Question

我正在用python开发一款基于文本的游戏。我目前停留在一个系统上，在该系统上您可以解锁某个功能，一旦解锁该功能，您必须获得一定数量的积分，您才能解锁下一个功能。但是我遇到了麻烦，这是我的代码：

if HS >= 100 and HS <= 199:
        print("You have enough HS points to unlock a new feature!")
        conf = input("Do you want to unlock a new feature? ")
        if conf == "yes":
            features.append("feature1")
            locked_features.remove("feature1")
            HS = HS - 100
            print("You have unlocked feature1!")
            print("Your available feature are: ", features)
            print("You now have", HS, "HS points. ")
    elif HS >= 200 and HS <= 299:
        features.append("feature2")
            locked_features.remove("feature2")
            HS = HS - 100
            print("You have unlocked feature2!")
            print("Your available feature are: ", features)
            print("You now have", HS, "HS points. ")

HS是解锁新功能所需的要点。有一个名为“功能”的列表，其中包含播放器的所有可用功能。锁定功能是玩家解锁之前无法使用的功能。但是，当我解锁功能1并获得100 HS积分时，它会问我是否要解锁新功能。但是，当我键入“是”时，它只会出现错误“ Feature1不在Locked_features中”。如果我已经解锁了功能1，该如何跳过功能1的行？

Answer 1

有几种方法可以做到这一点。一种可能是在if中添加其他条件。如果功能中没有“ feature1”，就这样。

对于第一个IF，您可以：

if HS >= 100 and HS <= 199 and "feature1" not in features:

Answer 2

我不知道这是否正是您想要的逻辑，但这是基本思想...只需在“功能”列表中查看以查看该功能是否已启用：

if HS >= 100 and HS <= 199:
    if not "feature1" in features:
        print("You have enough HS points to unlock a new feature!")
        conf = input("Do you want to unlock a new feature? ")
        if conf == "yes":
            features.append("feature1")
            locked_features.remove("feature1")
            HS = HS - 100
            print("You have unlocked feature1!")
            print("Your available feature are: ", features)
            print("You now have", HS, "HS points. ")
if HS >= 100 and HS <= 199:
    if not "feature2" in features:
        features.append("feature2")
        locked_features.remove("feature2")
        HS = HS - 100
        print("You have unlocked feature2!")
        print("Your available feature are: ", features)
        print("You now have", HS, "HS points. ")

我还更改了逻辑，以便如果您已经拥有第一个功能，则可以获得100 HS的第二个功能，如果您拥有200 HS，则仍可以同时获得两个功能。

Answer 3

首先，在我看来，您在elif语句上出现了缩进错误。关于您的问题，我认为您可以每次仅检查功能是否存在于数组中。

用以下代码替换上面显示的代码的第四行：

if(conf == "yes" and "feature1" in locked_features):

这样，每次用户选择“是”时，该语句仅在您要删除的功能（即“ feature1”）中处于锁定状态时运行。但是我认为程序的总体流程是错误的。您询问用户通常是否要解锁新功能，并且当他/她选择“是”时，您会自动选择功能1。我不确定这就是您想要的。您应该分别检查用户是否要解锁功能，然后询问有关功能编号或其他内容的输入。只是一点提示：）

Answer 4

我认为这就是您要寻找的。现在，if语句不再检查点，而是检查每个功能是否在locked_features中。

    if HS >= 100: # I removed the cap of 199.
        print("You have enough HS points to unlock a new feature!")
        conf = input("Do you want to unlock a new feature? ")
        if conf == "yes":
            if "feature1" in locked_features: # Added this check after conf == yes
                features.append("feature1")
                locked_features.remove("feature1")
                HS = HS - 100
                print("You have unlocked feature1!")
                print("Your available feature are: ", features)
                print("You now have", HS, "HS points. ")
            elif "feature2" in locked_features: # Similarly for feature2, and so on
                features.append("feature2")
                locked_features.remove("feature2")
                HS = HS - 100
                print("You have unlocked feature2!")
                print("Your available feature are: ", features)
                print("You now have", HS, "HS points. ")

如果元素在列表中，如何跳过几行代码？

4 个答案: