使用Jersey和Apache Tomcat用于Web Api的404错误

Question

I am trying to play around with Eclipse/Java and I am making a Dynamic Web Project in Eclipse. I have followed tutorials online and looked at several other Stackoverflow posts, but i'm still having the same problem.

When I run my server and try to hit my endpoint /test , I get a 404 error. Here is my code:

My Java Class:

package TestPackage;

import javax.ws.rs.Path;
import javax.ws.rs.GET;

@Path("/test")
public class Test {
    @GET
    public String Hello() {
        return "hello";
    }
}

My Pom.xml:

<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
  <modelVersion>4.0.0</modelVersion>
  <groupId>RestDemo</groupId>
  <artifactId>RestDemo</artifactId>
  <version>0.0.1-SNAPSHOT</version>
  <packaging>war</packaging>
  <build>
    <sourceDirectory>src</sourceDirectory>
    <plugins>
      <plugin>
        <artifactId>maven-compiler-plugin</artifactId>
        <version>3.5.1</version>
        <configuration>
          <source>1.8</source>
          <target>1.8</target>
        </configuration>
      </plugin>
      <plugin>
        <artifactId>maven-war-plugin</artifactId>
        <version>3.0.0</version>
        <configuration>
          <warSourceDirectory>WebContent</warSourceDirectory>
        </configuration>
      </plugin>
    </plugins>
  </build>
  <dependencies>
    <dependency>
        <groupId>org.glassfish.jersey.containers</groupId>
        <artifactId>jersey-container-servlet-core</artifactId>
        <version>2.25</version>
    </dependency>
  </dependencies>
</project>

I read that if you're using Jersey 2.+, you wouldn't have to fiddle around with your web.xml , so it's the standard format:

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://xmlns.jcp.org/xml/ns/javaee" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd" version="3.1">
  <display-name>RestDemo</display-name>
  <welcome-file-list>
    <welcome-file>index.html</welcome-file>
    <welcome-file>index.htm</welcome-file>
    <welcome-file>index.jsp</welcome-file>
    <welcome-file>default.html</welcome-file>
    <welcome-file>default.htm</welcome-file>
    <welcome-file>default.jsp</welcome-file>
  </welcome-file-list>
</web-app>

So when I try to call http://localhost:8080/RestDemo/test , I get a 404 error. Any idea what's going on?

Answer 1

In Servlet 2.5 environment, you have to explicitly declare the Jersey container Servlet in your Web application's web.xml deployment descriptor file.

As per https://jersey.java.net/documentation/latest/user-guide.html#deployment.servlet

Example 4.10. Hooking up Jersey as a Servlet

<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>