resteasy - 使用 undertow 和 resteasy 进行 HTTP 处理程序和 Resteasy 部署

Question

我正在尝试同时运行 HTTPServer 和 REST 处理程序。一次只能工作一个，不能同时工作。我需要提供 html 页面和 api。

这是我的代码。

    public class HttpServer {

    private final UndertowJaxrsServer server = new UndertowJaxrsServer();
    private static String rootPath = System.getProperty("user.dir");

    private final Undertow.Builder serverBuilder;

    public HttpServer(Integer port, String host) {
        serverBuilder = Undertow
                .builder()
                .addHttpListener(port, host)
                .setHandler(
                        Handlers.path().addPrefixPath(
                                "/",
                                Handlers.resource(
                                        new FileResourceManager(new File(
                                                rootPath + "/web"), 100))
                                        .addWelcomeFiles(
                                                rootPath + "/web/index.html")));
        server.start(serverBuilder);
    }

    public DeploymentInfo deployApplication(String appPath,
            Class<? extends Application> applicationClass) {
        ResteasyDeployment deployment = new ResteasyDeployment();
        deployment.setApplicationClass(applicationClass.getName());
        return server.undertowDeployment(deployment, appPath);
    }

    public void deploy(DeploymentInfo deploymentInfo) throws ServletException {
        server.deploy(deploymentInfo);
    }

    public static void main(String[] args) throws ServletException {
        HttpServer myServer = new HttpServer(8080, "0.0.0.0");

        DeploymentInfo di = myServer
                .deployApplication("/rest", MyApplication.class)
                .setClassLoader(HttpServer.class.getClassLoader())
                .setContextPath("/my").setDeploymentName("My Application");
        myServer.deploy(di);
    }
}

Answer 1

UndertowJaxrsServer正在覆盖您的构建器的文件处理程序:

public UndertowJaxrsServer start(Undertow.Builder builder)
{
    server = builder.setHandler(root).build();
    server.start();
    return this;
}

似乎没有办法将另一个处理程序传递给 UndertowJaxrsServer。可能的解决方法可能是:

使用一个资源类部署另一个应用程序，该资源类仅提供文件服务。

使用空白的 Undertow 并摆脱简单的 JAX-RS 部署带来的舒适感。