Creating a route with Spring in Netbeans - spring

I know this is a pretty basic issue, but I struggle hard with it...
I'd like to create a bunch of routes with the Spring framework in the Netbeans IDE, I have created a test #Controller class:
#Controller
public class HelloController {
#RequestMapping("/test")
public ModelAndView thisIsATest(HttpServletRequest request) {
return (new ModelAndView("myTestPage.jsp"));
}
}
myTestPage.jsp is a JSP file in the WEB-INF/jsp/ folder, and the HelloController class is in the Source Packages/ folder within a controller java package.
When I start the server, I can acces the root '/index.htm' that displays the index.jsp page (from the redirect.jsp file), but when I try to access '/test' or '/test.htm' I get a 404 error...
I really don't know how to make a Spring controller to work, and I did many tutorials without success.

In your applications web.xml make sure you have something like this. This will allow access to all your jsp pages. Add any other filters you may need here too.
<web-app version="3.0" 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">
<servlet>
<servlet-name>myapp</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
<async-supported>true</async-supported>
</servlet>
<welcome-file-list>
<welcome-file>index.html</welcome-file>
</welcome-file-list>
<servlet-mapping>
<servlet-name>myapp</servlet-name>
<url-pattern>*.jsp</url-pattern>
</servlet-mapping>
</web-app>

Related

My welcome-file is returning an empty page in Open Liberty

On Open Liberty 21.0.0.6., instead of presenting my welcome JSF page, my browser returns empty content when I point it to http://<host>:<port>/<ctxRoot>,
web.xml
<web-app id="WebApp_ID" version="4.0"
xmlns="http://xmlns.jcp.org/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_4_0.xsd">
<display-name>MyJSF</display-name>
<servlet>
<servlet-name>Faces Servlet</servlet-name>
<servlet-class>javax.faces.webapp.FacesServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Faces Servlet</servlet-name>
<url-pattern>*.xhtml</url-pattern>
</servlet-mapping>
<welcome-file-list>
<welcome-file>index.xhtml</welcome-file>
</welcome-file-list>
</web-app>
and a feature set of:
server.xml
<server>
<featureManager>
<feature>jaxrs-2.1</feature>
<feature>cdi-2.0</feature>
<feature>jpa-2.2</feature>
<feature>jdbc-4.3</feature>
<feature>jsf-2.3</feature>
<feature>mpHealth-3.0</feature>
</featureManager>
ADDITIONAL INFO
I also have some other stuff in my app like JPA and JAX-RS.
ROOT CAUSE - JAX-RS application with "/" application path
In my case the problem is that I had a JAX-RS Application configured to use: #ApplicationPath("/") which was colliding with my goal of having the "/" application path serve up the welcome-file.
SOLUTION
Move the JAX-RS app to its own path within the WAR, e.g:
import javax.ws.rs.core.Application;
#ApplicationPath("/api")
public class TestApp extends Application { }
THOUGHTS
This was easy to stumble into by taking an existing simple JAX-RS app using "/" as the app path, and then adding JSF to it. Then I thought I was having trouble with the JSF implementation.

Why does my Spring MVC application not work when changing the server mapping?

I'm working on the Spring MVC "FitnessTracker" application outlined on Pluralsight. Below is my "web.xml" file:
<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5"
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_2_5.xsd">
<servlet>
<servlet-name>fitTrackerServlet</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/config/servlet-config.xml</param-value>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>fitTrackerServlet</servlet-name>
<url-pattern>/FitnessTracker/*.html</url-pattern>
</servlet-mapping>
<display-name>Archetype Created Web Application</display-name>
</web-app>
The above makes Tomcat generate a bunch of exceptions, starting with
org.apache.catalina.LifecycleException: Failed to start component [StandardEngine[Catalina].StandardHost[localhost].StandardContext[]]
at org.apache.catalina.util.LifecycleBase.start(LifecycleBase.java:153)
at org.apache.catalina.core.ContainerBase.addChildInternal(ContainerBase.java:899)
But when I change what's between the <url-pattern> tag to *.html, it works fine. Why is that?
Note: My goal is to try to make my app's controller run when I type /FitnessTracker/greeting.html, instead of /greeting.html. I am using Intellij IDEA, and doing Maven project with Tomcat 7.0 as my server.
Application runs # http://localhost:9090/FitnessTracker/greeting.html URL . FitnessTracker is application root context and greeting.html is mapped to Hello controller method. Please see below.
Could you please post the web.xml and controller mapping .
The original code runs at the URL - http://localhost:8080/FitnessTracker/greeting.html. SO I am not sure why you need to change web.xml for that.
Also the URL pattern you are trying to use "/FitnessTracker/*.html" is not valid.
More details here.
https://stackoverflow.com/a/5441862/6352160

Spring Websocket, 404 error while connecting. App not using Spring MVC

I'm trying to get working the Spring Example tutorial:
https://spring.io/guides/gs/messaging-stomp-websocket/
But I need it working on an existing Application that I need to add WebSocket support. The idea is once I get the basic working, start building from there.
The differences I have with the example in the URL is:
I'm not using SpringBootApplication, but Tomcat instead (7.0.69)
That implies I do have a web.xml (included below)
I skipped the Application.java with the main... I'm building a WAR and deploying it manually into Tomcat.
Whem I start Tomcat I read the following which seems relevant:
03:06:52,908 INFO SimpleBrokerMessageHandler:157 - Starting...
03:06:52,908 INFO SimpleBrokerMessageHandler:260 - BrokerAvailabilityEvent[available=true, SimpleBrokerMessageHandler [DefaultSubscriptionRegistry[cache[0 destination(s)], registry[0 sessions]]]]
03:06:52,913 INFO SimpleBrokerMessageHandler:166 - Started.
The problem is the following:
When I click on connect on the index.html, I get a 404 when I'm trying to reach the server at the mapping url "/hello" (not sure why info is there, I guess is part of the protocol..) http://localhost:8080/hello/info
I was googling this, even found some answers in stackoverflow. Some have fixed this addding as prefix the DispatcherServlet mapping to the websocket url...
However my app isn't using Spring MVC, so I don't have a DispatcherServlet configured... and I looks like an overkill to include it just for the WebSockets.
I tried adding the annotation #EnableWebMvc to the WebSocketConfig class (in the link is the code, I have the same lines)
Any suggestion will be much appreciated !
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_3_0.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>App</display-name>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/classes/applicationContext.xml</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<listener>
<listener-class>org.springframework.web.context.request.RequestContextListener</listener-class>
</listener>
<welcome-file-list>
<welcome-file>index.jsp</welcome-file>
</welcome-file-list>
</web-app>
I give up trying to avoid having an MVC Dispatcher for an APP witch doesn't use Spring MVC.
This solved it:
web.xml
<servlet>
<!-- Only used by websockets -->
<servlet-name>SpringMVCDispatcherServlet</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/classes/websockets-application-config.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>SpringMVCDispatcherServlet</servlet-name>
<url-pattern>/ws/*</url-pattern>
</servlet-mapping>
And then, in the index.html I only added the prefix /ws/ when connecting (only there, sinding messages without the prefix is ok).

spring rest project with client side in angular js - how to approach the rest service with ajax

