i use tomcat 7.0.77, spring, now sockjs can not open info.
web.xml
<servlet>
<servlet-name>mvc-dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
<async-supported>true</async-supported>
</servlet>
<servlet-mapping>
<servlet-name>mvc-dispatcher</servlet-name>
<url-pattern>/cmd/*</url-pattern>
<url-pattern>/webSocketServer</url-pattern>
<url-pattern>/sockjs/webSocketServer/*</url-pattern>
<url-pattern>/sockjs/webSocketServer/info</url-pattern>
<url-pattern>/webSocketServerLogout</url-pattern>
<url-pattern>/webSocketServerSpAlarm</url-pattern>
java
public class WebSocketConfig extends WebMvcConfigurerAdapter implements WebSocketConfigurer {
#Override
public void registerWebSocketHandlers(WebSocketHandlerRegistry registry) {
// registry.addHandler(systemWebSocketHandler(),"/webSocketServer").addInterceptors(new WebSocketHandshakeInterceptor());
registry.addHandler(systemWebSocketHandler(), "/sockjsWebSocketServer").addInterceptors(new WebSocketHandshakeInterceptor())
.withSockJS();
catalina.out
INFO: Mapped URL path [/sockjs/webSocketServer/**] onto handler of type [class org.springframework.web.socket.sockjs.support.SockJsHttpRequestHandler]
Jul 17, 2017 1:01:46 PM org.springframework.web.servlet.handler.SimpleUrlHandlerMapping registerHandler
html
websocket = new SockJS("http://"+local+"/sockjs/webSocketServer");
firstly can not open /sockjs/webSocketServer/info when i add /sockjs/webSocketServer/info into web.xml like above, report
http://172.24.170.178/sockjs/webSocketServer/093/vl3vnd0p/websocket 404 error
thanks
add servlet-mapping /sockjs/* to web.xml
and change html to
websocket = new SockJS("http://"+local+"/sockjs/sockjsWebSocketServer");
ok now.
because for sockjs, in webConfig java map url can not directly put it to web.xml.
need add a prefix sockjs(example).
Related
Okay I've done a lot of googling and I can't seem to find a clear answer. Let's keep it as simple as possible. I have a web.xml file
<listener>
<listener-class>A</listener-class>
</listener>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>classpath*:springcontexts/*.xml</param-value>
</context-param>
<context-param>
<param-name>contextClass</param-name>
<param-value>org.springframework.web.context.support.AnnotationConfigWebApplicationContext</param-value>
</context-param>
<servlet>
<servlet-name>spring-ws</servlet-name>
<servlet-class>org.springframework.ws.transport.http.MessageDispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>classpath*:wsspringcontexts/*.xml</param-value>
</init-param>
</servlet>
<servlet>
<servlet-name>DispatcherServlet</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>classpath*:spring_mvc_contexts/*.xml</param-value>
</init-param>
</servlet>
I think I know how to migrate this to Spring Boot ...
#SpringBootApplication(exclude = DispatcherServletAutoConfiguration.class)
#ImportResource("classpath*:springcontexts/*.xml")
public class Application
{
public static void main(String[] args)
{
SpringApplication.run(Application.class, args);
}
}
Somewhere in a sub-package...
#Configuration
#EnableWebMvc
public class SpringMVCConfiguration
{
#Bean
public ServletRegistrationBean mvc()
{
AnnotationConfigWebApplicationContext applicationContext = new AnnotationConfigWebApplicationContext();
applicationContext.setConfigLocation("classpath*:spring_mvc_contexts/*.xml");
// the dispatcher servlet should automatically add the root context
// as a parent to the dispatcher servlet's applicationContext
DispatcherServlet dispatcherServlet = new DispatcherServlet(applicationContext);
ServletRegistrationBean servletRegistrationBean = new ServletRegistrationBean(dispatcherServlet, "/spring/*");
servletRegistrationBean.setName("DispatcherServlet");
return servletRegistrationBean;
}
}
...and we do the above again for the other Servlet
My first problem is how to add the listener "A" to Spring Boot and ensure it runs before the root application is refreshed? Some beans that get configured require some static fields to be setup (legacy code), and this setup is done in listener "A". This works fine when deployed as a standard war using the above web.xml
In addition does the above Spring Boot setup look correct?
Why not put your legacy initialisation in a postConstruct method on a bean ?
Failing that you can add a listener that implements
ApplicationListener<ContextRefreshedEvent>
and overrides
public void onApplicationEvent(final ContextRefreshedEvent event)
Does your Spring Boot setup look OK ? Difficult to tell, though I'd let Spring Boot autoconfigure things like the dispatcher servlet for you and get rid of any XML config if at all possible.
This is my controller that maps a request to this url http://localhost:8080/SpringMVCJSON/rest/kfc/brands
contoller file
#Controller
#RequestMapping("/kfc/brands")
public class JSONController {
#RequestMapping(value = "{name}", method = RequestMethod.GET)
public #ResponseBody
Shop getShopInJSON(#PathVariable String name) {
Shop shop = new Shop();
shop.setName(name);
shop.setStaffName(new String[] { "name1", "name2" });
return shop;
}
this is the web.xml with the servlet request that dispatches the request/response along with the url
<display-name>Spring Web MVC Application</display-name>
<servlet>
<servlet-name>mvc-dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>mvc-dispatcher</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/mvc-dispatcher-servlet.xml</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
Assuming that everything is alright, when I launch my app on this url it returns error 404 http://localhost:8080/SpringMVCJSON/rest/kfc/brands
My server console returns this warning
Apr 26, 2016 12:14:47 PM org.springframework.web.servlet.DispatcherServlet noHandlerFound
WARNING: No mapping found for HTTP request with URI [/SpringMVCJSON/rest/kfc/brands] in DispatcherServlet with name 'mvc-dispatcher'
Please why is tomcat not mapping request to the server?
You configured your controller to be available on /kfc/brands/{name} URL but trying to access it on /kfc/brands.
Here you can find more information about using #RequestMapping: http://docs.spring.io/autorepo/docs/spring/3.2.x/spring-framework-reference/html/mvc.html#mvc-ann-requestmapping
I am trying to integrate jersey to an existing Spring application (Spring 2.5.5).
Jersey is working fine, but however when I AutoWire an existing spring bean, the object is null.
Below is my web.xml
<servlet>
<servlet-name>fs3web</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet>
<servlet-name>jersey-servlet</servlet-name>
<servlet-class>com.sun.jersey.spi.spring.container.servlet.SpringServlet</servlet-class>
<init-param>
<param-name>com.sun.jersey.config.property.packages</param-name>
<param-value>com.fl.fs3.api;org.codehaus.jackson.jaxrs</param-value>
</init-param>
<init-param>
<param-name>com.sun.jersey.api.json.POJOMappingFeature</param-name>
<param-value>true</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>fs3web</servlet-name>
<url-pattern>/fs3/*</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>jersey-servlet</servlet-name>
<url-pattern>/api/*</url-pattern>
</servlet-mapping>
And, here my application context xml (obviously this is not complete, since this is a huge application, there is much more bean definitions):
TestPojo is my bean I would like to autowire to my jersey resource.
<context:annotation-config />
<aop:aspectj-autoproxy/>
<context:component-scan base-package="com.fl.fs3.api,com.fl.fs3.integration.*.web"/>
Both my jersey resource class and POJO class is in package com.fl.fs3.api
#Component
#Path("/v1/site")
public class SitesApiControllerV1 {
#Autowired TestPojo testPojo;
#GET
#Path("/{folderName}")
#Produces(MediaType.APPLICATION_JSON)
public Response getSite(#PathParam("folderName") String folderName) {
System.out.println("pojo obj:" + testPojo);
return Response.ok("info for " + folderName).build();
}
}
#Component
public class TestPojo {
}
When I start my tomcat, I do not see the expected line in logs:
INFO: Registering Spring bean, hello, of type ..... as a root resource class
When I invoke my service /v1/site/xyz, testPojo object is null.
However, before integrating this to my existing project, I did a sample jersey+spring application, and it worked perfectly. I was able to see 'Registering Spring bean' line in logs.
Any help is appreciated.
Try this, it may be more simplified:
Load spring through web.xml like shown below as normal spring confifuration:
<servlet>
<servlet-name>project-spring</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>classpath:project-spring-servlet.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>project-spring</servlet-name>
<url-pattern>*.htm</url-pattern>
</servlet-mapping>
Now load your jersey Resources through Application as shown below:
#ApplicationPath("/rest")
public class ResourceLoader extends Application
{
/* (non-Javadoc)
* #see javax.ws.rs.core.Application#getClasses()
*/
#Override
public Set<Class<?>> getClasses()
{
Set<Class<?>> classes = new HashSet<Class<?>>();
loadResourceClasses(classes);
return classes;
}
private void loadResourceClasses(Set<Class<?>> classes)
{
classes.add(StudentResource.class);
}
}
Then in your resource:
#Path("student")
class StudentResource
{
private StudentService studentService;
StudentResource(#Context ServletContext servletContext)
{
ApplicationContext applicationContext = WebApplicationContextUtils.getWebApplicationContext(servletContext);
this.transactionService = applicationContext.getBean(StudentService .class);
}
}
There you go.
