2013-10-11 129 views
2

我正在编写一个程序,它将数据从一个模式中提取出来,重新构造数据以适合新模式,然后将数据插入到具有新模式的新数据库中。问题是,在我的测试代码中,最后一条记录没有被插入到新数据库中。插入跳过记录集中的最后一条记录

我附上了一个大大简化的下面的代码版本,但仍然重现了这个问题。任何人都可以告诉我如何解决以下问题,以便将记录集中的所有记录插入到目标数据库中?目前,下面的不正确地打印出最后一个记录的System.out.println,但还没有那最后记录不存在目标表算账:

static void migrateDataTest(){ 
    try { 
     Class.forName("sun.jdbc.odbc.JdbcOdbcDriver"); 
     Connection sourceConn = DriverManager.getConnection("jdbc:odbc:source_data_test"); 
     Statement st = sourceConn.createStatement(); 
     Connection destinationConn = DriverManager.getConnection("jdbc:odbc:receive_data_test"); 
     int ClientNumber; String ClientsLastName; String ClientsFirstName; 
     ResultSet rest = st.executeQuery("SELECT ClientNumber, ClientsLastName, ClientsFirstName FROM sourceTable"); 
     PreparedStatement ps5 = null; 
     while(rest.next()){ 
      ClientNumber = rest.getInt(1); 
      ClientsLastName = rest.getString(2); 
      ClientsFirstName = rest.getString(3); 
      System.out.println(ClientNumber+", "+ClientsLastName+", "+ClientsFirstName); 
      ps5 = destinationConn.prepareStatement(
       "INSERT INTO destinationTable (" 
       + "ClientNumber, FirstName, LastName) VALUES (?, ?, ?)" 
      ); 
      ps5.setInt(1, ClientNumber); 
      ps5.setString(2, ClientsFirstName); 
      ps5.setString(3, ClientsLastName); 
      ps5.executeUpdate(); 
      destinationConn.commit(); 
     } 
     ps5.close(); 
    } 
catch (ClassNotFoundException cnfe){cnfe.printStackTrace();} 
catch (SQLException e) {e.printStackTrace();} 
} 

编辑:

按LOKESH的要求,我将整个代码块放在下面创建这个错误。我再次运行它以确认它在system.out.println中打印记录30,但是目标表不包含记录号30.使用system.out.println打印出跳过的记录会导致我相信下面的代码包含错误:

static void migrateDataTest(){ 
    try { 
     Class.forName("sun.jdbc.odbc.JdbcOdbcDriver"); 
     Connection sourceConn = DriverManager.getConnection("jdbc:odbc:source_test"); 
     Statement st = sourceConn.createStatement(); 
     Connection destinationConn = DriverManager.getConnection("jdbc:odbc:receive_data_test"); 
     int ClientNumber; 
     String ClientsLastName; 
     String ClientsFirstName; 
     String ClientsMiddleInitial; 
     Date DOB; 
     int GenderNumber; 
     int RaceNumber; 
     ResultSet rest = st.executeQuery("SELECT ClientNumber, ClientsLastName, ClientsFirstName, ClientsMiddleInitial, DOB, GenderNumber, RaceNumber FROM sourceTable"); 
     PreparedStatement ps5 = null; 
     while(rest.next()){ 
      ClientNumber = rest.getInt(1); 
      ClientsLastName = rest.getString(2); 
      ClientsFirstName = rest.getString(3); 
      ClientsMiddleInitial = rest.getString(4); 
      DOB = rest.getDate(5); 
      GenderNumber = rest.getInt(6); 
      RaceNumber = rest.getInt(7); 
      System.out.println(ClientNumber+", "+ClientsLastName+", "+ClientsFirstName+", "+ClientsMiddleInitial+", "+DOB+", "+GenderNumber+", "+RaceNumber); 
      ps5 = destinationConn.prepareStatement(
       "INSERT INTO destinationTable (" 
       + "ClientNumber, FirstName, MiddleInitial, LastName, DOB, GenderNumber, RaceNumber) " 
       +"VALUES (?, ?, ?, ?, ?, ?, ?)" 
      ); 
      ps5.setInt(1, ClientNumber); 
      ps5.setString(2, ClientsFirstName); 
      ps5.setString(3, ClientsMiddleInitial); 
      ps5.setString(4, ClientsLastName); 
      ps5.setDate(5, DOB); 
      ps5.setInt(6, GenderNumber); 
      ps5.setInt(7, RaceNumber); 
      ps5.executeUpdate(); 
      destinationConn.commit(); 
     } 
     ps5.close(); 
    } 
    catch (ClassNotFoundException cnfe){cnfe.printStackTrace();} 
    catch (SQLException e) {e.printStackTrace();} 
} 
+0

代码是OK我想,是System.out.println打印什么东西?任何exceptiosn? –

+0

@ user2310289 System.out.println打印每个记录,包括最后一条记录,这些记录似乎永远不会在目标表中结束。 – CodeMed

+0

1,2,4可以吗?你只插入3件东西..'ps5.setString(4,..)' – nachokk

回答

0

的解决方案,奇怪的是,是创造,那就是未能插入记录其最终值一前一后执行额外准备好的声明。一旦我添加了一个额外的准备好的声明后,第一个开始始终插入其所有值。

看起来像java代码中的一些细微差别,可能在我上面发布的代码示例中缺少代码示例。