我有一些二傳手。問題是，我想從用戶準確輸入錯誤輸入的地方啟動程序。例如：如果用戶提出了錯誤的輸入street問題，它將不會name再次從street. 我知道這個選項，但實施起來很糟糕。boolean isBadInput = true;    while (isBadInput) {        try {            System.out.print("name: ");            client.setName(input.next());            isBadInput = false;        } catch (InputMismatchException e) {            System.out.println("bad input, try again");        }    }    isBadInput = true;    while (isBadInput) {        try {            System.out.print("surname: ");            client.setSurname(input.next());            isBadInput = false;        } catch (InputMismatchException e) {            System.out.println("bad input, try again");        }    }    isBadInput = true;    // and so on    System.out.print("city: ");    client.setCity(input.next());    System.out.print("rent date: ");    client.setRentDate(input.next());    System.out.print("street: ");    client.setStreet(input.next());    System.out.print("pesel number: ");    client.setPeselNumber(input.nextLong());    System.out.print("house number: ");    client.setHouseNumber(input.nextInt());如您所見，我需要編寫大量的 try/catch 塊才能做到這一點。還有其他選擇嗎？我不想做這樣的事情：boolean isBadInput = true;    while (isBadInput) {        try {            System.out.print("name: ");            client.setName(input.next());            System.out.print("surname: ");            client.setSurname(input.next());            System.out.print("city: ");            client.setCity(input.next());            System.out.print("rent date: ");            client.setRentDate(input.next());            System.out.print("street: ");            client.setStreet(input.next());            System.out.print("pesel number: ");            client.setPeselNumber(input.nextLong());            System.out.print("house number: ");            client.setHouseNumber(input.nextInt());        } catch (InputMismatchException e) {            System.out.println("bad input, try again");        }    }因為程序name每次都會重復。