Deploy and launch a war file

(see code below). I can run it on Tomcat in an eclipse environment and it works as it should. I exported the following to a military file and created Manifest.MF with:

Manifest-Version: 1.0
Main-Class: com.process.Test

When the code runs in Eclipse, the server-side response is output to the console.

Now finally my question (Excuse my ignorance, I'm pretty new to this):

Once the war is deployed on my Tomcat server, how do I send a REST request or start a war and display the server response?

Which is equivalent: http: // localhost: 8080 / rest / xml / list on a real server?

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_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>com.process.Test</display-name>
  <servlet>
    <servlet-name>Jersey REST Service</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.process.Test</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
  </servlet>
  <servlet-mapping>
    <servlet-name>Jersey REST Service</servlet-name>
    <url-pattern>/rest/*</url-pattern>
  </servlet-mapping>
</web-app>

Client Code:

import java.net.URI; 
import javax.ws.rs.core.MediaType;
import javax.ws.rs.core.UriBuilder;
import com.sun.jersey.api.client.Client;
//import com.sun.jersey.api.client.ClientResponse;
import com.sun.jersey.api.client.WebResource;
import com.sun.jersey.api.client.config.ClientConfig;
import com.sun.jersey.api.client.config.DefaultClientConfig;

public class Test 
{       
public static void main(String[] args) {

    //Instead on using Apache Client 
    //used default Jersey client 
    //to send requests          
    ClientConfig config = new DefaultClientConfig();
    Client client = Client.create(config);      
    String _package = "api.amebatv.com";
    WebResource service = client.resource(getBaseURI(_package));

    //runRequest(service,"indexpath");
    runRequest(service,"list");
}

/*
 * For testing purposes the base URI is :http://localhost:8080/package name
 * Will change to domain name for production code
 */
private static URI getBaseURI(String _package){
    return UriBuilder.fromUri(
            "http://localhost:8080/api.process.com").build();
}


private static void runRequest(WebResource service,String path){
    String response = service.path("rest/xml/"+path).accept(MediaType.APPLICATION_XML).get(String.class);
    System.out.println("Post Response :"+response);
}

}

Server side:

@Path("/xml")

public class Service {

// List of video objects
private ArrayList<Video> videolist; 
//Parser object, parses xml  into video objects
private Parser parser = new Parser();


/*
 * Constructor
 * Parse XML and create video list on call
 */
public Service(){
    parser.createXML();
    videolist = parser.getList();       
}

/************************************************
 *GET
 * Path: /indexpath
 * list of all only <video> items
 * @return : ArrayList of Video objects
 ***********************************************/
@GET
@Path("/list")
@Produces(MediaType.APPLICATION_XML)
public List<Video> getCustomerInXML() 
{ 
    return videolist;
}
}
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1 answer

URL- - : , , web.xml .

", , " ( , " " ).

http://hostnameOrIP:8080/WARFILENAMEWITHOUTEXTENSION
  • hostnameOrIP: , IP-
  • 8080 (9001, 9090, 3000 ..)
  • WARFILENAMEWITHOUTEXTENSION, , ,

, - , ...

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