java.sql.SQLException:最后一行之后的结果集
java.sql.SQLException:最后一行之后的结果集
提问于 2019-03-12 04:17:54
我想我遇到问题是因为下面提到的RequestDaoImpl类中的rs.next();这一行。每当我尝试检索STATUS的值时，我总是得到以下错误：
java.sql.SQLException: Result set after last row
我在这里做错了什么？
@Component
public class GetStatus {
@JmsListener(destination = "Queue1")
public void processStatusMessage(String message) throws DaoException {
System.out.println("Message Retrieved is:" +message);
try {
RequestDao requestDao = (RequestDao) context.getBean("requestDao");
String receivedStatus = requestDao.getRequestStatus(message);
System.out.println("Testing March 11");
System.out.println(receivedStatus);
}
catch(Throwable th){
th.printStackTrace();
}
}
private static ClassPathXmlApplicationContext context = new ClassPathXmlApplicationContext("ApplicationContext.xml");
}我的RequestDao是：
public interface RequestDao {
public String getRequestStatus(String msg)throws DaoException;
}我的带有方法实现的RequestDaoImpl：
public class RequestDaoImpl implements RequestDao {
public void setDataSource(DataSource dataSource)
{
jdbcTemplate = new JdbcTemplate(dataSource);
}
@Override
public String getRequestStatus(String msg) throws DaoException {
DataSource ds = null;
Connection conn = null;
PreparedStatement pstmt = null;
ResultSet rs = null;
String requestStatus = null;
//List<String> mylist = new ArrayList<String>();
try {
ds = jdbcTemplate.getDataSource();
conn = ds.getConnection();
//I am receiving message like this hence splitting it : 123456#Tan#development
String[] parts = msg.split("#");
String requestID = parts[0].trim();
String userName = parts[1].trim();
String applicationName = parts[2].trim();
/*===========================================================================*/
/* Code to get the request status from Mytable */
/*===========================================================================*/
pstmt = conn.prepareStatement("SELECT STATUS FROM Mytable WHERE request_user= ? and app_name =? and request_id=?");
pstmt.setString(1,userName);
pstmt.setString(2,applicationName);
pstmt.setString(3, requestID);
rs = pstmt.executeQuery();
rs.next();
System.out.println("The status received is as follows:");
requestStatus = rs.getString("STATUS");
System.out.println(requestStatus);
}
catch(Throwable th) {
throw new DaoException(th.getMessage(), th);
}
finally {
if (rs != null) { try { rs.close(); } catch (SQLException e) { e.printStackTrace(); }}
if (pstmt != null) { try { pstmt.close(); } catch(SQLException sqe) { sqe.printStackTrace(); }}
if (conn != null) { try { conn.close(); } catch (SQLException sqle) { sqle.printStackTrace(); }}
}
return requestStatus;
}
private JdbcTemplate jdbcTemplate;
}我在here上看到了类似的错误，但他们的代码和我的不同。
回答 1
Stack Overflow用户
发布于 2019-03-12 04:24:08
您应该询问结果集是否有值，可能需要一个while
while(rs.next){ //your code here }
这样你就不会得到错误，而只是跳过return
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/55109741
复制相关文章
相似问题