Spring has been configured with all dependency injections with Jersey!
So I am tring to get a JAX-RS application working on my WebSphere 8.5 instance. I created the following interface...
#Path("service")
public class RestService {
#GET
#Produces("text/plain")
public int getCount(){
return 1;
}
}
And This is my Application...
public class RESTConfig extends Application{
#Override
public Set<Class<?>> getClasses() {
Set<Class<?>> classes = new Hashset<?>();
classes.add(RestService.class);
return classes;
}
}
And then this is my web.xml...
<servlet>
<servlet-name>Rest Servlet</servlet-name>
<servlet-class>com.ibm.websphere.jaxrs.server.IBMRestServlet</servlet-class>
<init-param>
<param-name>jaxrs.ws.rs.Application</param-name>
<param-value>com.company.rest.RESTConfig</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
....
<servlet-mapping>
<servlet-name>Rest Servlet</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
Then I have an EAR configured with the WAR as a module. But when I start everything and try going to http://localhost:[port]/war/rest/app/service I see..
[TIME] 00000115 RequestProces I org.apache.wink.server.internal.RequestProcessor logException The following error occurred during the invocation of the handlers chain: WebApplicationException (404 - Not Found) with message 'null' while processing GET request sent to http://localhost:[port]/war/rest/service
Please Help!
WAS8.5 supports v2.4 and v3 servlets. The reason removing your web.xml contents (and using 3.0 code) worked for you is because you had a mistake in the param-name tag of your web.xml. v2.4 servlet works fine in WAS8.5 when you use the correct param-name.
This is incorrect.
<param-name>jaxrs.ws.rs.Application</param-name>
This is correct:
<param-name>javax.ws.rs.Application</param-name>
Details:
http://pic.dhe.ibm.com/infocenter/wasinfo/v8r5/topic/com.ibm.websphere.nd.multiplatform.doc/ae/twbs_jaxrs_configwebxml.html
The RestConfig class (that is defined as the JAX-RS Application) should override getClasses to return the resources:
#Path("app")
public class RESTConfig extends Application{
#Override
public Set<Class<?>> getClasses() {
Set<Class<?>> classes = new Hashset<?>();
classes.add(RestService.class);
return classes;
}
}
The issue appears to be related to 8.5 only supporting v3 servlets. this seems to fix the issue....
#Path("service")
public class RestService {
#GET
#Produces("text/plain")
public String getCount(){
//Text-Plain cannot be int apparently
return String.valueOf(1);
}
}
#ApplicationPath("rest")
public class RESTConfig extends Application{
//Override no longer needed.
}
This should now deploy fine...
Here was my source IBM
Also, You can try buy changing the below web.xml File
<servlet>
<servlet-name>javax.ws.rs.core.Application</servlet-name>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>javax.ws.rs.core.Application</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
Also, In Project Facets - Change Web Module version to 3.0
For More Reference Visit: How to deploy a JAX-RS application?
I am working on a Spring application using Tomcat 6 and Spring 2.5. I'm trying to get my URL mapping correct. What I would like to have work is the following:
http://localhost:8080/idptest -> doesn't work
But instead, I have to reference the context name in my URL in order to resolve the mapping:
http://localhost:8080/<context_name>/idptest -> works
How can I avoid the requirement of referencing the context name in my URL without using a rewrite/proxy engine e.g. Apache?
Here is the servlet definition and mapping from my web.xml:
<servlet>
<servlet-name>idptest</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/conf/idptest.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>idptest</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
Here's the outline of my controller (showing annotations for request mappings):
#Controller
#RequestMapping("/idptest")
public class MyController {
#RequestMapping(method=RequestMethod.GET)
public String setupForm(Model model){
MyObject someObject = new MyObject();
model.addAttribute("someObject", someObject);
return "myform";
}
#RequestMapping(method = RequestMethod.POST)
public String processSubmit(#ModelAttribute("someObject") MyObject someObject) throws Exception {
// POST logic...
}
}
Thanks!
That's going to depend on your servlet container, for Tomcat - you pretty much have to deploy your webapp as the ROOT webapp, that is, under $CATALINA_HOME/webapps/ROOT/
More info here
Just rename your war file to ROOT.war, then the application runs in root context (i.e. with empty context name)