Question

我试图将值插入表中，但是它总是抛出“ java.sql.SQLException：无效的列索引”。据我所知，我已经验证了列索引。下面是我尝试过的。

stmt = con.prepareStatement("insert into test_table(ASSESMENT_YEAR,ASSESMENT_MON,CODE,KRA_CODE,BASE_VALUE,ACTUAL_VALUE,USERNAME,CHANGE_DATE)"
                + "values (?,?,?,?,?,?,?,SYSDATE)");

        stmt.setInt(1, Calendar.getInstance().get(Calendar.YEAR));
        stmt.setInt(2, Calendar.getInstance().get(Calendar.MONTH));
        stmt.setString(3, getcode(dm.getStr1(), dm.getStr2())); 
        stmt.setString(4, dm.getStr4());
        stmt.setInt(5, dm.getNum1());
        stmt.setInt(6, dm.getNum2());
        stmt.setString(7, "admin");

以上内容在第4个stmt中引发错误。但是，如果我按如下所示对值进行硬编码，则会正确插入数据。

        stmt.setInt(1, 2018);
        stmt.setInt(2, 12);
        stmt.setString(3, "1111"); 
        stmt.setString(4, "0000");
        stmt.setInt(5, 12);
        stmt.setInt(6, 1);
        stmt.setString(7, "admin");

任何帮助都会得到帮助！

编辑：下面是堆栈跟踪

java.sql.SQLException: Invalid column index
    at oracle.jdbc.driver.OraclePreparedStatement.setStringInternal(OraclePreparedStatement.java:5386)
    at oracle.jdbc.driver.OraclePreparedStatement.setString(OraclePreparedStatement.java:5374)
    at oracle.jdbc.driver.OraclePreparedStatementWrapper.setString(OraclePreparedStatementWrapper.java:282)
    at com.dao.KRADataDao.addData(KRADataDao.java:59)
    at com.service.Controller.service(Controller.java:56)
    at com.service.Dispatcher.service(Dispatcher.java:25)
    at javax.servlet.http.HttpServlet.service(HttpServlet.java:741)
    at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:231)
    at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:166)
    at org.apache.tomcat.websocket.server.WsFilter.doFilter(WsFilter.java:53)
    at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:193)
    at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:166)
    at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:199)
    at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:96)
    at org.apache.catalina.authenticator.AuthenticatorBase.invoke(AuthenticatorBase.java:475)
    at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:140)
    at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:80)
    at org.apache.catalina.valves.AbstractAccessLogValve.invoke(AbstractAccessLogValve.java:625)
    at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:87)
    at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:342)
    at org.apache.coyote.http11.Http11Processor.service(Http11Processor.java:498)
    at org.apache.coyote.AbstractProcessorLight.process(AbstractProcessorLight.java:66)
    at org.apache.coyote.AbstractProtocol$ConnectionHandler.process(AbstractProtocol.java:796)
    at org.apache.tomcat.util.net.NioEndpoint$SocketProcessor.doRun(NioEndpoint.java:1372)
    at org.apache.tomcat.util.net.SocketProcessorBase.run(SocketProcessorBase.java:49)
    at java.util.concurrent.ThreadPoolExecutor.runWorker(Unknown Source)
    at java.util.concurrent.ThreadPoolExecutor$Worker.run(Unknown Source)
    at org.apache.tomcat.util.threads.TaskThread$WrappingRunnable.run(TaskThread.java:61)
    at java.lang.Thread.run(Unknown Source)

以下部分也不起作用：

        stmt.setInt(1, Calendar.getInstance().get(Calendar.YEAR));
        stmt.setInt(2, Calendar.getInstance().get(Calendar.MONTH));
        stmt.setString(3, getcode(dm.getStr1(), dm.getStr2())); 
        stmt.setString(4, "0000");
        stmt.setInt(5, dm.getNum1());
        stmt.setInt(6, dm.getNum2());
        stmt.setString(7, "admin");

Answer 1

可能是stmt的重用。对本地实例使用try-with-resources语法，即使发生异常也会自动关闭：

try (PreparedStatement stmt = con.prepareStatement("insert into test_table"
            + "(ASSESMENT_YEAR,ASSESMENT_MON,CODE,KRA_CODE,BASE_VALUE,ACTUAL_VALUE,"
            + "USERNAME,CHANGE_DATE)"
            + "values (?,?,?,?,?,?,?,SYSDATE)")) {
    stmt.setInt(1, Calendar.getInstance().get(Calendar.YEAR));
    stmt.setInt(2, Calendar.getInstance().get(Calendar.MONTH));
    stmt.setString(3, getcode(dm.getStr1(), dm.getStr2())); 
    stmt.setString(4, dm.getStr4());
    stmt.setInt(5, dm.getNum1());
    stmt.setInt(6, dm.getNum2());
    stmt.setString(7, "admin");

    stmt.executeUpdate();
}

我的猜测是，在getcode中，stmt的重用发生了。

通常总是使用局部变量try-with-resources。某些PreparedStatement可能会受益于更长的范围，但使用固定的SQL，则名称不应为stmt。

Answer 2

如果该列不存在，或者您必须输入stmt.executeUpdate（）;。声明。

Answer 3

我不知道您使用的是哪个数据库系统。因此，请确认 CODE 不是保留字/关键字。

在这种情况下，请重命名第三列，然后重试。

java.sql.SQLException：插入数据时无效的列索引

3 个答案: