated methods, choose Tools | Templates.
log.info("doPost was called!!");
}
}
Then I my created Configuration like so:
#Configuration
public class ServletConfig {
#Bean //exposes the TestServlet at /test
public Servlet test() {
return new TestServlet();
}
}
Then I run the application within Tomcat7. I see in the logs:
ServletRegistrationBean: Mapping servlet: 'test' to [/test/]
Then I try to hit the endpoint with cUrl like so:
curl -v http://localhost:8080/test -data-binary '{"test":true}'
or
curl -XPOST -H'Content-type: application/json' http://localhost:8080/test -d '{"test":true}'
I've tried adding a #RequestMapping, but that didn't work either. Can anyone help me figure out how to support another Servlet inside my Spring Boot application?
You can find the sample application here: https://github.com/andrewserff/servlet-demo
Thanks!
From my previous experiences you have to call the servlet with a slash at the end (like http://localhost:8080/test/). If you don't put the slash at the end, the request is routed to the servlet mapped to /, which is by default the DispatcherServlet from Spring (your error message comes from that servlet).
The TestServlet#doPost() implementation calls the super.doPost() - which always sends a 40x error (either 405 or 400 depending on the HTTP Protocol used).
Here's the implementation:
protected void doPost(HttpServletRequest req, HttpServletResponse resp)
throws ServletException, IOException {
String protocol = req.getProtocol();
String msg = lStrings.getString("http.method_post_not_supported");
if (protocol.endsWith("1.1")) {
resp.sendError(HttpServletResponse.SC_METHOD_NOT_ALLOWED, msg);
} else {
resp.sendError(HttpServletResponse.SC_BAD_REQUEST, msg);
}
}
A Servlet can be registered using two ways:
registering the Servlet as a Bean (your approach - which should be fine) or
using a ServletRegistrationBean:
#Configuration
public class ServletConfig {
#Bean
public ServletRegistrationBean servletRegistrationBean(){
return new ServletRegistrationBean(new TestServlet(), "/test/*");
}
}
The slightly changed Servlet:
public class TestServlet extends HttpServlet {
private static final Logger log = LoggerFactory.getLogger(TestServlet.class);
#Override
protected void doPost(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
// super.doPost(req, resp);
log.info("doPost was called!!");
}
}
I am working on a filter, this code fails to execute/response.write if there is a 'forward' involved in the request. But it works fine for basic servlets that simply steam HTML content to the user. How can address "forwards" with this code.
For example, here is the filter that simple captures text content and attempts to manipulate that content.
public void doFilter(ServletRequest servletRequest, ServletResponse servletResponse, FilterChain chain) throws IOException, ServletException {
HttpServletRequest request = (HttpServletRequest) servletRequest;
HttpServletResponse response = (HttpServletResponse) servletResponse;
HttpSession session = request.getSession(false);
CharResponseWrapper responseWrapper = new CharResponseWrapper((HttpServletResponse) response);
chain.doFilter(request, responseWrapper);
final boolean commit1 = responseWrapper.isCommitted();
final boolean commit2 = response.isCommitted();
if (!commit2) {
final String res = responseWrapper.toString().replaceAll("(?i)</form>", "<input type=\"hidden\" name=\"superval\" value=\""+superval"\"/></form>");
response.getWriter().write(res);
}
return;
}
...
This works for most basic servlets, the goal is at the line with the "replaceAll".
Now, if I create a servlet with a 'forward' the code does not work, it fails at the line with 'if (!commit2)' because the stream is already committed apparently?
For example, if I make a request to this servlet and tie the filter to this servlet, then the filter does not execute completely.
public class TestCommitServlet extends HttpServlet {
private static final long serialVersionUID = 1L;
#Override
protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
req.getRequestDispatcher("TestCommitServlet2").forward(req, resp);
}
#Override
protected void doPost(final HttpServletRequest req, final HttpServletResponse resp) throws ServletException, IOException {
doGet(req, resp);
}
}
And here is the servlet that I am forwarding to:
public class TestCommitServlet2 extends HttpServlet {
private static final long serialVersionUID = 1L;
#Override
protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
final PrintWriter out = resp.getWriter();
resp.setContentType("text/html");
out.println("<html><body>(v-1)testing<form action='test'><input type='submit' value='Run' /> </form></body></html>");
}
#Override
protected void doPost(final HttpServletRequest req, final HttpServletResponse resp) throws ServletException, IOException {
doGet(req, resp);
}
}
Tl;dr : Do I have to make this call 'if (!commit2) {' The code works without it. Under this code, how would I avoid Response already committed or IllegalStateExceptions (around the line with response.getWriter().write(res);
More on the issue here:
https://codereview.stackexchange.com/questions/41260/capturing-content-within-a-j2ee-filter-and-then-replacing-this-text-request-for
I´m using Servlet API 3.0 to check this scenario.
What I found is the following. Using your code for the servlet and the filters when I call the TestCommitServlet2 , I´m able to see the following output.
http://localhost:8080/Question/TestCommitServlet2
(v-1)testing
Button here
com.koitoer.CharResponseWrapper#5b5b6746
When I call the servlet TestCommitServlet , Im able to see the following.
http://localhost:8080/Question/TestCommitServlet
(v-1)testing
Button here
this shown that filter is not apply to this forwarded request at all.
