How do I catch a MongoSecurityException?

问题

I'm trying to validate the login details of a specific user here.

This just won't work. I have no idea why, it never reaches the catch block even though there is a MongoSecurityException. Does someone know why?

try{
            MongoCredential credential = MongoCredential.createCredential("user", "admin",
                    "password".toCharArray());

            ServerAddress address = new ServerAddress("localhost", 27017);
            mongoClient = new MongoClient(address, Arrays.asList(credential));
            }catch (MongoSecurityException e) {
                e.printStackTrace();
            }

Update:

Stack trace:

May 24, 2017 1:01:08 AM com.mongodb.diagnostics.logging.JULLogger log
INFO: Cluster created with settings {hosts=[localhost:27017], mode=SINGLE, requiredClusterType=UNKNOWN, serverSelectionTimeout='30000 ms', maxWaitQueueSize=500}
May 24, 2017 1:01:08 AM com.mongodb.diagnostics.logging.JULLogger log
INFO: Exception in monitor thread while connecting to server localhost:27017
com.mongodb.MongoSecurityException: Exception authenticating MongoCredential{mechanism=null, userName='test', source='admin', password=<hidden>, mechanismProperties={}}
    at com.mongodb.connection.SaslAuthenticator.wrapInMongoSecurityException(SaslAuthenticator.java:157)
    at com.mongodb.connection.SaslAuthenticator.access$200(SaslAuthenticator.java:37)
    at com.mongodb.connection.SaslAuthenticator$1.run(SaslAuthenticator.java:66)
    at com.mongodb.connection.SaslAuthenticator$1.run(SaslAuthenticator.java:44)
    at com.mongodb.connection.SaslAuthenticator.doAsSubject(SaslAuthenticator.java:162)
    at com.mongodb.connection.SaslAuthenticator.authenticate(SaslAuthenticator.java:44)
    at com.mongodb.connection.DefaultAuthenticator.authenticate(DefaultAuthenticator.java:32)
    at com.mongodb.connection.InternalStreamConnectionInitializer.authenticateAll(InternalStreamConnectionInitializer.java:109)
    at com.mongodb.connection.InternalStreamConnectionInitializer.initialize(InternalStreamConnectionInitializer.java:46)
    at com.mongodb.connection.InternalStreamConnection.open(InternalStreamConnection.java:116)
    at com.mongodb.connection.DefaultServerMonitor$ServerMonitorRunnable.run(DefaultServerMonitor.java:113)
    at java.lang.Thread.run(Thread.java:745)
Caused by: com.mongodb.MongoCommandException: Command failed with error 18: 'Authentication failed.' on server localhost:27017. The full response is { "ok" : 0.0, "errmsg" : "Authentication failed.", "code" : 18, "codeName" : "AuthenticationFailed" }
    at com.mongodb.connection.CommandHelper.createCommandFailureException(CommandHelper.java:170)
    at com.mongodb.connection.CommandHelper.receiveCommandResult(CommandHelper.java:123)
    at com.mongodb.connection.CommandHelper.executeCommand(CommandHelper.java:32)
    at com.mongodb.connection.SaslAuthenticator.sendSaslStart(SaslAuthenticator.java:117)
    at com.mongodb.connection.SaslAuthenticator.access$000(SaslAuthenticator.java:37)
    at com.mongodb.connection.SaslAuthenticator$1.run(SaslAuthenticator.java:50)
    ... 9 more

回答1:

You cannot catch MongoSecurityException as it is thrown in a background thread.

You can wait for a MongoTimeoutException to handle 'synchronously':

  MongoClientOptions clientOptions = new MongoClientOptions.Builder().serverSelectionTimeout(500).build();
    mongoClient = new MongoClient(serverAddress, Collections.singletonList(credential), clientOptions);
    try {
        String address = mongoClient.getConnectPoint();
        System.out.println(address);
    }catch (Throwable e){
        System.out.println(e);
    }

Or you can implement a ServerListener and handle asynchronously

{ 
MongoClientOptions clientOptions = new MongoClientOptions.Builder().addServerListener(this).build();
mongoClient = new MongoClient(host1, Collections.singletonList(credential), clientOptions);
}

@Override
public void serverDescriptionChanged(ServerDescriptionChangedEvent event) {
    Throwable exception = event.getNewDescription().getException();
    handle(exception);
}

来源：https://stackoverflow.com/questions/44145425/how-do-i-catch-a-mongosecurityexception