jersey - 使用 web.xml、Servlet 3.0 和 Jersey 部署 JAX-RS 应用程序

Question

我正在尝试使用 web.xml、servlet 3.0 和 jersey API 部署我的应用程序。不幸的是，它不起作用。

这是 MyApplication.class ：

package com.example;

public class MyApplication extends Application {
   public Set<Class<?>> getClasses() {
       Set<Class<?>> s = new HashSet<Class<?>>();
       s.add(MyResource.class);
       return s;
   }
}

这是 MyResource ：

@Path("/helloworld")
@Produces(MediaType.TEXT_PLAIN)
public class MyResource {
    @GET
    public String getHello() {
        return "HelloWorld !";
    }
}

我的 web.xml ：

 <web-app>
     <servlet>
         <servlet-name>com.example.MyApplication</servlet-name>
     </servlet>
     <servlet-mapping>
         <servlet-name>com.example.MyApplication</servlet-name>
         <url-pattern>/webapi/*</url-pattern>
     </servlet-mapping>
 </web-app>

在客户端，我使用这个网址：http://localhost:8080/[projectname]/webapi/helloworld

我有这个错误：

java.lang.NullPointerException
    sun.misc.Launcher$AppClassLoader.loadClass(Launcher.java:294)
    java.lang.ClassLoader.loadClass(ClassLoader.java:247)
    org.apache.catalina.loader.WebappClassLoader.loadClass(WebappClassLoader.java:1629)
    org.apache.catalina.loader.WebappClassLoader.loadClass(WebappClassLoader.java:1559)
    org.apache.catalina.authenticator.AuthenticatorBase.invoke(AuthenticatorBase.java:491)
    org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:99)
    org.apache.catalina.valves.AccessLogValve.invoke(AccessLogValve.java:953)
    org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:408)
    org.apache.coyote.http11.AbstractHttp11Processor.process(AbstractHttp11Processor.java:1023)
    org.apache.coyote.AbstractProtocol$AbstractConnectionHandler.process(AbstractProtocol.java:589)
    org.apache.tomcat.util.net.AprEndpoint$SocketProcessor.run(AprEndpoint.java:1852)
    java.util.concurrent.ThreadPoolExecutor$Worker.runTask(ThreadPoolExecutor.java:895)
    java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:918)
    java.lang.Thread.run(Thread.java:662)

怎么了？：/我正在使用Tomcat 7。

PS：使用 servlet 2.x，它可以工作：

    <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.example</param-value>
        </init-param>
    </servlet>
    <servlet-mapping>
        <servlet-name>Jersey Web Application</servlet-name>
        <url-pattern>/webapi/*</url-pattern>
    </servlet-mapping>

但我稍后需要异步模式。

谢谢！

score 8 · Accepted Answer

Update: Since writing this answer, I've found out a way to avoid needing a web.xml on Tomcat using the official Glassfish Jersey implementation. Look here for details.

If you're using a standard Tomcat install (or some other servlet container), AFAIK you can't avoid explicitly telling it what servlets to start in the web.xml file*. Since you have to use web.xml anyway, the simplest way to get restful web services working is to forget extending javax.ws.rs.core.Application entirely and just specify the context path there. You can still use standard jax-rs annotations to declare the actual web services.

web.xml:

<?xml version="1.0" encoding="UTF-8"?>
<web-app
  xmlns="http://java.sun.com/xml/ns/javaee"
  xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
  xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
  version="3.0"
>
  <servlet>
    <servlet-name>rest-test</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.domain.mypackage</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
  </servlet>
  <servlet-mapping>
    <servlet-name> rest-test</servlet-name>
    <url-pattern>/rest/*</url-pattern>
  </servlet-mapping>
</web-app>

Two noteworthy points:

You will need to bundle a REST implementation in your WAR file, since servlet containers don't usually contain one. Since Jersey is the reference implementation for JAX-RS, that's the one I'm using in the servlet-class element above. You can replace this with Apache CXF implementation if you want.
The init-param element tells Jersey which of your packages to search for Java files with web service annotations. Edit this to point to your web services. Note that if you opt to use apache CXF instead of Jersey, the stuff needed in any init-param elements will be different. Someone who knows CXF please post what they would be.

If you're using Maven, just add a dependency to jersey-servlet in the dependencies section of your pom.xml file:

<dependencies>
  <dependency>
    <groupId>com.sun.jersey</groupId>
    <artifactId>jersey-servlet</artifactId>
    <version>1.18.2</version>
  </dependency>
  ...
</dependencies>

After this, declaring your web services is straight forward using the standard JAX-RS annotations in your Java classes:

package com.domain.mypackage;
import javax.ws.rs.Consumes;
import javax.ws.rs.Produces;
import javax.ws.rs.GET;
import javax.ws.rs.MatrixParam;
import javax.ws.rs.Path;

// It's good practice to include a version number in the path so you can have
// multiple versions deployed at once. That way consumers don't need to upgrade
// right away if things are working for them.
@Path("calc/1.0")
public class CalculatorV1_0 {
  @GET
  @Consumes("text/plain")
  @Produces("text/plain")
  @Path("addTwoNumbers")
  public String add(@MatrixParam("firstNumber") int n1, @MatrixParam("secondNumber") int n2) {
    return String.valueOf(n1 + n2);
  }
}

This should be all you need. If your Tomcat install is running locally on port 8080 and you deploy your WAR file to the context myContext, going to

http://localhost:8080/myContext/rest/calc/1.0/addTwoNumbers;firstNumber=2;secondNumber=3

...should produce the expected result (5).

Cheers!

* Someone please correct me if you know of a way to a add the Jersey servlet to the context in Tomcat without using web.xml--maybe by using a context or life cycle listener?

score 0 · Accepted Answer

您需要使用以下代码行在 WEB-INF 文件夹下创建 web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="3.0" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd">
    <servlet>
        <servlet-name>ServletAdaptor</servlet-name>
        <servlet-class>
            org.glassfish.jersey.servlet.ServletContainer
        </servlet-class>
        <init-param>
            <param-name>jersey.config.server.provider.packages</param-name>
            <param-value>package where MyResource  resides</param-value>
        </init-param>
        <init-param>
            <param-name>com.sun.jersey.api.json.POJOMappingFeature</param-name>
            <param-value>true</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>ServletAdaptor</servlet-name>
        <url-pattern>/webresources/*</url-pattern>
    </servlet-mapping>
    <session-config>
        <session-timeout>
            30
        </session-timeout>
    </session-config>
</web-app>

现在您可以删除您的 MyApplication 类。部署和测试 Web 服务。它会起作用的

score 0 · Accepted Answer

package com.example;

@ApplicationPath("/webapi")

public class MyApplication extends Application {

     public Set<Class<?>> getClasses() {
       Set<Class<?>> s = new HashSet<Class<?>>();
       s.add(MyResource.class);
       return s;
   }
}

score -1 · Accepted Answer

访问此链接可能会有所帮助： https ://jersey.java.net/documentation/latest/deployment.html#deployment.servlet.2

部分：4.7.1。Servlet 2.x 容器

<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>

jersey - 使用 web.xml、Servlet 3.0 和 Jersey 部署 JAX-RS 应用程序

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