需要返回类的Python函数，返回非

Question

所以伙计们，我有这门课。它是静态类的Treasure生成器，返回Spell类、Weapon类、法力值或健康药水值。 问题是，当I return返回值None时，如果我用print代替return，那么in会像在str函数中一样毫无问题地打印它。如何返回类型，使其不显示任何内容？你知道吗

from random import randint
from weapon_class import Weapon
from spell_class import Spell


class Treasure:

    @staticmethod
    def generate_random_index(limit):
        rand_index = randint(0, int(limit))
        return rand_index

    @staticmethod
    def return_generated_treasure(max_mana, max_health):
        # generate num from 0-3
        rand_gen_num = Treasure.generate_random_index(3)
        options = {
            0: Treasure.generate_spell,
            1: Treasure.generate_weapon,
            2: Treasure.generate_mana_pot,
            3: Treasure.generate_health_pot
        }
        # give arguments for mana and health potion functions only (for now)
        if rand_gen_num == 2:
            options[rand_gen_num](max_mana)
        elif rand_gen_num == 3:
            options[rand_gen_num](max_health)
        else:
            # call other functions, which doesn't need arguments,
            # like generate spell and weapon
            options[rand_gen_num]()

    @staticmethod
    def generate_spell():
        with open('spell_names.txt', 'r') as f:
            database_spell_names = f.read().replace('\n', '').split(',')

        lst_len = len(database_spell_names) - 1
        # generate number in range 0 - <spell names length>
        rand_gen_num = Treasure.generate_random_index(lst_len)

        spell_name = database_spell_names[rand_gen_num]
        spell_mana_cost = randint(5, 35)
        spell_damage = randint(5, 40)
        cast_range = randint(1, 3)
        # return spell
        return Spell(spell_name, spell_damage, spell_mana_cost, cast_range)

    @staticmethod
    def generate_weapon():
        with open('weapon_names.txt', 'r') as f:
            database_weapon_names = f.read().replace('\n', '').split(',')
        lst_len = len(database_weapon_names) - 1
        rand_gen_num = Treasure.generate_random_index(lst_len)

        weapon_name = database_weapon_names[rand_gen_num]
        weapon_damage = randint(5, 40)
        # return weapon
        return Weapon(weapon_name, weapon_damage)

    @staticmethod
    def generate_mana_pot(max_mana):
        max_possible_mana_limit = max_mana * 1/2
        mana_portion = randint(0, int(max_possible_mana_limit))
        return mana_portion

    @staticmethod
    def generate_health_pot(max_health):
        max_possible_health_limit = max_health * 1/3
        health_portion = randint(0, int(max_possible_health_limit))
        return health_portion

def main():
    award = Treasure.return_generated_treasure(100, 100)
    print (award)

if __name__ == '__main__':
    main()

Answer 1

return_generated_treasure不返回任何内容-那里没有return关键字。你知道吗

Answer 2

您需要使其他函数的输出成为全局的。当它们在函数中返回时，它们的返回就变成本地的，然后就不会通过类传播或进一步返回。你知道吗