java - 仅当没有抛出异常时如何执行代码块

Question

我有一部分代码，如果捕获到异常，我想在 JOptionPane 中显示包含异常详细信息的消息。该异常错误正在正确显示。

我还想显示“成功”消息，前提是预期的过程成功并且没有捕获异常。但即使捕获到异常，也会显示此成功消息。

观察以下代码中 catch block 内外的 JOptionPane 语句。

private void submitBtnActionPerformed(java.awt.event.ActionEvent evt) {                                                  
    saveToDataBase();        
}                                         

public void saveToDataBase(){
    try{
    String url = "jdbc:mysql://localhost:3306/employee?autoReconnect=true&useSSL=false";
    String uname = "root";
    String pass = "password";
    String query = "insert into emp values (";
    Connection con = DriverManager.getConnection(url,uname,pass);
    Statement st = con.createStatement();

    query += idTextField.getText() + "," + fNameTextField.getText() + "," + lNameTextField.getText() + ")";
     st.executeUpdate(query);
    } 

   catch (SQLException ex) {
        Logger.getLogger(AddFrame.class.getName()).log(Level.SEVERE, null, ex);
        JOptionPane.showMessageDialog(null, ex, "Error", JOptionPane.ERROR_MESSAGE);
    }

    JOptionPane.showMessageDialog(null, "The data has been saved successfully!!", "Success", JOptionPane.INFORMATION_MESSAGE);

}

Answer 1

将此行移至 try-catch block 中

JOptionPane.showMessageDialog(null, "The data has been saved successfully!!", "Success", JOptionPane.INFORMATION_MESSAGE);

由于此行已从您的 try-catch 中排除 - 在处理异常后，同一行将再次执行。