I am using Spring boot 2.6.3 and I have the following web.xml in my project. What would be the ideal way to move this to Java config?
<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">
<display-name>My Web Application</display-name>
<servlet>
<servlet-name>my</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.xyz.my.service.custom.config.CustomServiceConfig</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>my</servlet-name>
<url-pattern>/v1/</url-pattern>
</servlet-mapping>
</web-app>
You don't need all of that. Just ditch your web.xml. The AnnotationConfigWebApplicationContext (actually Spring Boot has a specialized sub-class for it) is the default and your configuration class just can be imported or automatically detected
If your #SpringBootApplication annotated class is in the com.xyz.my.service you don't need to do anything. Your CustomServiceConfig will be automatically detected. If it isn't you can add an #Import(CustomServiceConfig.class) to your #SpringBootApplication annotated class.
Related
I have been spending hours trying to figure out why I am gettin 404 error. I have checked similar postings in Stackoverflow, but I am still having a problem.
I know it runs on Tomcat and can display index file. But, when I try to test RESTful ws with the following url, http://localhost:8080/RestWs/rest/hello,
I get 404 error:
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>RestWs</display-name>
<servlet>
<servlet-name>Faces Servlet</servlet-name>
<servlet-class>javax.faces.webapp.FacesServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Faces Servlet</servlet-name>
<url-pattern>*.xhtml</url-pattern>
</servlet-mapping>
<servlet>
<servlet-name>Jersey REST Service</servlet-name>
<servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Jersey REST Service</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
<context-param>
<description>State saving method: 'client' or 'server' (=default). See JSF Specification 2.5.2</description>
<param-name>javax.faces.STATE_SAVING_METHOD</param-name>
<param-value>client</param-value>
</context-param>
<context-param>
<param-name>javax.servlet.jsp.jstl.fmt.localizationContext</param-name>
<param-value>resources.application</param-value>
</context-param>
<listener>
<listener-class>com.sun.faces.config.ConfigureListener</listener-class>
</listener>
</web-app>
My resource class in Java:
package DemoRest;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;
#Path("/hello")
public class HelloResource {
#GET
#Produces(MediaType.TEXT_PLAIN)
public String sayHello() {
return "Hello Jersey";
}
}
I downloaded Jersey jars from https://jersey.github.io/download.html#
and copied the jars in the WEB/INF/lib folder.
The extracted jaxrs-ri-2.26 jar has three folders - api, ext and lib. I copied all jars from these folders to the WEB/INF/lib folder.
I am wondering if I am missing some jars or they are not in a right location. However, I checked most of the related postings in the stackoverflow for solution, but I was not successful.
I appreciate any help.
I solved my problem. One solution I found worked. I had to download JaxWs jar from https://github.com/javaee/metro-jax-ws and copy jaxws-rt.jar into Tomcat lib folder.
I´m using spring mvc, I have got a project (Apliconta2Web) and your url is //Apliconta2Web/, but if i copy this project and change your name by (Apliconta3WebPrimefaces) when i push button run as i want url //Apliconta3WebPrimefaces/ but always i have got //Apliconta2Web/.
I have changed servlet-name. and add server the name is Apliconta3WebPrimefaces(Apliconta2Web)
How i can change it? Very thank. sorry by my english. i´m learning.
web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5" xmlns="http://java.sun.com/xml/ns/j2ee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:web="http://java.sun.com/xml/ns/j2ee/web-app_2_5.xsd"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/j2ee/web-app_2_5.xsd">
<display-name>Aplicacion Web Apliconta3WebPrimefaces</display-name>
<servlet>
<servlet-name>apliconta3webPrimeFaces</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet
</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/app-config.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>apliconta3webPrimefaces</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
<welcome-file-list>
<welcome-file>index.htm</welcome-file>
</welcome-file-list>
</web-app>
I'm trying to deploy my application, using web.xml, servlet 3.0, and jersey API. Unfortunately, it doesn't work.
This is MyApplication.class :
package com.example;
public class MyApplication extends Application {
public Set<Class<?>> getClasses() {
Set<Class<?>> s = new HashSet<Class<?>>();
s.add(MyResource.class);
return s;
}
}
This is MyResource :
#Path("/helloworld")
#Produces(MediaType.TEXT_PLAIN)
public class MyResource {
#GET
public String getHello() {
return "HelloWorld !";
}
}
And my web.xml :
<web-app>
<servlet>
<servlet-name>com.example.MyApplication</servlet-name>
</servlet>
<servlet-mapping>
<servlet-name>com.example.MyApplication</servlet-name>
<url-pattern>/webapi/*</url-pattern>
</servlet-mapping>
</web-app>
On client side, I'm using this url :
http://localhost:8080/[projectname]/webapi/helloworld
And i have this error :
java.lang.NullPointerException
sun.misc.Launcher$AppClassLoader.loadClass(Launcher.java:294)
java.lang.ClassLoader.loadClass(ClassLoader.java:247)
org.apache.catalina.loader.WebappClassLoader.loadClass(WebappClassLoader.java:1629)
org.apache.catalina.loader.WebappClassLoader.loadClass(WebappClassLoader.java:1559)
org.apache.catalina.authenticator.AuthenticatorBase.invoke(AuthenticatorBase.java:491)
org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:99)
org.apache.catalina.valves.AccessLogValve.invoke(AccessLogValve.java:953)
org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:408)
org.apache.coyote.http11.AbstractHttp11Processor.process(AbstractHttp11Processor.java:1023)
org.apache.coyote.AbstractProtocol$AbstractConnectionHandler.process(AbstractProtocol.java:589)
org.apache.tomcat.util.net.AprEndpoint$SocketProcessor.run(AprEndpoint.java:1852)
java.util.concurrent.ThreadPoolExecutor$Worker.runTask(ThreadPoolExecutor.java:895)
java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:918)
java.lang.Thread.run(Thread.java:662)
What's wrong ? :/ I'm using Tomcat 7.
PS : with a servlet 2.x, it works :
<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.example</param-value>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>Jersey Web Application</servlet-name>
<url-pattern>/webapi/*</url-pattern>
</servlet-mapping>
but i will need asynchronous mode later.
Thanks !
Update: Since writing this answer, I've found out a way to avoid needing a web.xml on Tomcat using the official Glassfish Jersey implementation. Look here for details.
If you're using a standard Tomcat install (or some other servlet container), AFAIK you can't avoid explicitly telling it what servlets to start in the web.xml file*. Since you have to use web.xml anyway, the simplest way to get restful web services working is to forget extending javax.ws.rs.core.Application entirely and just specify the context path there. You can still use standard jax-rs annotations to declare the actual web services.
web.xml:
<?xml version="1.0" encoding="UTF-8"?>
<web-app
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"
version="3.0"
>
<servlet>
<servlet-name>rest-test</servlet-name>
<servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>com.sun.jersey.config.property.packages</param-name>
<param-value>com.domain.mypackage</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name> rest-test</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
</web-app>
Two noteworthy points:
You will need to bundle a REST implementation in your WAR file, since servlet containers don't usually contain one. Since Jersey is the reference implementation for JAX-RS, that's the one I'm using in the servlet-class element above. You can replace this with Apache CXF implementation if you want.
The init-param element tells Jersey which of your packages to search for Java files with web service annotations. Edit this to point to your web services. Note that if you opt to use apache CXF instead of Jersey, the stuff needed in any init-param elements will be different. Someone who knows CXF please post what they would be.
If you're using Maven, just add a dependency to jersey-servlet in the dependencies section of your pom.xml file:
<dependencies>
<dependency>
<groupId>com.sun.jersey</groupId>
<artifactId>jersey-servlet</artifactId>
<version>1.18.2</version>
</dependency>
...
</dependencies>
After this, declaring your web services is straight forward using the standard JAX-RS annotations in your Java classes:
package com.domain.mypackage;
import javax.ws.rs.Consumes;
import javax.ws.rs.Produces;
import javax.ws.rs.GET;
import javax.ws.rs.MatrixParam;
import javax.ws.rs.Path;
// It's good practice to include a version number in the path so you can have
// multiple versions deployed at once. That way consumers don't need to upgrade
// right away if things are working for them.
#Path("calc/1.0")
public class CalculatorV1_0 {
#GET
#Consumes("text/plain")
#Produces("text/plain")
#Path("addTwoNumbers")
public String add(#MatrixParam("firstNumber") int n1, #MatrixParam("secondNumber") int n2) {
return String.valueOf(n1 + n2);
}
}
This should be all you need. If your Tomcat install is running locally on port 8080 and you deploy your WAR file to the context myContext, going to
http://localhost:8080/myContext/rest/calc/1.0/addTwoNumbers;firstNumber=2;secondNumber=3
...should produce the expected result (5).
Cheers!
* Someone please correct me if you know of a way to a add the Jersey servlet to the context in Tomcat without using web.xml--maybe by using a context or life cycle listener?
You need to create web.xml under WEB-INF folder with following line of code
<?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">
<servlet>
<servlet-name>ServletAdaptor</servlet-name>
<servlet-class>
org.glassfish.jersey.servlet.ServletContainer
</servlet-class>
<init-param>
<param-name>jersey.config.server.provider.packages</param-name>
<param-value>package where MyResource resides</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>ServletAdaptor</servlet-name>
<url-pattern>/webresources/*</url-pattern>
</servlet-mapping>
<session-config>
<session-timeout>
30
</session-timeout>
</session-config>
</web-app>
Now you can delete your MyApplication Class. Deploy and test web service. it will work
package com.example;
#ApplicationPath("/webapi")
public class MyApplication extends Application {
public Set<Class<?>> getClasses() {
Set<Class<?>> s = new HashSet<Class<?>>();
s.add(MyResource.class);
return s;
}
}
Visit this link may help:
https://jersey.java.net/documentation/latest/deployment.html#deployment.servlet.2
section: 4.7.1. Servlet 2.x Container
<web-app>
<servlet>
<servlet-name>MyApplication</servlet-name>
<servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
<init-param>
...
