Python 函数循环是真还是假

我现在正在学习python中的函数。我遇到一个练习，要求我编写一个脚本，告诉用户该学生是否在课堂上。以下是脚本应显示的结果：

Welcome to the student checker!
Please give me the name of a student (enter 'q' to quit): [student1]
No, that student is not in the class.
Please give me the name of a student (enter 'q' to quit): [student2]
Yes, that student is enrolled in the class!
Please give me the name of a student (enter 'q' to quit): q
Goodbye!

我已经写了一个符合要求的脚本。以下是脚本：

print "Welcome to the student checker!"

students = ['Andreas', 'Martina', 'Maria']

while True:
    name_student = raw_input("Please give me the name of a student (enter 'q' to quit): ")

    if name_student == 'q':
        print "Goodbye!"
        break

    if name_student in students:
        print "Yes, {} is enrolled in the class!".format(name_student)

    elif name_student not in students:
        print "No, {} is not in the class.".format(name_student)

但是这个练习说明，如果学生在场，应该有一个函数返回True，如果学生不在场，则返回False

有人能告诉我如何修改脚本以满足添加一个函数的要求吗？如果学生在场，该函数将返回True，如果学生不在场，则返回False

---

提前谢谢！！和平

试试这样做：

print "Welcome to the student checker!"

students = ['Andreas', 'Martina', 'Maria']

def is_present(name_student):
    if name_student in students:
        print "Yes, {} is enrolled in the class!".format(name_student)
        return True

    elif name_student not in students:
        print "No, {} is not in the class.".format(name_student)
        return False

while True:
    name_student = raw_input("Please give me the name of a student (enter 'q' to quit): ")

    if name_student == 'q':
        print "Goodbye!"
        break

    is_present(name_student)

将逻辑封装到函数中，这样它就可以返回一个值，并在每次无限循环迭代时调用它

@编辑

---

@barak manos是对的，我更新了代码，以下是将学生测试转换为函数的一种方法：

def student_present(student):
    return student in ['Andreas', 'Martina', 'Maria']

print "Welcome to the student checker!"

while True:
    name_student = raw_input("Please give me the name of a student (enter 'q' to quit): ")

    if name_student == 'q':
        print "Goodbye!"
        break

    if student_present(name_student):
        print "Yes, {} is enrolled in the class!".format(name_student)
    else:
        print "No, {} is not in the class.".format(name_student)

注[xxxxxx]中的

student会自动为您提供一个可直接返回的布尔结果。无需显式返回
True
和
False

另一点要考虑的是，如果输入了<玛蒂娜> <代码>，会发生什么？使用当前代码，它将声明
不在类中
。如果您按以下方式更改函数，它也会接受：

def student_present(student):
    return student.lower() in ['andreas', 'martina', 'maria']

我会把
raw_input
行放在函数外部，然后将输入传递给函数。啊哈..现在我知道应该在哪里以及如何使用函数了。谢谢大家！！！你们都是最棒的！！！谢谢马丁！！！你的解释很清楚，可以理解！！！为此干杯！！在我的编程过程中，我需要你和这个社区的专家给我很多建议！！谢谢大家：）