Question

错误：在函数外部'返回'

为什么？

 def get_all_students(self):
    database = open(self.database_path, 'r')
    list_of_students = list()
    for idx, l in enumerate(database.readlines()):
        params = list()
        params.append(idx)
        params += l.split(self.data_delimiter)
        student = Student(*self.item_un_escape(params))
        list_of_students.append(student)
    return list_of_students

编辑：我会发布课程的其余部分，我很确定它是正确缩进的。似乎无法找到错误，我敢打赌它可能是愚蠢的东西。道歉，第一次在python编码，但这个错误让我烦恼了一个小时左右！哈哈

Edit2：在params.append（idx）上发出意外缩进的错误？

from student import Student

database_path = 'C:/Users/Alan/Desktop/flask/flask/app/database'


class Database(object):
    data_delimiter = ','

    @staticmethod
    def escape_new_lines(value):
        if type(value) == str:
            value = value.replace('\n', '@n-nl@')
            value = value.replace('\r', '@r-nl@')
        return value

    @staticmethod
    def un_escape_new_lines(value):
        if type(value) == str:
            value = value.replace('@n-nl@', '\n')
            value = value.replace('@r-nl@', '\r')
        return value

    def __init__(self, database_path=database_path):
        self.database_path = database_path

    def get_all_students(self):
        database = open(self.database_path, 'r')
        list_of_students = list()
        for idx, l in enumerate(database.readlines()):
            params = list()
            params.append(idx)
            params += l.split(self.data_delimiter)
            student = Student(*self.item_un_escape(params))
            list_of_students.append(student)
        return list_of_students

Answer 1

您发布的代码很好。您可能的错误是在返回语句之前缺少缩进。 编辑：正如@Anonymous在评论中指出的那样，另一种可能性是你将制表符缩进与空格缩进混合。检查你的缩进。

错误：在函数外部'返回'

1 个答案: