在 Python 中遇到冒号语法错误 [英] Getting a Colon Syntax Error in Python

问题描述

输入

yourChoice = True

while yourChoice:
    inf_or_ltd = input('Would you like to play infinte or a limited game?: '):
        if inf_or_ltd = ((('Limited') or ('limited')) and (compScore == ('3'))):
            exit = True
            print("You lost, Goodbye")
            break
            time.sleep(1)

为什么冒号是无效语法?我已经检查了括号平衡和前面的行(它们只是变量并导入了一些模块).它说 a limited game?: '): 是无效的语法.

Why is the colon a invalid syntax? I've checked the bracket balance and the previous lines (which are just variables and importing a few modules). It says that a limited game?: '): is invalid syntax.

推荐答案

在 Python 中，一个冒号 : 定义了作用域块的开始，它本质上与 { 相同> 在大多数语言中.但是冒号是不够的，还需要识别作用域块.标识的数量取决于父块.

In Python, a colon : defines the start of a scope block, it's essentially the same as { in most languages. But the colon isn't enough, scope blocks also need to be idented. The amount of identation depends on the parent block.

scope1:
    scope2:
        scope3:

块在其标识结束时结束，即:

A block ends when its identation ends, i.e.:

scope1:
    scope1_statement
    scope2:
        scope3:
            scope3_statement
        scope2_statement
    scope1_statement

现在，你什么时候创建一个新的作用域块?好吧，当您需要定义新范围时创建它们，例如:

Now, when would you create a new scope block? Well, you create them when you need to define a new scope, such as:

• 声明新方法(def、lambda、...)
• 声明新的代码分支(if 条件，try-catch...)

• Declaring new methods (def, lambda, ...)
• Declaring new code branches (if conditions, try-catch...)

在您的场景中，您试图在语句(指令)之后创建一个新的作用域，在这种情况下是一个 赋值 inf_or_ltd = input('...').

In your scenario, you are trying to create a new scope after a statement (an instruction) which in this case is an assignment inf_or_ltd = input('...').

Python 指令无法创建新的作用域块.

Python instructions cannot create a new scope block.

你需要把你的 if 放在与上面赋值相同的标识上；

You need to place your if on the same identation as the above assignment;

inf_or_ltd = input('...')
if inf_or_ltd == anything:
     # do something
else:
     # do something else

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