JDBC错误：结果集开始之前[重复]答案

【问题标题】：JDBC Error: Before start of result set [duplicate]JDBC错误：结果集开始之前[重复]
【发布时间】：2014-05-05 21:37:23
【问题描述】：

我在 Java eclipse 中有一条错误消息。我在 MySql 中有一个数据库，它有列字符串 user_name、int id_time、int id_desk、int user_password。我想使用一行的列数据

public ArrayList showReservation(String user_name) throws SQLException{

    // array list keeps the information
    ArrayList<String> showReservation = new ArrayList<String>();
    try{

        String getReservationSql = "SELECT * FROM reservation " 
                +"WHERE user_name ='" + user_name + "';";
        rs = stmt.executeQuery(getReservationSql);

        // first element of the array list keeps user name
        showReservation.add("" + rs.getString("user_name")); 

        // second element of the array list keeps id of the desk which is selected by user
        showReservation.add("" + rs.getInt("id_time")); 

        // third element of the array list keeps id of the time interval which is selected by user
        showReservation.add("" + rs.getInt("id_desk"));  

        // forth element of the array list keeps user's password which is generated automatically in a controller class
        showReservation.add("" + rs.getInt("user_password")); 

    }catch (SQLException ex) {
        System.out.println("Error: " + ex.getMessage());
    }
    return showReservation;
}

但是当我运行此代码时出现错误：在结果集开始之前。我该如何解决这个错误？谢谢

【问题讨论】：

查看此帖stackoverflow.com/questions/2120255/…

标签： java mysql jdbc

【解决方案1】：

您永远不会前进到ResultSet 中的第一个结果。使用the next() method 前进到下一条记录，当没有更多记录时返回false。

rs = stmt.executeQuery(getReservationSql);
while (rs.next())
{
   // Make your calls to getString and getInt here
}

【讨论】：

【解决方案2】：

在尝试调用任何ResultSet getter 方法之前使用next

rs = stmt.executeQuery(getReservationSql);
if (rs.next()) {
   ...
}

旁白：考虑Using Prepared Statements

【讨论】：