JDBC to hive connection fails on invalid operation isValid()

Question

I have followed this doc to try to set up a jdbc connection to hive. But eclipse shows this error. Not seem to figure out what it exactly means and the connection with appropriate password and username works in beeline so its not the problem of authentication.Below is the error i'm facing:

> 15/11/27 13:15:41 INFO jdbc.Utils: Supplied authorities: localhost:10000
> 15/11/27 13:15:41 INFO jdbc.Utils: Resolved authority: localhost:10000
> 15/11/27 13:15:41 INFO jdbc.HiveConnection: Will try to open client transport with JDBC Uri: jdbc:hive2://localhost:10000/default
> Exception in thread "main" java.sql.SQLException: Method not supported
    at org.apache.hive.jdbc.HiveConnection.isValid(HiveConnection.java:1026)
    at HiveJDBC.main(HiveJDBC.java:21)

here is the code:

import java.sql.SQLException;
import java.sql.Connection;
import java.sql.ResultSet;
import java.sql.Statement;
import java.sql.DriverManager;

public class HiveJDBC {
  private static String driverName = "org.apache.hive.jdbc.HiveDriver";

  public static void main(String[] args) throws SQLException {
    try {
      Class.forName(driverName);

    } catch (ClassNotFoundException e) {
      // TODO Auto-generated catch block
      e.printStackTrace();
      System.exit(1);
    }
    Connection con = DriverManager.getConnection("jdbc:hive2://localhost:10000/default", "hive", "PASSWORD");
    if(con.isValid(0)){
        System.out.println("success");
    }else{
        System.out.println("fail");
    }

    Statement stmt = con.createStatement();
    String tableName = "tabledriver";
    //stmt.executeQuery("create database " + tableName);

  }
}

Answer 1

Your call to

if(con.isValid(0)){

is legal - but not implemented by the Hive JDBC Driver.

See Hive Source:

@Override
public boolean isValid(int timeout) throws SQLException {
  // TODO Auto-generated method stub
  throw new SQLException("Method not supported");
}

Replace the check with a simple if(con != null) and you'll be fine.