无法在Eclipse中创建Jersey RESTful服务

Question

Im trying to create a RESTful Service on Eclipse using Jersey.

I repeatedly get the following error when I'm trying to run in localhost within Eclipse.

SEVERE: Servlet [Jersey Web Application] in web application [/TestPrject] threw load() exception
java.lang.ClassNotFoundException: org.glassfish.jersey.servlet.ServletContainer

My web.xml looks something like this :

<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>TestPrject</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>
    <servlet>
        <servlet-name>Jersey Web 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.eviac.blog.MyTestSvc</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>Jersey Web Application</servlet-name>
        <url-pattern>/rest/*</url-pattern>
    </servlet-mapping>
</web-app>

I have all the Jersey 2.0 jars under WEB-INF/lib as well. I'm running Apache Tomcat 8.0 with JDK 1.7.0_79 .

The Class used for Service entry is as follows:

@Path("/")
public class MyTestSvc {

    @GET
    @Path("/callService")
    @Produces({ "text/plain" })
    public String userName() {

        return "hello";
    }
}

Any ideas where I might be going wrong?

Answer 1

确保您具有jersey-container-servlet-core依赖性。

Answer 2

if you are using maven, add the following dependency to your pom

<dependency>
            <groupId>org.glassfish.jersey.containers</groupId>
            <artifactId>jersey-container-servlet-core</artifactId>
            <version>{your version here (2.0)?}</version>
</dependency>

Answer 3

As per service class or controller your url seems to be like this

ProjName/rest//callService

So make in service class

@Path("/service")
public class MyTestSvc {

    @GET
    @Path("/callService")
    @Produces({ "text/plain" })
    public String userName() {

        return "hello";
    }
}

Then url will be ProjName/rest/service/callService

Answer 4

There are a couple of issues in your application:

Fixing the dependencies

If you are using Maven, ensure you have the following dependency:

<dependency>
    <groupId>org.glassfish.jersey.containers</groupId>
    <artifactId>jersey-container-servlet</artifactId>
    <version>2.23.2</version>
</dependency>

If not, ensure you have all the jersey-container-servlet artifact dependencies on the classpath.

Fixing the web.xml

The jersey.config.server.provider.packages parameter defines one or more packages that contain application-specific resources and providers. Change:

<init-param>
    <param-name>jersey.config.server.provider.packages</param-name>
    <param-value>com.eviac.blog.MyTestSvc</param-value>
</init-param>

To:

<init-param>
    <param-name>jersey.config.server.provider.packages</param-name>
    <param-value>com.eviac.blog</param-value>
</init-param>

If you need to add more packages, use ; or , to separate them.

Once Tomcat 8.x supports Servlet 3.x, you don't need a web.xml deployment descriptor for simple applications. For more details, check the Jersey documentation regarding deployment in Servlet 3.x containers .

Improving the resource path design

Avoid @Path annotations with an empty string or just a slash as a value. It smells bad.