java - HashMap always returning null? -

for odd reason hashmap returning null value if shouldn't null. row isn't null in mysql database. i'm getting no error did debug test see it's printing out , it's returing null.

auth.java

private string uuid; private string name; private int bits; private string gang; private string rank;    public auth(string uuid, string name, int bits, string gang, string rank) {     this.uuid = uuid;     this.name = name;     this.bits = bits;     this.gang = gang;     this.rank = rank; } 

authmanager.java

public hashmap<string, auth> auth = new hashmap<>();    public void saveuser(string uuid, string name, int bits, string gang, string rank) {      this.auth.put(uuid, new auth(uuid, name, bits, gang, rank));  } 

establishconnection.java

public void establishprofile(player p){      string uuid = p.getuniqueid().tostring();     string name = p.getname();      try     {           resultset query = sql.querysql("select * `profiles` `uuid`= '" + uuid + "';");          preparedstatement create = c.preparestatement("insert `profiles` (`uuid`,`name`, `bits`, `bans`, `gang`, `rank`) values (?, ?, ?, ?, ?, ?) ");         preparedstatement load = c.preparestatement("select * `profiles` `uuid`= ?");          if ( query.next() )         {             load.setstring(1, uuid);             plugin.authmanager.saveuser(query.getstring("uuid"), query.getstring("name"), query.getint("bits"), query.getstring("gang"), query.getstring("rank"));             load.close();             query.close();             p.sendmessage(tables.profile_loaded);             bukkit.getconsolesender().sendmessage(chatcolor.green + "[sql]" +                     " has loaded profile " + name + "(" + uuid + ")"  );           } else          {             create.setstring(1, uuid);             create.setstring(2, name);             create.setint(3, 0);             create.setint(4, 0);             create.setstring(5, null);             create.setstring(6, "default");              create.executeupdate();             create.close();             p.sendmessage(tables.profile_created);             bukkit.getconsolesender().sendmessage(chatcolor.yellow + "[sql]" +                     " executed new query " + name + "(" + uuid + "}" +  plugin.authmanager.auth.get(0) );          }       }   catch (exception e)      {         e.printstacktrace();              bukkit.broadcastmessage("someone's profile has failed load!\n error: " + e);     }  } 

if you're sticking auth objects hashmap, need properly override equals , hashcode in auth class.

also, should name class auth instead of auth follow standard java naming conventions. rename auth.java auth.java, , define follows (requires java 7 or higher):

auth.java

import java.util.objects;  public class auth {     private string uuid;     private string name;     private int bits;     private string gang;     private string rank;      public auth() {     }      public auth(string uuid, string name, int bits, string gang, string rank) {         this.uuid = uuid;         this.name = name;         this.bits = bits;         this.gang = gang;         this.rank = rank;     }      public string getrank() {         return rank;     }      @override     public boolean equals(object o) {         if (this == o) {             return true;         }         if (!(o instanceof auth)) {             return false;         }         auth auth = (auth) o;         return objects.equals(uuid, auth.uuid);     }      @override     public int hashcode() {         return objects.hash(uuid);     }      @override     public string tostring() {         return "auth{" +             "uuid='" + uuid + '\'' +             ", name='" + name + '\'' +             ", bits=" + bits +             ", gang='" + gang + '\'' +             ", rank='" + rank + '\'' +             '}';     } } 

i included uuid , not other fields in equals , hashcode means if 2 auths have same uuid treated being equal. including other fields trip if modify fields after object has been put hashmap. thought easiest approach you, know include in equals , hashcode persistent objects has been subject of debate time.