</init-param>
</servlet>
...
<servlet-mapping>
<servlet-name>MyApplication</servlet-name>
<url-pattern>/myApp/*</url-pattern>
</servlet-mapping>
...
</web-app>
Hi guys I'm working with Spring 3.2.3, hibernate 4.2.2 and org.springframework.security 3.0.5. Before I start working with security in spring, my context file was called servlet-context.xml and everything worked fine. Since I start using org.springframework.security 3.0.5 when I try to run my application I get the error:
java.io.FileNotFoundException: Could not open ServletContext resource [/WEB-INF/appServlet-servlet.xml]
My web.xml is:
<?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">
<!-- Spring MVC -->
<servlet>
<servlet-name>appServlet</servlet-name>
<servlet-class>
org.springframework.web.servlet.DispatcherServlet
</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>appServlet</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
<listener>
<listener-class>
org.springframework.web.context.ContextLoaderListener
</listener-class>
</listener>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
/WEB-INF/servlet-context.xml,
/WEB-INF/spring-security.xml
</param-value>
</context-param>
<!-- Spring Security -->
<filter>
<filter-name>springSecurityFilterChain</filter-name>
<filter-class>
org.springframework.web.filter.DelegatingFilterProxy
</filter-class>
</filter>
<filter-mapping>
<filter-name>springSecurityFilterChain</filter-name>
<url-pattern>/*</url-pattern>
</filter-mapping>
Notice that I'm specifying my servlet-context.xml:
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
/WEB-INF/servlet-context.xml,
/WEB-INF/spring-security.xml
</param-value>
</context-param>
If I leave my web.xml just like this and copy/paste my servlet-context.xml content in a new file called appServlet-servlet.xml, everything works fine. This is confusing for me since I told web.xml that my context file name was servlet-context.xml. Am I forced to call my context file as appServlet-servlet.xml??. Of course, if I delete servlet-context.xml and leave appServlet-servlet.xml making the specficiation in web.xml, works fine.
I just wanna know if is obligatory to call my context file as appServlet-servlet.xml if I want to use spring security in my app.
This is not strange. Spring works this way. Since your DispatchServlet name is appServlet, Spring automatically tries to find the servlet context that has the same name, here "appServlet-servlet.xml".
Here is a page that may help : http://syntx.io/difference-between-loading-context-via-dispatcherservlet-and-contextloaderlistener/
This the file web.xml in WEB-INF
<?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>/WEB-INF/applicationContext.xml</param-value>
</context-param>
<filter>
<filter-name>LoginFilter</filter-name>
<filter-class>glpi.filter.LoginFilter</filter-class>
</filter>
<filter-mapping>
<filter-name>LoginFilter</filter-name>
<url-pattern>/index.jsp</url-pattern>
</filter-mapping>
<servlet>
<servlet-name>context</servlet-name>
<servlet-class>org.springframework.web.context.ContextLoaderServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet>
<servlet-name>dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet>
<servlet-name>action</servlet-name>
<servlet-class>org.apache.struts.action.ActionServlet</servlet-class>
<init-param>
<param-name>config</param-name>
<param-value>/WEB-INF/struts-config.xml</param-value>
</init-param>
<init-param>
<param-name>debug</param-name>
<param-value>2</param-value>
</init-param>
<init-param>
<param-name>detail</param-name>
<param-value>2</param-value>
</init-param>
<load-on-startup>2</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>dispatcher</servlet-name>
<url-pattern>*.htm</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>action</servlet-name>
<url-pattern>*.do</url-pattern>
</servlet-mapping>
<session-config>
<session-timeout>
30
</session-timeout>
</session-config>
<welcome-file-list>
<welcome-file>/login.jsp</welcome-file>
</welcome-file-list>
</web-app>
I think you are missing the context loader listener(to pick your spring context file(s)).
Add this to your web.xml
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
You could also check out the Initial web configuration section # http://static.springsource.org/spring/docs/2.0.x/reference/beans.html
You have both ContextLoaderServlet and DispatcherServlet set to load-on-startup = 1. That means either one of them could start first, and you need the ContextLoaderServlet to start first, since that's what creates the root WebApplicationContext that your error says is missing. So leave ContextLoaderServlet's load-on-startup at 1, and change the DispatcherServlet's to 2 or higher.
Actually, it's preferred to use ContextLoaderListener instead of the Servlet unless you're on a really old container where the Listener doesn't work properly.
Add following code in web.xml file, bcs it looks for contex to load so we have to declare it initially.
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/HelloWeb-servlet.xml</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
I've recently stumbled upon the same issue, and I knew for sure it couldn't be caused by misconfiguration because I've copied the entire working Tomcat installation from another machine. Yet I kept getting the same exception:
java.lang.IllegalStateException: No WebApplicationContext found: not in a DispatcherServlet request and no ContextLoaderListener registered?
As I've eventually figured out, it was a wrong JVM version that broke the application: this one used Java 7, whereas the working instance (and the webapp) was on Java 8.
Hope it helps someone struggling with this counter-intuitive error message.