Question

当我尝试运行此操作时，我收到语法错误，并且它没有说明它是什么行或什么。我不知道还能说些什么..以下是获得错误的代码：

if "q" in attack:
        if random.randint(1,100) != range(1,21):
            print("You hit with a quick attack!")
            ehp -= 20
            print("The",enam,"loses 20 damage! It now has",ehp,"health.")
        else:
            print("You missed.. :(")
    elif "p" in attack:
        if random.randint(1,100) != range(1,51):
            print("You hit with a power attack!")
            ehp -= 50
            print("The",enam,"loses 50 damage! It now has",ehp,"health.")
        else:
            print("You missed.. :(")
    elif "1" in attack:
        if mana >= skill1[2]:
            print("You hit with",skill1[0])
            ehp -= skill1[1]
            mana -= skill1[2]
            print("The",enam,"loses",skill1[1],"damage! It now has",ehp,"health.")
            print("You now have",mana,"mana.")
    elif "2" in attack:
        if mana >= skill2[2]:
            print("You hit with",skill2[0])
            ehp -= skill2[1]
            mana -= skill2[2]
            print("The",enam,"loses",skill2[1],"damage! It now has",ehp,"health.")
            print("You now have",mana,"mana.")
    elif "3" in attack:
        if mana >= skill3[2]:
            print("You hit with",skill3[0])
            ehp -= skill3[1]
            mana -= skill3[2]
            print("The",enam,"loses",skill3[1],"damage! It now has",ehp,"health.")
            print("You now have",mana,"mana.")
    else:
        print("You typed something wrong.")

顺便说一下，技能1，技能2和技能3都是我制作的游戏中不同技能的列表，技能1 [0]是技能的名称，技能[1]是技能＆＃39;攻击力和技能[2]是习惯使用技能的法术力。

skill1 = []
skill2 = []
skill3 = []

skill1.append("Very Weak Fireball")
skill1.append(20)
skill1.append(30)
skill2.append("Weak Fireball")
skill2.append(30)
skill2.append(40)
skill3.append("Average Fireball")
skill3.append(40)
skill3.append(50)

Answer 1

你不能将elif嵌入if：

中

if "q" in attack: # in line with the elif's
    if random.randint(1,100) > 21: # cannot compare to range use > 
        print("You hit with a quick attack!")
        ehp -= 20
        print("The",enam,"loses 20 damage! It now has",ehp,"health.")
    else:
        print("You missed.. :(")
elif "p" in attack:
    if random.randint(1,100)> 51: # greater that 51
        print("You hit with a power attack!")
        ehp -= 50
        print("The",enam,"loses 50 damage! It now has",ehp,"health.")
    else:
        print("You missed.. :(")
elif "1" in attack:

您的其余代码语法很好，只需将范围更改为>。

Answer 2

你的代码存在很多问题。例如：

    if random.randint(1,100) != range(1,21):

你在这里做的是比较一个整数（1到100之间的随机数）和一个列表（范围的输出，[1,2, ..., 20）。你的意思可能是not (... in range(...));这是好的，但是检查数字是否在两个其他数字之间的最耗时和耗费内存的方法。但是，这不是语法错误。

然而，这里的要点是你没有正确地缩进;您的elif必须与相应的if具有相同的缩进深度。

Answer 3

作为IDLE的输出对我来说：

unindent与任何外部缩进级别都不匹配

您没有正确使用缩进。

检查以下内容：

if "q" in attack:
    if random.randint(1,100) != range(1,21):
        print("You hit with a quick attack!")
        ehp -= 20
        print("The",enam,"loses 20 damage! It now has",ehp,"health.")
    else:
        print("You missed.. :(")
elif "p" in attack:
    if random.randint(1,100) != range(1,51):
        print("You hit with a power attack!")
        ehp -= 50
        print("The",enam,"loses 50 damage! It now has",ehp,"health.")
    else:
        print("You missed.. :(")
elif "1" in attack:
    if mana >= skill1[2]:
        print("You hit with",skill1[0])
        ehp -= skill1[1]
        mana -= skill1[2]
        print("The",enam,"loses",skill1[1],"damage! It now has",ehp,"health.")
        print("You now have",mana,"mana.")
elif "2" in attack:
    if mana >= skill2[2]:
        print("You hit with",skill2[0])
        ehp -= skill2[1]
        mana -= skill2[2]
        print("The",enam,"loses",skill2[1],"damage! It now has",ehp,"health.")
        print("You now have",mana,"mana.")
elif "3" in attack:
    if mana >= skill3[2]:
        print("You hit with",skill3[0])
        ehp -= skill3[1]
        mana -= skill3[2]
        print("The",enam,"loses",skill3[1],"damage! It now has",ehp,"health.")
        print("You now have",mana,"mana.")
    else:
        print("You typed something wrong.")

之后如果你没有定义攻击，你将收到另一个错误：

NameError：名称'攻击'未定义

如果攻击是字符串而不是变量，则必须将其替换为“attack”（添加引号）

正如@Padraic Cunningham在评论中所说，攻击显然是一个变量！所以你必须定义它。：）

我尝试运行它时出现语法错误

3 个答案: