Question

我正在编写这个简单的程序，并且在数据验证部分，程序必须拒绝用户输入，如果它包含一个数字，使用ASCII，但使用(ch >= 'a' && ch <= 'z') || (ch >= 'A' && ch <= 'Z'))似乎不工作

import java.text.SimpleDateFormat;
import java.util.Calendar;
import java.util.Scanner;

public class Liu_YiJun_A2Q4 {
public static void main(String[] args) {

    String fName = "", lName = "", type = "", add = "", post = "", use = "c", use2 = "";
    int age = 0, phone = 0, numOfU = 0;
    char ch = 0;

    while (use.equalsIgnoreCase("c")) {
        System.out.println(numOfU);
        Scanner input = new Scanner(System.in);

        System.out.print("Please enter your FIRST name:   ");
        fName = input.nextLine();
        while (fName.length() >= 15 && (ch >= 'a' && ch <= 'z') || (ch >= 'A' && ch <= 'Z')){
            System.out.println("The characters in the name are too long OR The value is not a letter");
            System.out.print("Please enter your FIRST name:   ");
            fName = input.nextLine();
        }

        System.out.print("Please enter your LAST name:    ");
        lName = input.nextLine();
        System.out.print("Please enter your AGE:          ");
        age = input.nextInt();
        input.nextLine();
        System.out.print("Please either FT ot PT:         ");
        type = input.nextLine();
        System.out.print("Please enter the ADDRESS:       ");
        add = input.nextLine();
        System.out.print("Please enter the POSTAL CODE:   ");
        post = input.next();
        System.out.print("Please enter your PHONE NUMBER: ");
        phone = input.nextInt();

        System.out.println();

        System.out.println("*********************************************");

        String val = "" + ((int) (Math.random() * 9000) + 1000);
        System.out.println("Membership ID: " + val);
        System.out.println("Name of new member: " + fName + " " + lName);
        System.out.println("Age: " + age);
        System.out.println("User is: " + type);
        System.out.println("Address: " + add + " " + post);
        Calendar cal = Calendar.getInstance();
        SimpleDateFormat sdf = new SimpleDateFormat("dd/MM/yyy");
        System.out.println("Member Since: " + sdf.format(cal.getTime()));

        System.out.println();
        System.out.print("Enter (C) to enter a new user or X to Exit: ");
        use = input.next();

        if (use.equals("x")) {
            System.out.println("Thanks for using ICMSG Membership System. You have entered " + numOfU + " users in this session. Have a nice day!");
        }
    }

}

}

Answer 1

由于学习目的，我不会直接回答，但是你应该如何做到这一点：

      test1 = x

然后你会知道，如果给定的字母是字母。正如@VGR在评论中指出的那样，这不是Regex的任务。

另外，将final String alphabet = "abcdefghijklmnoprstuwxyz"; // define allowed chars char inputChar = 'c'; // get the input from scanner or whatever boolean allowed = alphabet.toUpperCase().contains(Character.toUpperCase(inputChar) + "");变量作为类成员来阻止循环中的常量内存分配。

Answer 2

您检查过字符ch是否为ASCII字母。但是，您必须检查输入中的每个字符都是一个字母。读取输入后。

要做到这一点，你需要遍历字符串fName中的所有字符，依此类推。

for (int i = 0; i < fName.length(); ++i) {
    char ch = fName.charAt(i);
    if (is not a letter) {
        break; // Error, jump out of the for-loop.
    }
}

fName和所有其他输入。

如果仍未处理for循环，请使用while循环。

int i = 0;
while (i < fName.length()) {
    ...
    ++i;
}

ASCII检查输入是否包含字母或数字不起作用？

2 个答案: