Question

我正在尝试通过点击按钮

来更新学生数据库

JButton NewUpdate = new JButton("UPDATE");
    NewUpdate.setFont(new Font("Lucida Grande", Font.BOLD, 14));
    NewUpdate.addActionListener(new ActionListener() {
        public void actionPerformed(ActionEvent e) {
            String  regNo   =   RegNoCombo.getSelectedItem().toString();
            String  subjectCode =   SubjectCombo.getSelectedItem().toString(); //gets value selected from subject code JCombobox
            double  score   =   Double.parseDouble(scoreTextField.getText().toString());
            String  testType    =   AssesmentCombo.getSelectedItem().toString();//gets value selected from assessment code JCombobox
            String  queryFromScoresTable    =   "SELECT CONCAT(student_id,subject_id) FROM scores;";

下面的switch语句确定了教师选择的选项。并使用＆＃34; queryFromScoresTable＆＃34;从数据库中检索信息。上述

            switch (testType) {
            case "Class Work":
                try {
                    conn3   =   DriverManager.getConnection("jdbc:mysql://localhost:3306/resultchecker_db","%username%","%password%");
                    ps3 =   conn3.prepareStatement(queryFromScoresTable);
                    rs5 =   ps3.executeQuery();
                } catch (SQLException e1) {
                    JOptionPane.showMessageDialog(null, e1);
                }
                try {
                    rs5.next();
                    String  compareData =   rs5.getString(1);

从DB检索到的查询将转换为带有变量名称的字符串＆＃34; compareData＆＃34;

＆＃34; if＆＃34;下面的陈述应该检查学生是否＆＃34; regNo＆＃34;和＆＃34; testType＆＃34;存在，如果是，则通过更新表中的信息来执行该块

                    if((regNo + subjectCode).equals(compareData)){
                        String  queryUpdateStudRecord   =   "UPDATE scores SET class_work=? WHERE student_id=? and subject_id = ?;";
                        conn4   =   DriverManager.getConnection("jdbc:mysql://localhost:3306/resultchecker_db","%username%","%password%");
                        ps4 =   conn4.prepareStatement(queryUpdateStudRecord);
                        ps4.setDouble(1, score);
                        ps4.setString(2, regNo);
                        ps4.setString(3, subjectCode);
                        ps4.execute();
                        rs5.next();

此外，通过将学生regNo和所选testType的分数插入数据库来执行此块。

                    }else{
                        String  queryInsertRecord   =   "INSERT INTO scores(student_id, subject_id, class_work) VALUES (?,?,?)";
                        conn4   =   DriverManager.getConnection("jdbc:mysql://localhost:3306/resultchecker_db","%username%","%password%");
                        ps4 =   conn4.prepareStatement(queryInsertRecord);
                        ps4.setString(1, regNo);
                        ps4.setString(2, subjectCode);
                        ps4.setDouble(3, score);
                        ps4.execute();
                        rs5.next();
                    }

但问题是只有＆＃34;否则＆＃34;语句执行时不评估＆＃34; if＆＃34;声明，这意味着如果已存在regNo，则无法更新。因为它将返回现有的主键错误。

需要帮助，谢谢你

Answer 1

不知道为什么if没有触发，但你只能使用一个查询：

INSERT ...
ON DUPLICATE KEY UPDATE student_id=VALUES(student_id), subject_id=VALUES(subject_id)

然后您无需关心是应该进行更新还是插入 - 让DB自动为您决定。

Answer 2

小心equals（）方法。请尝试调试并查看您的内容（regNo + subjectCode）和compareData。

如果在比较从MySql DB

2 个答案: