Question

我是java的新手，当我尝试从SQL DB中获取数据并在登录时显示数据时，我遇到了这个问题。

sql = "select nom from adherent where id_adherent=3";
try {
    pst = con.prepareStatement(sql);
    ResultSet rs = pst.executeQuery();
} catch (SQLException e) {
    JOptionPane.showMessageDialog(null, "here 1");
}

try {
    if (rs.next()) {
        String sum = rs.getString("select *");
        nom.setText(" " + sum);
    }
} catch (SQLException e) {
    JOptionPane.showMessageDialog(null, "here 2");
}

谢谢。

Answer 1

你不需要两次尝试捕捉，因为你在第一个尝试块中定义了ResultSet，所以你不能在另一次尝试中使用它，所以请改为：

sql = "select nom from adherent where id_adherent=3";
try {
    pst = con.prepareStatement(sql);
    ResultSet rs = pst.executeQuery();
    if (rs.next()) {
        String sum = rs.getString("nom");
        nom.setText(" " + sum);
    }
} catch (SQLException e) {
    JOptionPane.showMessageDialog(null, "here 1");
}

如果你想要完美的东西，任何语法错误或SQL注入，你必须使用PreparedStatement的运算符?，例如：

sql = "select nom from adherent where id_adherent = ?";
//--------------------------------------------------^
try {
    pst = con.prepareStatement(sql);
    pst.setInt(1, id_adherent);//you can set any number in id_adherent in your case 3
    ....

Answer 2

我认为这会对你有所帮助：

sql = "select nom from adherent where id_adherent=3";

try {
    pst = con.prepareStatement(sql);
    ResultSet rs = pst.executeQuery();

    if (rs.next()) {
        String sum = rs.getString("nom");
        nom.setText(" " + sum);
    }  
}
catch (SQLException e) {
    JOptionPane.showMessageDialog(null, "here 1");
}

Java + SQL executeQuery并返回结果

2 个答案: