从java插入查询到mysql数据库无法正常工作

时间:2015-11-15 07:37:34

标签: java mysql jdbc

我遇到以下方法的问题。该方法生成有效查询,但无法插入表监视器。

方法:

public void upLoadData(String table, String[] dbCols,List<LinkedHashMap<String,String>> values){
        String wq = "INSERT INTO "+table+" (";
        for (int j = 0; j < dbCols.length; j++) {
            wq+=dbCols[j]+",";
        }
        wq = wq.replaceAll(",$", ""); //Remove comma from the end of line.
        wq+=") VALUES";

        for (LinkedHashMap<String, String> val : values) {
            wq+="(";
            for (int i = 0; i < dbCols.length; i++) {
                wq+="'"+val.get(dbCols[i])+"',";
            }
            wq = wq.replaceAll(",$","");
            wq+="),";
        }
        wq = wq.replaceAll(",$", ";");

        System.out.println(wq);
        myDB.DBInsert(wq, MyDBPool.getConnections());
}

插入方法:

public void DBInsert(String query, Connection conn){
    Statement stm;
    try{
        stm = conn.createStatement(); // conn from db connection pool
        stm.executeUpdate(query);
    }catch(SQLException ex){ex.getLocalizedMessage();}
}

输出结果为:

INSERT INTO monitor (id,arr_date,company,disp_size,disp_type,producer,prod_type,color,cond_cat,comments) 
VALUES('H13/2:3445','2015-11-15','Valami','jó','21','Dell','T32','fekete','B','sadsadasdasd'),
('H14:/3:5567','2015-11-15','Nincs','TFT','19','HP','B32','piros','A','sadsadasd'),
('H13/8:3321','2015-11-15','CCCP','CRT','19','nincs','T24','fehér','D','sadsadsad');

Manualy(PhPMyAdmin)插入不是问题。 我使用JDBC和Hikari dbpool,XAMPP v3.2.1

任何帮助都将不胜感激。谢谢!

2 个答案:

答案 0 :(得分:2)

您不是commit,而不是close Statement(或Connection)。您可以使用try-with-resources close并添加commit之类的

public void DBInsert(String query, Connection conn) {
    try (Statement stm = conn.createStatement()){
        stm.executeUpdate(query);
        conn.commit();
    } catch (SQLException ex) {
        ex.getLocalizedMessage();
    } finally {
        try {
            conn.close();
        } catch (SQLException e) {
            e.printStackTrace();
        }
    }
}

此外,我强烈建议您使用PreparedStatement ?占位符(或绑定参数)来性能安全原因而不是将您的插入内容构建为String

答案 1 :(得分:0)

感谢您的帮助和建议,最后我发现了问题,我的旧插入方法工作正常(但是我按照Elliott Frisch建议更新和重构),问题是我只交换了两个不同类型的列,所以PhPMyAdmin是处理错误(将0赋予不匹配值),但Java JDBC无法解析,因此请删除插入任务。

再一次,非常感谢! :)