Tip: Dynamic Web Project - - > RightClick - - > Java EE Tools - - > Generate Deployment Descriptor Stub. This would create Web.xml
Starting all over again from the question:
If you are doing Java programming to create some Service using Jersey or any other framework, you will many times land into situation where you need Web.xml to be created. I had once wondered where this file would be present. Or, better say, how should i create the Web.xml file.
Note that the web.xml file should be listed right below the last line in your screenshot and resides in WebContent/WEB-INF.
If it is missing you might have missed to check the "Generate web.xml deployment descriptor" option on the third page of the Dynamic web project wizard.
Tip: Dynamic Web Project - - > RightClick - - > Java EE Tools - - > Generate Deployment Descriptor Stub. This would create Web.xml
Tip: Dynamic Web Project - - > RightClick - - > Java EE Tools - - > Generate Deployment Descriptor Stub. This would create Web.xml
A sample content of Web.xml can be as follows:
<?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"
id="WebApp_ID" version="3.0">
<display-name>User Management</display-name>
<servlet>
<servlet-name>Jersey RESTful 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.tutorialspoint</param-value>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>Jersey RESTful Application</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
</web-app>
Take Care.
- Mohd Anwar Jamal Faiz