Question

对于您是否输入.hasNextInt()之类的数字而言，Java很容易，但是String呢？

do {
    System.out.println("Please enter the students name: ");
    String name = s.nextLine();

    if (name.equalsIgnoreCase("Done")) {
        finished = true;
    } else {
        System.out.println("Please input the students number");
        int studentNo = s.nextInt();
        System.out.println("Please enter that students subject:  ");
        String subject = s.next();
        System.out.println("Pleae enter that students level- ");
        int level = s.nextInt();

        StudentRecords newStudent = new StudentRecords(name, studentNo, subject, level);
        studentRecordList.add(newStudent);
    }
} while (!finished);

我如何让它循环以确保用户为学生姓名输入String，而不是int等。

谢谢。

Answer 1

使用Regex尝试以下循环：

String name="";
while(!name.matches("[a-zA-Z \'\-ÄäÖöÜüßÉéæø]+")){
    System.out.println("Please enter the student name: ");
    name = s.nextLine();
}

这将询问用户输入，直到给出有效名称。

Answer 2

关于我的评论，这样的事情将会起作用：

if(!name.matches(".*\\d.*")){
    //It is correct name, because it does not contain any integer.
} else{
    //retry getting name from user because it contains integer.
}

编辑评论问题：上面的代码会检测输入是否包含任何整数，但是如果你想检查它是否包含字符，你可以使用它：

!name.matches("[a-zA-Z ]+")

Answer 3

你可以使用只允许字母和空格的regular expression：

    boolean nameIsValid = false;
    while (!nameIsValid)//while name is not valid
    {
    System.out.println("Please enter the students name: ");
    String name = s.nextLine();
    if (name.matches("[A-Za-z ]*"))//match only letters/spaces
        nameIsValid = true;//we have a valid name, we may now break out of the loop
    else
        System.out.println("Please enter a valid name!");//invalid name, loop again
    }

如何判断用户是否输入了字符串？

3 个答案: