我正在使用泽西岛,Maven;并且可以使用 Jetty、Tomcat 或 J2EE Preview(可以嵌入吗?).
I am using Jersey, Maven; and could use Jetty, Tomcat or J2EE Preview (is that embeddable?).
将我的 REST API 移植为独立/可执行 JAR 的最简单方法是什么?不使用 Spring Boot 可以吗?按照以下步骤使用 Jersey 和 Tomcat 创建一个独立的应用程序:
Follow these steps to create a standalone application with Jersey and Tomcat:
将以下依赖项和属性添加到您的 pom.xml
:
Add the following dependencies and properties to your pom.xml
:
<properties>
<tomcat.version>8.5.23</tomcat.version>
<jersey.version>2.26</jersey.version>
<maven.compiler.source>1.8</maven.compiler.source>
<maven.compiler.target>1.8</maven.compiler.target>
<project.build.sourceEncoding>UTF-8</project.build.sourceEncoding>
</properties>
<dependencies>
<dependency>
<groupId>org.apache.tomcat.embed</groupId>
<artifactId>tomcat-embed-core</artifactId>
<version>${tomcat.version}</version>
</dependency>
<dependency>
<groupId>org.glassfish.jersey.containers</groupId>
<artifactId>jersey-container-servlet</artifactId>
<version>${jersey.version}</version>
</dependency>
<dependency>
<groupId>org.glassfish.jersey.inject</groupId>
<artifactId>jersey-hk2</artifactId>
<version>${jersey.version}</version>
</dependency>
</dependencies>
定义您的 JAX-RS 资源类.以下只是一个例子:
Define your JAX-RS resource class(es). The following is just an example:
@Path("hello")
public class HelloWorldResource {
@GET
@Produces(MediaType.TEXT_PLAIN)
public Response helloWorld() {
return Response.ok("Hello World").build();
}
}
创建一个类来配置您的 Jersey 应用程序:
Create a class to configure your Jersey application:
public class JerseyConfiguration extends ResourceConfig {
public JerseyConfiguration() {
packages("com.example");
}
}
创建一个类来启动 Tomcat 并部署您的应用程序:
Create a class to launch Tomcat and deployment your application:
public class Launcher {
private static final String JERSEY_SERVLET_NAME = "jersey-container-servlet";
public static void main(String[] args) throws Exception {
new Launcher().start();
}
void start() throws Exception {
String port = System.getenv("PORT");
if (port == null || port.isEmpty()) {
port = "8080";
}
String contextPath = "";
String appBase = ".";
Tomcat tomcat = new Tomcat();
tomcat.setPort(Integer.valueOf(port));
tomcat.getHost().setAppBase(appBase);
Context context = tomcat.addContext(contextPath, appBase);
Tomcat.addServlet(context, JERSEY_SERVLET_NAME,
new ServletContainer(new JerseyConfiguration()));
context.addServletMappingDecoded("/api/*", JERSEY_SERVLET_NAME);
tomcat.start();
tomcat.getServer().await();
}
}
最后添加 Maven Shade 插件以创建可执行 JAR,其中 mainClass
属性引用启动类:
<build>
<plugins>
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-shade-plugin</artifactId>
<version>2.4.3</version>
<configuration>
<finalName>tomcat-embedded-example-${project.version}</finalName>
</configuration>
<executions>
<execution>
<phase>package</phase>
<goals>
<goal>shade</goal>
</goals>
<configuration>
<transformers>
<transformer implementation="org.apache.maven.plugins.shade.resource.ManifestResourceTransformer">
<mainClass>com.example.Launcher</mainClass>
</transformer>
</transformers>
</configuration>
</execution>
</executions>
</plugin>
</plugins>
</build>
要编译和运行应用程序,请按以下步骤操作:
To compile and run the application, follow these steps:
打开命令行窗口或终端.导航到pom.xml
所在的项目根目录.编译项目:mvn clean compile
.打包应用程序:mvn package
.查看目标目录.您应该会看到一个具有以下或类似名称的文件:tomcat-embedded-example-1.0-SNAPSHOT.jar
.切换到目标目录.执行 JAR:java -jar tomcat-embedded-example-1.0-SNAPSHOT.jar
.应用程序应该在 http://localhost:8080/api/hello
上可用.
Open a command line window or terminal.
Navigate to the root directory of the project, where the pom.xml
resides.
Compile the project: mvn clean compile
.
Package the application: mvn package
.
Look in the target directory. You should see a file with the following or a similar name: tomcat-embedded-example-1.0-SNAPSHOT.jar
.
Change into the target directory.
Execute the JAR: java -jar tomcat-embedded-example-1.0-SNAPSHOT.jar
.
The application should be available at http://localhost:8080/api/hello
.
如何在 Java SE 环境中部署 JAX-RS 应用程序?