So, I remember that some filters can act in diverse DispatcherTypes as FORWARD, INCLUDE, ERROR, ASYNC and the commong REQUEST, what I decide is change the filter declaration to.
#WebFilter(filterName = "/MyFilter", urlPatterns = { "/TestCommitServlet2" }, dispatcherTypes = {
DispatcherType.FORWARD, DispatcherType.REQUEST })
public class MyFilter implements Filter {
Then when I excecute a GET over the servlet TestCommitServlet I got:
(v-1)testing
Button
com.koitoer.CharResponseWrapper#1b3bea22
the above shown that Filter is now applied to the forward request.
Also if I remove or comment lines for if (!commit2) { code still works, so there is no IllegalStateException as request need to pass over the filter which is who invoke the doChain method.
One note more, if you try to replace the content of the response using this.
responseWrapper.toString().replaceAll
You are doing it wrong as responseWrapper.toString() returns something like this CharResponseWrapper#5b5b6746, not the content, if you want to modify the response use a Wrapper that extends from HttpServletResponseWrapper and override the correct methos to manipulate the outpustream.
I've got this issue, recently I read about the REST arquitecture and it makes a perfect sense, so I'd like to achieve a RESTful web application.
Now, I'm following the Front Controller pattern that means that all of the URL mappings go to the controller.java servlet, I map the by specific URLs, not by using the /* wildcard,
the controller implements the four HTTP methods POST,GET,PUT,DELETE, each method calls the controllers service method and there I determine based on the HttpServletRequest and pathInfo the action to execute.
Controller.java
#Override
protected void service(HttpServletRequest req, HttpServletResponse resp)
throws ServletException, IOException {
IAction action;
View view;
try {
action = ActionFactory.produceAction(req);
view = action.execute(req, resp);
switch (view.getDispatchMethod()) {
case REDIRECT:
resp.sendRedirect(resp.encodeURL(view.getResource()));
break;
case FORWARD:
req.getRequestDispatcher(view.getResource()).forward(req, resp);
break;
case INCLUDE:
req.getRequestDispatcher(view.getResource()).include(req,resp);
break;
default:
}
} catch (ActionFailedException uae) {
req.setAttribute("ActionName", "Action");
req.setAttribute("FailCause", uae.getMessage());
req.getRequestDispatcher(VIEW_FAIL.getResource()).forward(req, resp);
}
}
#Override
protected void doPost(HttpServletRequest req, HttpServletResponse resp)
throws ServletException, IOException {
this.service(req, resp);
}
#Override
protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
this.service(req, resp);
}
#Override
protected void doPut(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
this.service(req, resp);
}
#Override
protected void doDelete(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
this.service(req, resp);
}
I've run into a particular issue when loading a specific order by the URI /orders/*, it is mapped to the controller servlet, the the action is executed and I load the appropriate order the action returns a View.java class
//ommited accessors and mutators for brevety.
public class View {
public enum DispatchMethod {
INCLUDE, FORWARD, REDIRECT
}
private DispatchMethod dispatchMethod;
private String resource;
public View(DispatchMethod dispatchMethod, String resource) {
this.dispatchMethod = dispatchMethod;
this.resource = resource;
}
}
Then the request is dispatched according to the getDispatchMethod() of the returned view.
Now, here is where the loop gets triggered, I use the following URL, myapp/orders/78965 /orders/* gets mapped to controller.java the appropriate action is executed and the correct order is found by the pathInfo() the returned view is new View(View.DispatchMethod.FORWARD,"order_details.jsp") the problem is that with the three available dispatch methods REDIRECT,FORWARD and INCLUDE a request is re-triggered on the URL and so on and on and on I never reach the order_details.jsp that renders the data.
So, how would you avoid the looping, as I'd like to preserve the URI displaying the order number I use the forward method, also, I'd like to do it using servlets, I've heard of the UrlRewriteFilter maybe in the future, but right now, how would it be done using "Plain Vanilla" since I'm using the Front Controller pattern, will it be necessary to add an additional servlet in the /orders/ URI ?
Any help or insights is truly appreciated.
EDIT 1:
Pasted the source code of the controller, a very basic one, I have my suspicions that the way the service method calls all of the overriden do[Method] of the servlet is triggering the loop and that it may be solved by splittig them.
Implementing a RESTful HTTP interface in Java is a lot easier using a JAX-RS implementation like RESTEasy or Jersey.
Using a Front Controller to dispatch requests to the right resource is a good approach, it's exactly the approach taken by these JAX-RS frameworks. I fear you may be re-inventing the wheel here by writing a bespoke URL parsing and dispatching mechanism when this can be taken off-the-shelf.
JAX-RS is a lightweight way to expose resources. By using a couple of simple annotations you can expose a REST interface without any plumbing required. For example:
public class Order {
#GET
#Path("/orders/{orderId}")
#Produces("text/html")
public void getOrder(#Context HttpServletResponse response,
#Context HttpServletRequest request,
#PathParam("orderId") String orderId) throws ServletException, IOException {
// ... create view and add to request here
request.getRequestDispatcher("orders.jsp").forward(request, response);
}
}
You can see how simple it is to attach this class to a URL path (using the #Path annotation), and how easily you can parse values from the URL using #PathParam. Since you get all the plumbing/dispatching/parsing off-the-shelf, you can concentrate on the bits of your app that are specific to your domain (such as what an order contains).