Java MongoDB client as a singleton Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Request for Comments: Singleton pattern implemented in JavaSingleton interface in JavaSimple Java SingletonMulti-player game server accessing and querying MongoDBEnum Singleton implementation in JavaClient to connect to server and show a text menuReturning a database connectionSpring MVC, Mongodb and javaJava Singleton getter/setterSecurity Helper Class According To NIST SP800-63B Guidelines

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Java MongoDB client as a singleton



Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Request for Comments: Singleton pattern implemented in JavaSingleton interface in JavaSimple Java SingletonMulti-player game server accessing and querying MongoDBEnum Singleton implementation in JavaClient to connect to server and show a text menuReturning a database connectionSpring MVC, Mongodb and javaJava Singleton getter/setterSecurity Helper Class According To NIST SP800-63B Guidelines



.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








0












$begingroup$


Is below code a valid Singleton design pattern for Mongodb in Java?



Please suggest any improvements.



public class Mongo extends MongoClient 

private static Mongo mongoClient = null;
private static MongoDatabase mogoDB = null;

private Mongo(ServerAddress server, List<MongoCredential> mongoCredentialList)
super(server, mongoCredentialList);


public static Mongo getMongoClient()
return getMongoClient("userName", "password", "127.0.0.1", 27017, "SampleDB");


public static Mongo getMongoClient(String userName, String pwd, String ip, int port, String dbName)

if (mongoClient == null)

logger.debug("******* Trying to connect to Mongo Server *******");

ServerAddress server = new ServerAddress(ip, port);
List<MongoCredential> mongoCredentialList = new ArrayList();
char[] password = pwd.toCharArray();
mongoCredentialList.add(MongoCredential.createCredential(userName, dbName, password));

mongoClient = new Mongo(server, mongoCredentialList);

logger.debug("******* Connected to the Mongo server Successfully *******");
else
logger.debug("Already connected to Mongo server...");

return mongoClient;


public static MongoDatabase getMongoDatabase()

if (mogoDB == null)
logger.debug("******* Trying to get Mongo DB - SampleDB - connection. *******");
mogoDB = getMongoClient().getDatabase("SampleDB");
logger.debug("******* Successfully Connected to - SampleDB - Mongo DB *******");

return mogoDB;



public static MongoDatabase getMongoDatabase(String dbName)

if (mogoDB == null)
logger.debug("******* Trying to get Mongo DB - " + dbName + " - connection. *******");
mogoDB = getMongoClient().getDatabase(dbName);
logger.debug("******* Successfully Connected to - " + dbName + " - Mongo DB *******");

return mogoDB;












share|improve this question











$endgroup$




bumped to the homepage by Community 7 mins ago


This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.










  • 1




    $begingroup$
    You should avoid the Java Singelton Pattern and pass around the (single) instance manually or use a DI-framwork taking care of the instance being a singelton!
    $endgroup$
    – Timothy Truckle
    Oct 25 '17 at 15:04

















0












$begingroup$


Is below code a valid Singleton design pattern for Mongodb in Java?



Please suggest any improvements.



public class Mongo extends MongoClient 

private static Mongo mongoClient = null;
private static MongoDatabase mogoDB = null;

private Mongo(ServerAddress server, List<MongoCredential> mongoCredentialList)
super(server, mongoCredentialList);


public static Mongo getMongoClient()
return getMongoClient("userName", "password", "127.0.0.1", 27017, "SampleDB");


public static Mongo getMongoClient(String userName, String pwd, String ip, int port, String dbName)

if (mongoClient == null)

logger.debug("******* Trying to connect to Mongo Server *******");

ServerAddress server = new ServerAddress(ip, port);
List<MongoCredential> mongoCredentialList = new ArrayList();
char[] password = pwd.toCharArray();
mongoCredentialList.add(MongoCredential.createCredential(userName, dbName, password));

mongoClient = new Mongo(server, mongoCredentialList);

logger.debug("******* Connected to the Mongo server Successfully *******");
else
logger.debug("Already connected to Mongo server...");

return mongoClient;


public static MongoDatabase getMongoDatabase()

if (mogoDB == null)
logger.debug("******* Trying to get Mongo DB - SampleDB - connection. *******");
mogoDB = getMongoClient().getDatabase("SampleDB");
logger.debug("******* Successfully Connected to - SampleDB - Mongo DB *******");

return mogoDB;



public static MongoDatabase getMongoDatabase(String dbName)

if (mogoDB == null)
logger.debug("******* Trying to get Mongo DB - " + dbName + " - connection. *******");
mogoDB = getMongoClient().getDatabase(dbName);
logger.debug("******* Successfully Connected to - " + dbName + " - Mongo DB *******");

return mogoDB;












share|improve this question











$endgroup$




bumped to the homepage by Community 7 mins ago


This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.










  • 1




    $begingroup$
    You should avoid the Java Singelton Pattern and pass around the (single) instance manually or use a DI-framwork taking care of the instance being a singelton!
    $endgroup$
    – Timothy Truckle
    Oct 25 '17 at 15:04













0












0








0





$begingroup$


Is below code a valid Singleton design pattern for Mongodb in Java?



Please suggest any improvements.



public class Mongo extends MongoClient 

private static Mongo mongoClient = null;
private static MongoDatabase mogoDB = null;

private Mongo(ServerAddress server, List<MongoCredential> mongoCredentialList)
super(server, mongoCredentialList);


public static Mongo getMongoClient()
return getMongoClient("userName", "password", "127.0.0.1", 27017, "SampleDB");


public static Mongo getMongoClient(String userName, String pwd, String ip, int port, String dbName)

if (mongoClient == null)

logger.debug("******* Trying to connect to Mongo Server *******");

ServerAddress server = new ServerAddress(ip, port);
List<MongoCredential> mongoCredentialList = new ArrayList();
char[] password = pwd.toCharArray();
mongoCredentialList.add(MongoCredential.createCredential(userName, dbName, password));

mongoClient = new Mongo(server, mongoCredentialList);

logger.debug("******* Connected to the Mongo server Successfully *******");
else
logger.debug("Already connected to Mongo server...");

return mongoClient;


public static MongoDatabase getMongoDatabase()

if (mogoDB == null)
logger.debug("******* Trying to get Mongo DB - SampleDB - connection. *******");
mogoDB = getMongoClient().getDatabase("SampleDB");
logger.debug("******* Successfully Connected to - SampleDB - Mongo DB *******");

return mogoDB;



public static MongoDatabase getMongoDatabase(String dbName)

if (mogoDB == null)
logger.debug("******* Trying to get Mongo DB - " + dbName + " - connection. *******");
mogoDB = getMongoClient().getDatabase(dbName);
logger.debug("******* Successfully Connected to - " + dbName + " - Mongo DB *******");

return mogoDB;












share|improve this question











$endgroup$




Is below code a valid Singleton design pattern for Mongodb in Java?



Please suggest any improvements.



public class Mongo extends MongoClient 

private static Mongo mongoClient = null;
private static MongoDatabase mogoDB = null;

private Mongo(ServerAddress server, List<MongoCredential> mongoCredentialList)
super(server, mongoCredentialList);


public static Mongo getMongoClient()
return getMongoClient("userName", "password", "127.0.0.1", 27017, "SampleDB");


public static Mongo getMongoClient(String userName, String pwd, String ip, int port, String dbName)

if (mongoClient == null)

logger.debug("******* Trying to connect to Mongo Server *******");

ServerAddress server = new ServerAddress(ip, port);
List<MongoCredential> mongoCredentialList = new ArrayList();
char[] password = pwd.toCharArray();
mongoCredentialList.add(MongoCredential.createCredential(userName, dbName, password));

mongoClient = new Mongo(server, mongoCredentialList);

logger.debug("******* Connected to the Mongo server Successfully *******");
else
logger.debug("Already connected to Mongo server...");

return mongoClient;


public static MongoDatabase getMongoDatabase()

if (mogoDB == null)
logger.debug("******* Trying to get Mongo DB - SampleDB - connection. *******");
mogoDB = getMongoClient().getDatabase("SampleDB");
logger.debug("******* Successfully Connected to - SampleDB - Mongo DB *******");

return mogoDB;



public static MongoDatabase getMongoDatabase(String dbName)

if (mogoDB == null)
logger.debug("******* Trying to get Mongo DB - " + dbName + " - connection. *******");
mogoDB = getMongoClient().getDatabase(dbName);
logger.debug("******* Successfully Connected to - " + dbName + " - Mongo DB *******");

return mogoDB;









java object-oriented singleton mongodb






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 24 '17 at 21:09









200_success

131k17157422




131k17157422










asked Oct 25 '17 at 10:54









Pure'ajaxPure'ajax

12




12





bumped to the homepage by Community 7 mins ago


This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.







bumped to the homepage by Community 7 mins ago


This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.









  • 1




    $begingroup$
    You should avoid the Java Singelton Pattern and pass around the (single) instance manually or use a DI-framwork taking care of the instance being a singelton!
    $endgroup$
    – Timothy Truckle
    Oct 25 '17 at 15:04












  • 1




    $begingroup$
    You should avoid the Java Singelton Pattern and pass around the (single) instance manually or use a DI-framwork taking care of the instance being a singelton!
    $endgroup$
    – Timothy Truckle
    Oct 25 '17 at 15:04







1




1




$begingroup$
You should avoid the Java Singelton Pattern and pass around the (single) instance manually or use a DI-framwork taking care of the instance being a singelton!
$endgroup$
– Timothy Truckle
Oct 25 '17 at 15:04




$begingroup$
You should avoid the Java Singelton Pattern and pass around the (single) instance manually or use a DI-framwork taking care of the instance being a singelton!
$endgroup$
– Timothy Truckle
Oct 25 '17 at 15:04










1 Answer
1






active

oldest

votes


















0












$begingroup$

Simple answer: No. Unless if your application only ever runs on 1 thread.



The easiest way to lazy initialise a singleton thread safe would be with a holder pattern. Example from wikipedia:



public class Something 
private Something()

private static class LazyHolder
static final Something INSTANCE = new Something();


public static Something getInstance()
return LazyHolder.INSTANCE;




This is thread safe because the java class loader ensures that the INSTANCE is initialised exactly once, when the LazyHolder class is used for the first time.



