从Java中的方法返回之前，需要等待异步api回调 [英] Need to wait for asynchronous api callback before I return from method in Java
问题描述
import java.util.concurrent.CountDownLatch;
import quickfix.Initiator;
public class UserSession {
private final CountDownLatch latch = new CountDownLatch(1);
public String await() {
try {
System.out.println("waiting...");
if (latch.await(5, TimeUnit.SECONDS))
System.out.println("released!");
else
System.out.println("timed out");
return secret;
} catch (InterruptedException e) {
// TODO Auto-generated catch block
System.out.println(e.getMessage());
e.printStackTrace();
}
return null;
}
public void countdown(String s) {
System.out.println("In countdown: "+s+ ". Latch count: "+latch.getCount());
secret = s;
latch.countDown();
System.out.println("Latch count: "+latch.getCount());
}
}
public class LogonHandler extends AbstractHandler {
public void handle(String target, Request baseRequest, HttpServletRequest request, HttpServletResponse response)
throws IOException, ServletException
{
Map<String,String[]> query = request.getParameterMap();
if (query.containsKey("method")) {
if (query.get("method")[0].compareTo(method) == 0) {
baseRequest.setHandled(true);
response.getWriter().println(logon(query));
}
}
else
baseRequest.setHandled(false);
}
private String logon(Map<String,String[]> query) {
if (query.containsKey("username") && query.containsKey("password") && query.containsKey("sendercompid")) {
app.mapUser(query.get("sendercompid")[0], new UserSession(query.get("username")[0], query.get("password")[0]));
SessionID session = new SessionID(new BeginString("FIX.4.4"), new SenderCompID(query.get("sendercompid")[0]), new TargetCompID("PARFX"));
try {
ThreadedSocketInitiator tsi = new ThreadedSocketInitiator(app, app.getFileStoreFactory(), settings, app.getLogFactory(), app.getMessageFactory());
UserSession userSession = new UserSession(query.get("username")[0], query.get("password")[0]);
userSession.setInitiator(tsi);
tsi.start();
return userSession.await();
} catch (ConfigError e) {
// TODO Auto-generated catch block
e.printStackTrace();
return e.toString();
}
}
return "fail";
}
}
public class QuickfixjApplication implements Application {
private Map<String,UserSession> users = new HashMap<String,UserSession>();
public void mapUser(String s, UserSession u) {
users.put(s, u);
}
public void toAdmin(Message message, SessionID sessionId) {
try {
if (message.getHeader().getField(new StringField(MsgType.FIELD)).valueEquals(Logon.MSGTYPE)) {
UserSession user = users.get(sessionId.getSenderCompID());
message.setField(new Username(user.getUsername()));
message.setField(new Password(user.getPassword()));
}
} catch (FieldNotFound e) {
e.printStackTrace();
}
}
public void fromAdmin(Message message, SessionID sessionId)
throws FieldNotFound, IncorrectDataFormat, IncorrectTagValue, RejectLogon {
if (message.getHeader().getField(new StringField(MsgType.FIELD)).valueEquals(Logon.MSGTYPE)) {
System.out.println(message.toString());
UserSession user = users.get(sessionId.getSenderCompID());
user.countdown(message.toString());
}
}
}
好的，我尝试只在此处包括最少的代码.有三个有趣的类，UserSession是Jetty处理程序和QuickFix/j应用程序之间的内部粘合剂.
Ok, I've tried to only include the minimum amount of code here. There are three interesting classes, UserSession is the internal glue between the Jetty handler and the QuickFix/j application.
LogonHandler接收HTTP登录请求，并尝试将用户登录到QuickFix/j应用程序会话.
The LogonHandler receives an HTTP logon request and tries to log a user onto a QuickFix/j application session.
QuickFix/j正在向FIX服务器发送登录消息，该登录请求/响应是异步的. HTTP登录请求当然是同步的.因此，我们必须等待FIX服务器的回复，然后才能从HTTP请求返回.我使用CountDownLatch和此UserSession对象进行此操作.
QuickFix/j is sending a logon message to a FIX server, this logon request / response is asynchronous. The HTTP logon request is of course synchronous. So we have to wait for the reply from the FIX server before we return from the HTTP request. I do this using CountDownLatch and this UserSession object.
当我创建QuickFix/j会话对象时，我还创建了一个UserSession对象，并将其添加到地图中(这发生在LogonHandler登录方法中).
When I create the QuickFix/j session object I also create a UserSession object and add it to a map (that happens in the LogonHandler logon method).
QuickFix/j应用程序对象中有两个回调，分别为toAdmin()和fromAdmin().在fromAdmin()中，我检查消息是否为登录响应，如果是，则调用UserSession的方法对闩锁进行倒计时.在调试代码时，我看到点击了fromAdmin()方法，在映射中找到了UserSession对象，并调用了countdown()方法，latch.getCount()从1变为0，但是latch.await( )UserSession await()中的方法永远不会返回.它总是超时.
There are two callbacks in the QuickFix/j application object, toAdmin() and fromAdmin(). In fromAdmin() I check if the message is a logon response and if it is I call a method of UserSession to countdown the latch. In debugging the code I see that the fromAdmin() method is hit, the UserSession object is found in the map and the countdown() method is called and the latch.getCount() goes from 1 to 0, but the latch.await() method in UserSession await() never returns. It always times out.
推荐答案
您可以像这样使用CountDownLatch:
public class LogonHandler implements Handler {
private final CountDownLatch loginLatch = new CountDownLatch (1);
private boolean callbackResults;
public void serverResponseCallback(boolean result) {
callbackResults = result;
loginLatch.countDown ();
}
public boolean tryLogon(Credentials creds) throws InterruptedException {
SomeServer server = new SomeServer(address);
server.tryLogon (creds.getName (), creds.getPass ());
loginLatch.await ();
return callbackResults;
}
}
如果要将等待时间限制为例如5秒，请使用以下命令代替loginLatch.await ():
If you want to limit waiting time by, for example, 5 seconds, then instead of loginLatch.await () use the following:
if (loginLatch.await (5L, TimeUnit.SECONDS))
return callbackResults;
else
return false; // Timeout exceeded
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