Can I initialize Spring via Tomcats web xml (../webapps/WEB-INF/web.xml)? - spring

Thanks to the UNUSUAL deployment of Tomcat6 I need to deploy my web app that is using the Spring & Hibernate on a server where access to the WAR deployment's web.xml at /webapps/MyDeployment/WEB-INF/web.xml is not allowed.
So I need to know if such a deployment is even possible, Where we will initialize the spring framework in the serever's web.xml at /webapps/WEB-INF/web.xml and not use the WAR's web.xml ?
Below is the WAR's web.xml that I am currently using at my environment,
<?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_2_5.xsd"
id="WebApp_ID" version="2.5">
<display-name>abcd</display-name>
<!--Here we specify about the DispatcherServlet class in the Web Deployment
Descriptor -->
<servlet>
<servlet-name>dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>dispatcher</servlet-name>
<url-pattern>*.abcd</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>dispatcher</servlet-name>
<url-pattern>*.xyz</url-pattern>
<url-pattern>*.pqr</url-pattern>
</servlet-mapping>
Unfortunately changes in the Tomcat configuration is not allowed,
Any suggestions will be highly appreciated.

Thank you for all your help, but we ended up solving the problem by asking the server administrator to lift some of the restrictions that had been put regarding file access to the tomcat user.
Now we can access the domain web xml aswell as the server web xml.

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).

<async-supported>true</async-supported> in web.xml

Someone help me plss, i got this error when i put async-supported tag in web.xml:
cvc-complex-type.2.4.a: Invalid content was found starting with element 'async-supported'. One of '{"http://java.sun.com/xml/ns/javaee":run-as, "http://java.sun.com/xml/ns/javaee":security-role-ref}' is expected.
this is my web.xml
<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>Jersey Web Application</servlet-name>
<servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>jersey.config.server.provider.packages</param-name>
<param-value>com.yeditepeim.messenger.resources</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
<async-supported>true</async-supported>
</servlet>
<servlet-mapping>
<servlet-name>Jersey Web Application</servlet-name>
<url-pattern>/webapi/*</url-pattern>
</servlet-mapping>
The web.xml goes of an XML schema. If you're not familiar with XML schemas, they describe what elements and attributes an XML document can contain in order to be a valid instance of that schema.
That being said, you can see in the schema location the version of the schema file being used, i.e. ...web-app_2_5.xsd. This means that your web.xml is going to be based on that version of the schema, which maps to that version of the servlet specification, which in your case is 2.5. The problem with this is that async is not introduced to the servlet spec until 3.0. So there is no element specification in the 2.5 schema. So when the xml is being validated, it's saying that not such element <async-supported> is allowed in the document, as it doesn't comply to the schema.
To fix it, just change the version to 3.0 and the schema file to 3_0
<!-- change to 3.0 -->
<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">
<!-- change to 3_0 -->

Integrating Crystal Report with Spring Framework - JSP

I am integrating my Crystal Report code with Spring Framework.
Initially I had developed simple web application with only jsps (without spring feature) and CRystal Reports are rendering properly. I had tested it with parameters and DB connection Also.
Now I am trying to integrate it with Spring framework. I did all necessary set-up.
Reports are rendering properly, with parameters and DB connection. But when I click on subreport link I am getting error.
"The viewer is unable to connect with the CrystalReportViewerServlet
that handles asynchronous requests. Please ensure that the Servlet and
Servlet-Mapping have been properly declared in the application's
web.xml file."
Also, images on reports are not displaying (showing cross).I tried to search on different forums but no luck.
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_2_5.xsd" id="WebApp_ID" version="2.5"> <display-name>CRWeb1</display-name>
<welcome-file-list>
<welcome-file>index.jsp</welcome-file>
</welcome-file-list>
<context-param>
<param-name>crystal_image_uri</param-name>
<param-value>/reports/crystalreportviewers</param-value>
</context-param>
<context-param>
<param-name>crystal_image_use_relative</param-name>
<param-value>reports</param-value>
</context-param>
<servlet>
<description></description>
<display-name>reports</display-name>
<servlet-name>reports</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet- class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>reports</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
<servlet>
<display-name>reports</display-name>
<servlet-name>CrystalReportViewerServlet</servlet-name>
<servlet-class>com.crystaldecisions.report.web.viewer.CrystalReportViewerServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>CrystalReportViewerServlet</servlet-name>
<url-pattern>/CrystalReportViewerHandler</url-pattern>
</servlet-mapping>
</web-app>
Thanks,
Sarika

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