我正在编写一个基于示例的Heroku托管的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();
}
}
我的主要是:
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")+"/";
}
}
我的问题是如何在我的服务中找到请求来自的IP地址和端口组合?我已经阅读了@Context上的东西,它注入了javax.ws.rs.core.HttpHeaders,javax.ws.rs.core.Request等。但是,没有传入的IP或端口信息。
我知道如果你实现了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();
}
并访问
request.remoteAddr()
但我真的想以更加结构化的方式分离我的RESTful API,就像我的第一次实现一样。
非常感谢任何帮助。谢谢!
答案 0 :(得分:51)
您可以注入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();
}
答案 1 :(得分:11)
正如Luke所说,使用Heroku时,远程主机是AWS应用层,因此是EC2 IP地址。
“X-Forwarded-For”标题就是答案:
String ip = "0.0.0.0";
try {
ip = req.getHeader("X-Forwarded-For").split(",")[0];
} catch (Exception ignored){}
答案 2 :(得分:5)
基于@ user647772和@Ethan fusion。感谢他们;)
注入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();
}