Question

我正在尝试绕过try / except块。我有带两个表的sqlite db，要在不同的数据框中加载。

存在三种可能的结果，表a或表b不在，或者都不存在。

我不清楚如何写try / except块？

例如

try:
    df = pd.read_sql("select * from table_a", db)
    try:
        df2 = pd.read_sql("select * from table_b", db)
    except DatabaseError:
        print('no table b')
except DatabaseError:
    print('no table a')

我该如何嵌套以说找不到表a或b？

Answer 1

通过取消嵌套：）

public static void main(String[] args) {
        // Moved scanner outside of the try-catch block so that it can be closed in the finally block.
        Scanner input = new Scanner(System.in);
        try {
            long storage = 99999;
            int choice;
            int secondChoice;
            boolean exit = true;
            int i = 0;

            System.out.println("Would you like to add a contact?" + " Type 1 for yes");
            choice = input.nextInt();

            // The TreeMap must go outside of the while loop, otherwise a new map is created each time.
            TreeMap<String, List<String>> lastName = new TreeMap<String, List<String>>();
            if (choice == 1) {

                // Should this be "while (exit == false)" and you initialize exit above to be false?
                while (exit == true) {
                    List<String> array = new ArrayList<String>();
                    System.out.println("Plese add first name");
                    array.add(input.next());
                    System.out.println("Plese add phone number");
                    array.add(input.next());
                    System.out.println("Plese add email address");
                    array.add(input.next());
                    System.out.println("Plese add last name");
                    String last = input.next();
                    lastName.put(last, array);

                    // The format of this output is perhaps not what you want.
                    System.out.println("This is what you just entered: " + lastName);

                    System.out.println("\nWould you like to add another contact?" + " Type 1 for yes - 2 for no");
                    secondChoice = input.nextInt();
                    if (secondChoice == 1) {
                        exit = true;
                        String hold = lastName.toString();
                        System.out.println(hold);
                    }
                    else {

                        // Using "Map.Entry. etc..." isn't what you want here, and also isn't type-specific.
                        for (Entry<String, List<String>> product : lastName.entrySet()) {
                            System.out.println(product.getKey() + "   :   " + product.getValue());
                        }

                        // You should have an exit condition for the loop here. Currently it continues infinitely.
                        // exit = false;
                    }
                }
            }
            else {
                System.out.println("Thank you for using my program");
            }
        }
        catch (Exception e) {
            System.out.println("Error: " + e.getMessage());
        }
        finally {
            input.close();
        }
    }

Answer 2

您现有的代码可能是处理此问题的最佳方法。

但是，如果您要更改它以不区分这两个失败，则只需删除内部的try：

try:
    df = pd.read_sql("select * from table_a", db)
    df2 = pd.read_sql("select * from table_b", db)
except DatabaseError:
    print('either no table a, or no table b')

或者，如果您想同时检测到个故障，只需依次执行以下操作：

df = df2 = None
try:
    df = pd.read_sql("select * from table_a", db)
except DatabaseError:
    pass
try:
    df2 = pd.read_sql("select * from table_b", db)
except DatabaseError:
    pass
if not df and not df2:
    print('neither a nor b exists')

…，然后，如果需要也可以添加elif not df:和elif not df2:。

Python：多个try / except块，如果两个异常都失败，则打印错误？

2 个答案: