Python 如何将变量从一个函数调用到另一个函数？

注意：这是一个很长的问题。

我正在制作一个游戏来练习python。我需要从另一个函数（我称之为

spawn（）

）中获取一个变量，并在另一个函数（我称之为

damage\u take（）

）中使用

这是文件

敌方.py

。它的主要任务是产生一个敌人：

import random
import player

class Enemy(object):

    def types(self):
        type = ["slime", "ghost", "demon"]
        enemy = random.choice(type)
        return enemy


class Slime(Enemy):

    def types(self):
        colour = ["red", "green", "blue"]
        type = random.choice(colour)
        return type

    def health(self):
        health = random.randint(1,5)
        return health


class Ghost(Enemy):

    def types(self):
        form = ["spirit", "spectre", "phantom"]
        type = random.choice(form)
        return type

    def health(self):
        health = random.randint(10,30)
        return health


class Demon(Enemy):

    def types(self):
        being = ["demon", "hell hound", "wendigo"]
        type = random.choice(being)
        return type

    def health(self):
        health = random.randint(15,35)
        return health

这是重要的代码。我需要从这个函数中获取变量

health

，并在另一个函数中使用它

def spawn():
    enemy = Enemy()
    bad = enemy.types()

    if bad == "slime":
        slime = Slime()
        target = slime.types()
        health = slime.health()
        print(f"A {target} {bad} has appeared. It has {health} HP")
        return health
    elif bad == "ghost":
        ghost = Ghost()
        target = ghost.types()
        health = ghost.health()
        print(f"A {target} has appeared. It has {health} HP")
        return health
    elif bad == "demon":
        demon = Demon()
        target = demon.types()
        health = demon.health()
        print(f"A {target} has appeared. It has {health} HP")
        return health

def damage_taken():
    spawn()
    health = spawn.health - player.fight()
    return health

damage_taken()

这就是我正在努力的地方。我试图从函数

spawn（）

中获取变量

health

，并在下面的函数中使用它。然而，它不断地告诉我健康是不存在的。如何从另一个函数中获取变量并在该函数中使用它

def spawn():
    enemy = Enemy()
    bad = enemy.types()

    if bad == "slime":
        slime = Slime()
        target = slime.types()
        health = slime.health()
        print(f"A {target} {bad} has appeared. It has {health} HP")
        return health
    elif bad == "ghost":
        ghost = Ghost()
        target = ghost.types()
        health = ghost.health()
        print(f"A {target} has appeared. It has {health} HP")
        return health
    elif bad == "demon":
        demon = Demon()
        target = demon.types()
        health = demon.health()
        print(f"A {target} has appeared. It has {health} HP")
        return health

def damage_taken():
    spawn()
    health = spawn.health - player.fight()
    return health

damage_taken()

代码

spawn.health

尝试将变量调用到函数中失败

def spawn():
    enemy = Enemy()
    bad = enemy.types()

    if bad == "slime":
        slime = Slime()
        target = slime.types()
        health = slime.health()
        print(f"A {target} {bad} has appeared. It has {health} HP")
        return health
    elif bad == "ghost":
        ghost = Ghost()
        target = ghost.types()
        health = ghost.health()
        print(f"A {target} has appeared. It has {health} HP")
        return health
    elif bad == "demon":
        demon = Demon()
        target = demon.types()
        health = demon.health()
        print(f"A {target} has appeared. It has {health} HP")
        return health

def damage_taken():
    spawn()
    health = spawn.health - player.fight()
    return health

damage_taken()

守则：

player.fight()

---

正在从另一个名为

player.py

的文件调用函数。它的主要目的是处理与玩家相关的机制，例如创建角色和决定他们造成的伤害。

如果我完全理解你的问题，你只需要将spawn函数的返回值指定到一个变量中：

def damage_taken():
    spawn_health = spawn()
    health = spawn_health - player.fight()
    return health

---

我猜您实际上更希望使用

生成

函数返回敌人的对象，而不是敌人的

健康

，因此我建议如下：

def spawn():
    enemy = Enemy()
    bad = enemy.types()

    if bad == "slime":
        enemy = Slime()  # each of these now is assigned to the same variable name, "enemy"
        target = slime.types()
        health = slime.health()
        print(f"A {target} {bad} has appeared. It has {health} HP")
    elif bad == "ghost":
        enemy = Ghost()
        target = ghost.types()
        health = ghost.health()
        print(f"A {target} has appeared. It has {health} HP")
    elif bad == "demon":
        enemy = Demon()
        target = demon.types()
        health = demon.health()
        print(f"A {target} has appeared. It has {health} HP")

    return enemy

def damage_taken():
    enemy = spawn()
    health = enemy.health - player.fight()
    return health
damage_taken()

现在，仍然存在一个问题，即在调用

damage\u take（）

后，敌人的物体将完全掉落。如果你想让敌人继续存在，你可能需要在

受到的伤害

功能范围之外初始化玩家和敌人。更像这样：

enemy = spawn()
damage_taken(enemy)

---

你真正的问题是什么还不清楚。将其简化为。当

spawn

是一个函数时，

spawn.health

意味着什么？在文件敌方.py中，将其放入damage_take函数？@Hameda169非常清楚他的意思是能够访问spawn（）下的生命值，只需调用spawn（）并将其返回值存储在variable下