java 启动嵌入式 Jetty 服务器的最短代码
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Shortest code to start embedded Jetty server
提问by PeterMmm
I'm writing some example code where an embedded Jetty server is started. The server must load exactly one servlet, send all requests to the servlet and listen on localhost:80
我正在编写一些示例代码，其中启动了嵌入式 Jetty 服务器。服务器必须只加载一个 servlet，将所有请求发送到 servlet 并侦听 localhost:80
My code so far:
到目前为止我的代码：
static void startJetty() {
try {
Server server = new Server();
Connector con = new SelectChannelConnector();
con.setPort(80);
server.addConnector(con);
Context context = new Context(server, "/", Context.SESSIONS);
ServletHolder holder = new ServletHolder(new MyApp());
context.addServlet(holder, "/*");
server.start();
} catch (Exception ex) {
System.err.println(ex);
}
}
Can i do the same with less code/lines ? (Jetty 6.1.0 used).
我可以用更少的代码/行做同样的事情吗？（使用 Jetty 6.1.0）。
回答by workmad3
static void startJetty() {
try {
Server server = new Server();
Connector con = new SelectChannelConnector();
con.setPort(80);
server.addConnector(con);
Context context = new Context(server, "/", Context.SESSIONS);
context.addServlet(new ServletHolder(new MyApp()), "/*");
server.start();
} catch (Exception ex) {
System.err.println(ex);
}
}
Removed unnecessary whitespace and moved ServletHolder creation inline. That's removed 5 lines.
删除了不必要的空格并内联移动了 ServletHolder 创建。删除了 5 行。
回答by Jon
You could configure Jetty declaratively in a Spring applicationcontext.xml, e.g:
您可以在 Spring applicationcontext.xml 中以声明方式配置 Jetty，例如：
http://roopindersingh.com/2008/12/10/spring-and-jetty-integration/
http://roopindersingh.com/2008/12/10/spring-and-jetty-integration/
then simply retrieve the server bean from the applicationcontext.xml and call start... I believe that makes it one line of code then... :)
然后简单地从 applicationcontext.xml 中检索服务器 bean 并调用 start...我相信这使它成为一行代码然后...:)
((Server)appContext.getBean("jettyServer")).start();
It's useful for integration tests involving Jetty.
它对于涉及 Jetty 的集成测试很有用。
回答by Janus Troelsen
Works with Jetty 8:
适用于 Jetty 8：
import org.eclipse.jetty.server.Server;
import org.eclipse.jetty.webapp.WebAppContext;
public class Main {
public static void main(String[] args) throws Exception {
Server server = new Server(8080);
WebAppContext handler = new WebAppContext();
handler.setResourceBase("/");
handler.setContextPath("/");
handler.addServlet(new ServletHolder(new MyApp()), "/*");
server.setHandler(handler);
server.start();
}
}
回答by Renato
I've written a library, EasyJetty, that makes it MUCH easier to embed Jetty. It is just a thin layer above the Jetty API, really light-weight.
我编写了一个库EasyJetty，它使嵌入 Jetty 变得更加容易。它只是 Jetty API 之上的一个薄层，真的很轻量级。
Your example would look like this:
您的示例如下所示：
import com.athaydes.easyjetty.EasyJetty;
public class Sample {
public static void main(String[] args) {
new EasyJetty().port(80).servlet("/*", MyApp.class).start();
}
}
回答by sendon1982
Server server = new Server(8080);
Context root = new Context(server, "/");
root.setResourceBase("./pom.xml");
root.setHandler(new ResourceHandler());
server.start();