So I have a spring rest project that includes client side app.
I can run the service on a local tomcat and get responses querying "http://:8080/books" for example.
I can set app an apache server and go to "http://" to see my client app (the apache htdocs dir points to the project client app dir).
What I can't manage to do is send ajax rest queries to the service.
I'm using angular js so it looks like:
$http.get("http://:8080/books").success(...).error(...);
and it always enters the error callback method.
In the debugger/network tab I see that the request status is "canceled". Looking at the request's details I see next to the "Request headers" title the message: "CAUTION: Provisional headers are shown".
Here is my 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"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
version="3.0">
<display-name>Library Application</display-name>
<context-param>
<param-name>contextClass</param-name>
<param-value>org.springframework.web.context.support.AnnotationConfigWebApplicationContext</param-value>
</context-param>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>com.library.application</param-value>
</context-param>
<context-param>
<param-name>spring.profiles.active</param-name>
<param-value>prod</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<!-- Processes application requests -->
<servlet>
<servlet-name>BooksServlet</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextClass</param-name>
<param-value>org.springframework.web.context.support.AnnotationConfigWebApplicationContext</param-value>
</init-param>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>com.library.service.config.ControllerConfig</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>BooksServlet</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
</web-app>
My project structure is:
src
main
java
resources
webapp
resources
WEB-INF
test
Is it because the origin is different (mainly the port)?
Should I add some code to my web.xml to serve the client through tomcat? If so how?
thanks.
I guess you have a problem with cross-domain requests(yes, different port is another domain for the browser and it will cancel the request for security purposes if server will not include cross-domain headers). Try to add this headers to your REST response:
responseHeaders.add("Access-Control-Allow-Origin", "*");
if you need cookies you might need this as well:
responseHeaders.add("Access-Control-Allow-Credentials", "true");
in the second case you have to set an origin, '*' will not work in this case
more info here
Angular http uses relative urls.
For example using : www.stackoverflow.com
And
$http.get('/api/search?'
would call : www.stackoverflow.com/api/search
You can tests your rest api using your browser for gets requests, and one of the many plugins for post/put/delete.
(they are jsut made up urls btw)
Found it.
Thanks for all the answers, I was looking for the spring solution.
+1 JohnnyAW for your answer.
In my controller config class I've extended WebMvcConfigurerAdapter and overridden an addResourceHandlers method pointing to my client web app dir:
#Configuration
#EnableWebMvc
#ComponentScan("com.library.service")
public class ControllerConfig extends WebMvcConfigurerAdapter {
#Override
public void addResourceHandlers(ResourceHandlerRegistry registry) {
registry.addResourceHandler("/resources/**").addResourceLocations("/resources/");
}
}
Note that it works because configuration is annotated with #EnableWebMvc.

Spring 3, Flex 4 Integration with SpringFlex 1.5.0.M2 api + configuration

We are working on a project where we are using Spring 3 to create a web platform and using Flex 4 to create a specific client side application. Currently, we need to integrate Spring project with Flex.
We are using Spring-Flex integration library version: 1.5.0.M2
I checked the older questions but the integration configurations defined at those entries are generally for previous versions of BlazeDS and Spring. And as I understad, there may be some differencies.
Can anybody tell me how to do the configuration in web.xml and any other xml files needed,and how the folder structures will be. Any up-to-date tutorial links will be appreciated.
Our business requirements are:
Two servlets should exist: 1) projectServlet that has mappings /.html
2) flexServlet that has mappings /messageBroker/
Our service classes that can be used in Flex side will be like:
package com.ecognitio.service;
import org.springframework.flex.remoting.RemotingDestination;
import org.springframework.flex.remoting.RemotingInclude;
import org.springframework.stereotype.Service;
#Service
#RemotingDestination
public class Foo {
#RemotingInclude
public void sayHello(String name){
System.out.println("Hello: "+name);
}
}
Regards,
Ugur
My Flex 4, Hibernate 3, and Spring 3 Integration Refcard walks through the process of setting everything up and should work fine with 1.5.0.M2 (assuming you have changed the namespace in the Spring config file). But here is a basic web.xml example:
<?xml version="1.0" encoding="UTF-8"?>
<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 http://java.sun.com/xml/ns
/j2ee/web-app_2_4.xsd"
version="2.4">
<display-name>Project Template</display-name>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</liste
ner-class>
</listener>
<listener>
<listener-class>flex.messaging.HttpFlexSession</listener-class>
</listener>
<servlet>
<servlet-name>flex</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value></param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>flex</servlet-name>
<url-pattern>/messagebroker/*</url-pattern>
</servlet-mapping>
</web-app>
That should be enough to get you started.

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