Since the database you initialise isn't known at compile time in this class you cannot use this pattern. What you can do is a pattern called double-checked locking:



public final class Singleton 
private static volatile Singleton instance = null;

private Singleton()

public static Singleton getInstance()
if (instance == null)
synchronized(Singleton.class)
if (instance == null)
instance = new Singleton();



return instance;




Important here are the synchronized(...) and volatile keywords.
The synchronised part makes sure only 1 thread can execute the construction code. And the volatile ensures that all threads always see all changes to the variable.






share|improve this answer









$endgroup$













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    1






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    active

    oldest

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    active

    oldest

    votes









    0












    $begingroup$

    Simple answer: No. Unless if your application only ever runs on 1 thread.



    The easiest way to lazy initialise a singleton thread safe would be with a holder pattern. Example from wikipedia:



    public class Something 
    private Something()

    private static class LazyHolder
    static final Something INSTANCE = new Something();


    public static Something getInstance()
    return LazyHolder.INSTANCE;




    This is thread safe because the java class loader ensures that the INSTANCE is initialised exactly once, when the LazyHolder class is used for the first time.



    Since the database you initialise isn't known at compile time in this class you cannot use this pattern. What you can do is a pattern called double-checked locking:



    public final class Singleton 
    private static volatile Singleton instance = null;

    private Singleton()

    public static Singleton getInstance()
    if (instance == null)
    synchronized(Singleton.class)
    if (instance == null)
    instance = new Singleton();



    return instance;




    Important here are the synchronized(...) and volatile keywords.
    The synchronised part makes sure only 1 thread can execute the construction code. And the volatile ensures that all threads always see all changes to the variable.






    share|improve this answer









    $endgroup$

















      0












      $begingroup$

      Simple answer: No. Unless if your application only ever runs on 1 thread.



      The easiest way to lazy initialise a singleton thread safe would be with a holder pattern. Example from wikipedia:



      public class Something 
      private Something()

      private static class LazyHolder
      static final Something INSTANCE = new Something();


      public static Something getInstance()
      return LazyHolder.INSTANCE;




      This is thread safe because the java class loader ensures that the INSTANCE is initialised exactly once, when the LazyHolder class is used for the first time.



      Since the database you initialise isn't known at compile time in this class you cannot use this pattern. What you can do is a pattern called double-checked locking:



      public final class Singleton 
      private static volatile Singleton instance = null;

      private Singleton()

      public static Singleton getInstance()
      if (instance == null)
      synchronized(Singleton.class)
      if (instance == null)
      instance = new Singleton();



      return instance;




      Important here are the synchronized(...) and volatile keywords.
      The synchronised part makes sure only 1 thread can execute the construction code. And the volatile ensures that all threads always see all changes to the variable.






      share|improve this answer









      $endgroup$















        0












        0








        0





        $begingroup$

        Simple answer: No. Unless if your application only ever runs on 1 thread.



        The easiest way to lazy initialise a singleton thread safe would be with a holder pattern. Example from wikipedia:



        public class Something 
        private Something()

        private static class LazyHolder
        static final Something INSTANCE = new Something();


        public static Something getInstance()
        return LazyHolder.INSTANCE;




        This is thread safe because the java class loader ensures that the INSTANCE is initialised exactly once, when the LazyHolder class is used for the first time.



        Since the database you initialise isn't known at compile time in this class you cannot use this pattern. What you can do is a pattern called double-checked locking:



        public final class Singleton 
        private static volatile Singleton instance = null;

        private Singleton()

        public static Singleton getInstance()
        if (instance == null)
        synchronized(Singleton.class)
        if (instance == null)
        instance = new Singleton();



        return instance;




        Important here are the synchronized(...) and volatile keywords.
        The synchronised part makes sure only 1 thread can execute the construction code. And the volatile ensures that all threads always see all changes to the variable.






        share|improve this answer









        $endgroup$



        Simple answer: No. Unless if your application only ever runs on 1 thread.



        The easiest way to lazy initialise a singleton thread safe would be with a holder pattern. Example from wikipedia:



        public class Something 
        private Something()

        private static class LazyHolder
        static final Something INSTANCE = new Something();


        public static Something getInstance()
        return LazyHolder.INSTANCE;




        This is thread safe because the java class loader ensures that the INSTANCE is initialised exactly once, when the LazyHolder class is used for the first time.



        Since the database you initialise isn't known at compile time in this class you cannot use this pattern. What you can do is a pattern called double-checked locking:



        public final class Singleton 
        private static volatile Singleton instance = null;

        private Singleton()

        public static Singleton getInstance()
        if (instance == null)
        synchronized(Singleton.class)
        if (instance == null)
        instance = new Singleton();



        return instance;




        Important here are the synchronized(...) and volatile keywords.
        The synchronised part makes sure only 1 thread can execute the construction code. And the volatile ensures that all threads always see all changes to the variable.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Oct 25 '17 at 13:02









        ImusImus

        3,468324




        3,468324



























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