java.lang.NullPointerException - ResultSet

Question

我得到以下错误：

java.lang.NullPointerException
    at oracle.jdbc.driver.ScrollableResultSet.cacheRowAt(ScrollableResultSet.java:2086)
    at oracle.jdbc.driver.ScrollableResultSet.isValidRow(ScrollableResultSet.java:2060)
    at oracle.jdbc.driver.ScrollableResultSet.next(ScrollableResultSet.java:347)
    at website.web.InboxReader.getkeywordImportance(InboxReader.java:832)
    at website.web.InboxReader.main(InboxReader.java:54)

Mail表中有53行，keyword表中有1行。在调试时，一旦执行了kstmt.executeUpdate("UPDATE KEYWORD SET IMPORTANCE = IMPORTANCE + 1.0 WHERE SKEYWORD = '" + s2 + "'");行，它就会再次转到keyword set.next()并抛出异常。

这是代码：

Connection connection = connectToDatabase();
        Statement mstmt = connection.createStatement(ResultSet.TYPE_SCROLL_INSENSITIVE, ResultSet.CONCUR_UPDATABLE);
        String s1,s2;
        ResultSet mailset = mstmt.executeQuery("SELECT * from MAIL");
        System.out.println("hello in getImportance beg"); 
        //mailset.beforeFirst();
        while(mailset.next())
        {               
            System.out.println("hello in first while"); 
            Statement kstmt = connection.createStatement(ResultSet.TYPE_SCROLL_INSENSITIVE, ResultSet.CONCUR_UPDATABLE);
            ResultSet keywordset = kstmt.executeQuery("SELECT * FROM KEYWORD");
            while(keywordset.next())
            {
                s1=mailset.getString("SUBJECT");
                System.out.println("Subject: "+s1);
                s2=keywordset.getString("SKEYWORD");
                System.out.println("Keyword: "+s2);
                if(s1.contains(s2))
                {
                    System.out.println("hello in if");  
                    kstmt.executeUpdate("UPDATE KEYWORD SET IMPORTANCE = IMPORTANCE + 1.0 WHERE SKEYWORD = '" + s2 + "'");  
                }
            }
            keywordset.close();
        }
        mailset.close();
        connection.close();

谢谢！

Answer 1

如果你看一下这行似乎有问题：

kstmt.executeUpdate("UPDATE KEYWORD SET IMPORTANCE ='" + (keywordset.getFloat("IMPORTANCE") + 1.0) + "'");

你应该仔细检查通话

keywordset.getFloat("IMPORTANCE")

这很可能是由于记录的Importance列为空，或者该列可能不存在。只是为了调试你尝试

keywordset.getFloat(0);

或者您知道某个列的哪个id存在，以查看调用是否有效。

您是否尝试删除COMMIT语句？因为通常情况下，除非您另行指定，否则数据将自动提交，因此无需在语句后调用提交。如果您已将auto_commit指定为false，则要提交，您应该执行con.commit()，而不是通过更新语句

Answer 2

请参阅此处的文档http://download.oracle.com/javase/tutorial/jdbc/basics/retrieving.html

之前没必要打电话。首先，正如你所期望的那样，下一个会表现得很好。

public static void viewTable(Connection con) throws SQLException {
    Statement stmt = null;
    String query = "select COF_NAME, SUP_ID, PRICE, SALES, TOTAL from " + dbName + ".COFFEES";
    try {
      stmt = con.createStatement();
      ResultSet rs = stmt.executeQuery(query);
      while (rs.next()) {
        String coffeeName = rs.getString("COF_NAME");
        int supplierID = rs.getInt("SUP_ID");
        float price = rs.getFloat("PRICE");
        int sales = rs.getInt("SALES");
        int total = rs.getInt("TOTAL");
        System.out.println(coffeeName + "t" + supplierID + "t" + price + "t" + sales + "t" + total);
      }
    } catch (SQLException e ) {
      JDBCTutorialUtilities.printSQLException(e);
    } finally {
      if (stmt != null) { stmt.close(); }
    }
  }

Answer 3

您是否尝试过在没有关键字wordset.foreFirst((和mailset.foreFirst((的情况下运行它？这不是必需的，它可能会在ResultSet中产生问题。

Answer 4

kstmt.executeUpdate("UPDATE KEYWORD SET IMPORTANCE = IMPORTANCE + 1.0 WHERE SKEYWORD = '" + s2 + "'");  

当这一行在代码中执行时，您的关键字集ResultSet对象将发生更改，因为所有执行方法都会对ResultSet产生影响。Statement接口中的所有执行方法都会隐式关闭语句的当前ResultSet对象(如果存在打开的对象(。您要做的是创建一个单独的语句对象来运行此查询，这样关键字集ResultSet就不会受到影响……：(