如何在嵌入式Tomcat上部署Web应用程序 - JavaEE

Question

I want to deploy a web app on tomcat embbeded server which i load from maven: My rest endpoint:

@Path("/test")
public class RestTestController {

@GET
public String doNothing(){
    return "A";
}
}

And maven setup:

<build>
    <plugins>
        <plugin>
            <artifactId>maven-compiler-plugin</artifactId>
            <version>2.3.2</version>
            <configuration>
                <source>1.7</source>
                <target>1.7</target>
            </configuration>
        </plugin>

        <plugin>
            <groupId>org.apache.tomcat.maven</groupId>
            <artifactId>tomcat7-maven-plugin</artifactId>
            <version>2.2</version>
            <configuration>
                <port>8080</port>
                <path>/</path>
            </configuration>
        </plugin>
    </plugins>
</build>

After performing mvn clean install and later mvn tomcat7:run I recieve message that war tomcat was started, however when i enter localhost:8080/test i constantly recieve 404 Not Fount Http response. What am I possibly doing wrong? Should I add some kind of mapping into web.xml?

Logs on tomcat shows that I certainly deployed my app:

[INFO] ---------------------< pl.wachkar:take-restaurant >---------------------
[INFO] Building take-restaurant 1.0-SNAPSHOT
[INFO] --------------------------------[ war ]---------------------------------
[INFO]
[INFO] >>> tomcat7-maven-plugin:2.2:run (default-cli) > process-classes @ take-restaurant >>>
[INFO]
[INFO] --- maven-resources-plugin:2.6:resources (default-resources) @ take-restaurant ---
(...)
[INFO] --- tomcat7-maven-plugin:2.2:run (default-cli) @ take-restaurant ---
[INFO] Running war on http://localhost:8080/

@EDIT Empty web.xml file:

<?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_3_0.xsd" id="WebApp_ID" version="3.0">

</web-app>

Answer 1

What does your web.xml look like? You will need at least a welcome file list so that tomcat knows which file to serve at startup

For example:

<welcome-file-list>
    <welcome-file>index.html</welcome-file>
</welcome-file-list>

Answer 2

i think you entered a wrong url in the brower(maybe incomplete).without using a web.xml you should create a class that extends Application class and annotate it by @ApplicationPath("anyString") for example:

  import javax.ws.rs.core.Application;

     @ApplicationPath("myWebservice")
     public class MyConfig extends Application{

    }

you don't need to implement any method it works fine.this class can be in any package. and for testing the right url is:

localhost:port/contextpath/string1/string2

string1 is what you specify in Application annotation (here myWebservice) string2 is what you specify in Path annotation on your resource class(in your case test) so we have

localhost:8080/contextpath/myWebservice/test