4

我在我的应用程序中使用带有 jdk 1.6 的 postgresql-8.3-603.jdbc4.jar 来执行数据库操作。我有时会遇到以下异常,重新启动有助于暂时避免这种异常。

org.postgresql.util.PSQLException:在此 ResultSet 中找不到列名 sender_id。在 org.postgresql.jdbc2.AbstractJdbc2ResultSet.findColumn(AbstractJdbc2ResultSet.java:2502) 在 org.postgresql.jdbc2.AbstractJdbc2ResultSet.getString(AbstractJdbc2ResultSet.java:2345) 在 org.apache.commons.dbcp.DelegatingResultSet.getString(DelegatingResultSet.java :225) 在 org.apache.commons.dbcp.DelegatingResultSet.getString(DelegatingResultSet.java:225) 在 com.netcore.bulkrequest.db.FeedDAO.setFeedDetails(FeedDAO.java:142) 在 com.netcore.bulkrequest.feed。 Feed.getInstance(Feed.java:37) 在 com.netcore.bulkrequest.core.BulkRequestTask.(BulkRequestTask.java:86) 在 com.netcore.bulkrequest.core.BulkRequestValidate.getBulkRequestTaskObject(BulkRequestValidate.java:104) 在 com。网核。

这是代码片段:

public class FeedDAO { /** * 数据库连接池对象 */ private final DBContext dbc;

private final Feed feed;

public static final String SENDER_ID_ATTRIBUTE = "sender_id";

/**
 * Constructor
 * 
 * @param dbc
 * @param feed
 */
public FeedDAO(DBContext dbc, Feed feed) {
    this.dbc = dbc;
    this.feed = feed;
}

公共无效 setFeedDetails() 抛出 SQLException {

    String feedDetailsQuery = "SELECT a.priority, b.keyword, b.welcome " +
            "   FROM feed AS a, pub_feed_info AS b " +
            "   WHERE a.resource_id = b.resource_id AND b.resource_id = ?";

    String senderIdQuery = "SELECT b.attribute_value AS " +
            SENDER_ID_ATTRIBUTE + " FROM " +
            "attribute_master AS a, feed_attributes AS b " +
            "WHERE a.attribute_id = b.attribute " +
            "   AND a.attribute_name='" + SENDER_ID_ATTRIBUTE + "' " +
            "   AND feed_id = ?";

    Connection con = null;
    PreparedStatement fdStmt = null;
    PreparedStatement siStmt = null;

    try {
        con = dbc.getConnection();

        //Get the feed details
        fdStmt = dbc.getPreparedStatement(con, feedDetailsQuery);

        fdStmt.setInt(1, this.feed.getFeedId());
        fdStmt.execute();

        ResultSet fdResults = fdStmt.getResultSet();

        while (fdResults.next()) {
            String keyword = fdResults.getString("keyword");
            String welcomeMsg = fdResults.getString("welcome");
            int priority = fdResults.getInt("priority");

            if(null != keyword) {
                this.feed.setKeyword(keyword);
            } else {
                this.feed.setKeyword(String.valueOf(this.feed.getFeedId()));
            }
            this.feed.setWelcomeMsg(welcomeMsg);
            this.feed.setPriority(priority);
        }

        //Get the sender id
        siStmt = dbc.getPreparedStatement(con, senderIdQuery);
        siStmt.setInt(1, this.feed.getFeedId());

        if(siStmt.execute()) {
            ResultSet siResults = siStmt.getResultSet();

            while(siResults.next()) {
                String senderId = siResults.getString(SENDER_ID_ATTRIBUTE);

                this.feed.setSenderId(senderId);
            }

        } else {
            this.feed.setSenderId(Feed.DEFAULT_SENDER_ID);
        }

    } catch (SQLException ex) {
        throw ex;
    } finally {
        if (fdStmt != null) { fdStmt.close(); }
        if (siStmt != null) { siStmt.close(); }
        if (con  != null) { con.close(); }
    }
}

}

谁能帮我找到永久修复?

谢谢,玛尼

4

1 回答 1

1

错误的关键部分是“在此 ResultSet 中找不到列名 sender_id”——第一行。那么,如何向我们展示正在寻找不存在的列的查询,以及在 pgsql 中以交互方式执行该查询的结果、架构的相关部分等?当然,除了异常回溯之外,您不能指望我们帮助您进行调试,而关于您的代码和数据库的线索为零!

于 2009-08-01T05:08:52.507 回答