I'm using Spring Boot on weblogic 10.3.6; however we usually use the embedded tomcat during development.
I'd like to enable a specific Spring profile when on weblogic to accomodate for different configuration/beans to be used when on wl. Since I'm forced to use a web.xml on that ancient version of the application server, I'd like to use web.xml to activate the weblogic profile I need.
This is my web.xml
<?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">
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>my.class.annotated.with.SpringBootApplication</param-value>
</context-param>
<listener>
<listener-class>org.springframework.boot.legacy.context.web.SpringBootContextLoaderListener</listener-class>
</listener>
<servlet>
<servlet-name>appServlet</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextAttribute</param-name>
<param-value>org.springframework.web.context.WebApplicationContext.ROOT</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>appServlet</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
</web-app>
So far I've seen suggestion to use
<context-param>
<param-name>spring.profiles.active</param-name>
<param-value>weblogic</param-value>
</context-param>
or
<init-param>
<param-name>spring.profiles.active</param-name>
<param-value>weblogic</param-value>
</init-param>
in the DispatcherServlet configuration.
I even tried configuring an ApplicationContextInitializer, but SpringBootContextLoaderListener provides his own initializer.
Is there a proper way to accomplish what I want? I can't put weblogic configuration in main application.yml because that would break the embedded tomcat configuration we are using for development.
Related
Basically I have this jetty server running at my local. But I can't access my index.jsp file I see like that
this is my web.xml file, You see I use Apache CXF, and also use Spring, Hibernate and Jetty
<?xml version="1.0" encoding="UTF-8"?>
<web-app 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_3_1.xsd"
version="3.1">
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>classpath:applicationContext.xml,classpath:Spring-Security.xml</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<!-- Spring Security Start -->
<listener>
<listener-class>
org.springframework.web.context.request.RequestContextListener
</listener-class>
</listener>
<filter>
<filter-name>springSecurityFilterChain</filter-name>
<filter-class>org.springframework.web.filter.DelegatingFilterProxy
</filter-class>
</filter>
<!-- Projenin ana url'inden itibaren spring security aktif ediliyor -->
<filter-mapping>
<filter-name>springSecurityFilterChain</filter-name>
<url-pattern>/*</url-pattern>
</filter-mapping>
<!-- Spring Security End -->
<servlet>
<servlet-name>CXFServlet</servlet-name>
<servlet-class>org.apache.cxf.transport.servlet.CXFServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>CXFServlet</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
</web-app>
How can I resolve it. Where I am doing wrong ?
This has been answered in a few places on StackOverflow. You simply need to set the dirAllowed parameter to false on the default servlet. This can be done either in the WEB-INF/web.xml of the servlet descriptor or by providing a modified etc/webdefault.xml file (via the deploy module in Jetty, for example) which is loaded before any of the contexts.
In either file this would look like:
<servlet>
<servlet-name>default</servlet-name>
<servlet-class>org.eclipse.jetty.servlet.DefaultServlet</servlet-class>
....
<init-param>
<param-name>dirAllowed</param-name>
<param-value>false</param-value>
</init-param>
....
</servlet>
As user Eng.Fouad points out this can also be defined as a context parameter:
<context-param>
<param-name>org.eclipse.jetty.servlet.Default.dirAllowed</param-name>
<param-value>false</param-value>
</context-param>
I have a Vaadin application which uses ru.xpoft.vaadin.SpringVaadinServlet as its single servlet. Now I am trying to change the servlet class to com.vaadin.spring.server.SpringVaadinServlet.
But I don't know how to specify the UI class for com.vaadin.spring.server.SpringVaadinServlet inside web.xml. I know it is possible to specify the servlet and UI class without a web.xml(Using #WebServlet and #VaadinServletConfiguration annotations). But still I require to configure the servlet using web.xml.
Please help.
You can spedify servlet and UI like this in the web.xml (from the Vaadin 7 Documentation):
<?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>myservlet</servlet-name>
<servlet-class>
com.vaadin.server.VaadinServlet
</servlet-class>
<init-param>
<param-name>UI</param-name>
<param-value>com.ex.myprj.MyUI</param-value>
</init-param>
<!-- If not using the default widget set-->
<init-param>
<param-name>widgetset</param-name>
<param-value>com.ex.myprj.AppWidgetSet</param-value>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>myservlet</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
</web-app>
See also the Vaadin docs about Deploying an Application.
To publish one java-web application in tomcat, I copy the project that named 'demo-mvc' to the tomcat's webapps folder.Then I visit "http://localhost:8080/demo-mvc/xx.jsp" in the chorme browser after starting tomcat but it prompts "The requested resource is not available".I tried editing the server.xml as follows
<Context docBase="D:\apache-tomcat-7.0.57\webapps\demo-mvc" path="/demo-mvc" reloadable="true" source="org.eclipse.jst.jee.server:website"/>
Finally it still does no effect.I am confused about where the problem is.
Try to clean your project from Project clean option and make sure you are mapping all your resources
In your web.xml the code should be like this:
<?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">
<!-- The definition of the Root Spring Container shared by all Servlets and Filters -->
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring/root-context.xml</param-value>
</context-param>
<!-- Creates the Spring Container shared by all Servlets and Filters -->
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<!-- Processes application requests -->
<servlet>
<servlet-name>appServlet</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring/appServlet/servlet-context.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>appServlet</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
</web-app>
I am creating a webapp with Mule 3 and Spring mvc 3, the problem is I am not able to get mule beans in spring controllers. My application context load with mule context and is not available in spring context, when I load application context in spring context then mule and spring both have different reference to a single bean. How can I get mule context in spring mvc controller so that I can refer to the same bean object. For ref my web.xml is defined below:
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
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"
id="return-label-service" version="2.5">
<display-name>return</display-name>
<context-param>
<param-name>org.mule.config</param-name>
<param-value>mule-return-label-flow.xml,applicationContext.xml </param-value>
</context-param>
<context-param>
<param-name>mule.serverId</param-name>
<param-value>return-label-service</param-value>
</context-param>
<listener>
<listener-class>
org.mule.config.builders.MuleXmlBuilderContextListener
</listener-class>
</listener>
<!-- <context-param>
<param-name>contextConfigLocation</param-name>
<param-value>classpath://applicationContext-web.xml</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener> -->
<servlet>
<servlet-name>spring</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>classpath:applicationContext-web.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>spring</servlet-name>
<url-pattern>*.html</url-pattern>
</servlet-mapping>
</web-app>
This is a possible duplicate of this question. Moreover, I think there is good documentation available here and here
TL;DR: <load-on-startup> does not seem to be causing my Jersey web services to load on startup
I’m trying to set up a simple RESTful web service that will act as a registry for the other RESTful services in the container.
To do this I had planned on registering each other web service with the registry service in a static initialize block. Everything was going well until I remembered that servlets aren’t loaded until their first access.
A quick web search turned up the <load-on-startup> tag that can be added to the servlet definition in web.xml. Unfortunately, this was already set in my configuration and adding some logging to the static blocks verified that the classes were indeed not being loaded until the first access of each service.
I’m developing these servlets using Jersey annotations and deploying them to Apache Tomcat (both 7.0.22 embedded in netbeans and 6.0.33 deployed remotely fail in the same way).
Besides this perceived misbehavior, the servlets have been performing correctly.
Any ideas on what I could be doing wrong?
My web.xml
<?xml version="1.0" encoding="UTF-8"?>
<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">
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/applicationContext.xml</param-value>
</context-param>
<filter>
<filter-name>CORSFilter</filter-name>
<filter-class>com.foo.bar.CORSFilter</filter-class>
</filter>
<filter-mapping>
<filter-name>CORSFilter</filter-name>
<url-pattern>/*</url-pattern>
</filter-mapping>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<servlet>
<servlet-name>dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>2</load-on-startup>
</servlet>
<servlet>
<servlet-name>ServletAdaptor</servlet-name>
<servlet-class>com.sun.jersey.spi.spring.container.servlet.SpringServlet</servlet-class>
<init-param>
<description>Multiple packages, separated by semicolon(;), can be specified in param-value</description>
<param-name>com.sun.jersey.config.property.packages</param-name>
<param-value>com.foo.bar.webservices</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>dispatcher</servlet-name>
<url-pattern>*.htm</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>ServletAdaptor</servlet-name>
<url-pattern>/resources/*</url-pattern>
</servlet-mapping>
<session-config>
<session-timeout>
30
</session-timeout>
</session-config>
<welcome-file-list>
<welcome-file>redirect.jsp</welcome-file>
</welcome-file-list>
</web-app>
It was a couple years ago and a couple years of Java Web Stack knowledge later.
I believe this was a case of mistaken identity. I had a non container managed resource that I was trying to get to load on startup. The jersey servlet container was most assuredly loading on start up as the web.xml was specifying. The classes that I wanted to preload were not being referenced by the container itself so were not being loaded with the container.