部署和运行war文件

Question

(请参阅下面的代码代码)我可以在eclipse环境中在Tomcat上运行它,它可以正常工作.我已将以下内容导出到war文件并创建了Manifest.MF:

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

当代码在Eclipse中运行时,来自服务器端的响应将输出到控制台.

最后我的问题(原谅我的无知,我对此很新):

在我的Tomcat服务器上部署战争后,如何发送REST请求或运行war并显示服务器响应？

什么是相当于:http:// localhost:8080/rest/xml/list在实时服务器上？

在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>

客户端代码:

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);
}

}

服务器端:

@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;
}
}

Answer 1

Web应用程序的URL取决于许多因素,即应用程序服务器,配置文件,web.xml和特定于应用程序服务器的配置.

但在某些情况下,这是"如果你不知道的话,首先要尝试"(不是官方的,但是"它有效"的经验法则)

http://hostnameOrIP:8080/WARFILENAMEWITHOUTEXTENSION

• hostnameOrIP:嗯,只需替换为您的主机名或IP
• 将8080替换为应用服务器的默认端口(9001,9090,3000等...)
• 将WARFILENAMEWITHOUTEXTENSION替换为war文件名,而不是扩展名

如果你没有其他方法可以告诉我,这又是一种非官方的,先试试的方式......