如何找出使用传入Heroku上JAX-RS的RESTful请求的IP?(How to find ou

2019-08-02 05:00发布

我根据一个例子写托管在Heroku一个Java RESTful服务- > https://api.heroku.com/myapps/template-java-jaxrs/clone

我的示例服务是:

package com.example.services;

import com.example.models.Time;

import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;

@Path("/time")
@Produces(MediaType.APPLICATION_JSON)
public class TimeService {

    @GET
    public Time get() {
        return new Time();
    }

}

我主要是:

public class Main {

    public static final String BASE_URI = getBaseURI();

    /**
     * @param args
     */
    public static void main(String[] args) throws Exception{
        final Map<String, String> initParams = new HashMap<String, String>();
        initParams.put("com.sun.jersey.config.property.packages","services.contracts"); 

        System.out.println("Starting grizzly...");
        SelectorThread threadSelector = GrizzlyWebContainerFactory.create(BASE_URI, initParams);
        System.out.println(String.format("Jersey started with WADL available at %sapplication.wadl.",BASE_URI, BASE_URI));
    }

    private static String getBaseURI() 
    {
        return "http://localhost:"+(System.getenv("PORT")!=null?System.getenv("PORT"):"9998")+"/";      
    }

}

我的问题是我怎么能找到我的服务的IP地址和端口组合的请求是哪里来的? 我读过@Context东西中注入javax.ws.rs.core.HttpHeaders,javax.ws.rs.core.Request,等等。但是,没有进入的IP和端口信息是存在的。

我知道,如果你实现com.sun.grizzly.tcp.Adapter,你可以这样做:

public static void main(String[] args) {
    SelectorThread st = new SelectorThread();
    st.setPort(8282);
    st.setAdapter(new EmbeddedServer());
    try {
        st.initEndpoint();
        st.startEndpoint();
    } catch (Exception e) {
        System.out.println("Exception in SelectorThread: " + e);
    } finally {
        if (st.isRunning()) {
            st.stopEndpoint();
        }
    }
}

public void service(Request request, Response response)
        throws Exception {
    String requestURI = request.requestURI().toString();

    System.out.println("New incoming request with URI: " + requestURI);
    System.out.println("Request Method is: " + request.method());

    if (request.method().toString().equalsIgnoreCase("GET")) {
        response.setStatus(HttpURLConnection.HTTP_OK);
        byte[] bytes = "Here is my response text".getBytes();

        ByteChunk chunk = new ByteChunk();
        response.setContentLength(bytes.length);
        response.setContentType("text/plain");
        chunk.append(bytes, 0, bytes.length);
        OutputBuffer buffer = response.getOutputBuffer();
        buffer.doWrite(chunk, response);
        response.finish();
    }
}

public void afterService(Request request, Response response)
        throws Exception {
    request.recycle();
    response.recycle();
}

和访问

    request.remoteAddr()

但我真的很想我的RESTful API以更结构化的方式就像在我的第一个实现分离。

任何帮助将不胜感激。 谢谢!

Answer 1:

卢克表示,采用的Heroku时,远程主机是AWS应用程序层,因此EC2 IP地址。

“X - 转发,对于”头是答案:

String ip = "0.0.0.0";
try {
    ip = req.getHeader("X-Forwarded-For").split(",")[0];
} catch (Exception ignored){}


Answer 2:

你可以注入HttpServletRequest

@GET
@Produces(MediaType.TEXT_PLAIN)
public Response getIp(@Context HttpServletRequest req) {
    String remoteHost = req.getRemoteHost();
    String remoteAddr = req.getRemoteAddr();
    int remotePort = req.getRemotePort();
    String msg = remoteHost + " (" + remoteAddr + ":" + remotePort + ")";
    return Response.ok(msg).build();
}


Answer 3:

Based on @user647772 and @Ethan fusion. Thanks to them ;)

inject HttpServletRequest:

@GET
@Produces(MediaType.TEXT_PLAIN)
public Response getFromp(@Context HttpServletRequest req) {
    String from = _getUserFrom(req);
    return Response.ok(from).build();
}

private String _getUserFrom(HttpServletRequest req) {
    String xForwardedFor = req.getHeader("X-Forwarded-For");
    xForwardedFor = xForwardedFor != null && xForwardedFor.contains(",") ? xForwardedFor.split(",")[0]:xForwardedFor;
    String remoteHost = req.getRemoteHost();
    String remoteAddr = req.getRemoteAddr();
    int remotePort = req.getRemotePort();
    StringBuffer sb = new StringBuffer();
    if (remoteHost != null 
    && !"".equals(remoteHost)
    && !remoteHost.equals(remoteAddr)) {
        sb.append(remoteHost).append(" ");
    }
    if (xForwardedFor != null 
    && !"".equals(xForwardedFor)) {
        sb.append(xForwardedFor).append("(fwd)=>");
    }
    if (remoteAddr != null || !"".equals(remoteAddr)) {
        sb.append(remoteAddr).append(":").append(remotePort);
    }
    return sb.toString();
}


文章来源: How to find out incoming RESTful request's IP using JAX-RS on Heroku?