Question

我很困惑，if语句应该运行imo，但当然这是编程所以它不会起作用。

def create_player():
    name = str()
    names = []
    print("hey there")
    if numberofplayers == 2:
        name = input("Please enter Player 1's name: ")       
        names.append(name)
        name = input("Please enter Player 2's name: ")
        names.append(name)
    elif numberofplayers == 3:
        name = input("Please enter Player 1's name: ")       
        names.append(name)
        name = input("Please enter Player 2's name: ")
        names.append(name)
        name = input("Please enter Player 3's name: ")
        names.append(name)
    elif numberofplayers == 4:
        name = input("Please enter Player 1's name: ")       
        names.append(name)
        name = input("Please enter Player 2's name: ")
        names.append(name)
        name = input("Please enter Player 3's name: ")
        names.append(name)
        name = input("Please enter Player 4's name: ")
        names.append(name)
        print("Welcome to the game " + names[0] + ", " +names[1]+ ", "+names[2]+ " and "+names[3]+"!")

print("test")
numberofplayers = int(0)
numberofplayers = input("# of players")
create_player()
print ("aha")

（顺便说一句，这段代码不是任何接近完成的方法）随机打印操作在那里进行测试，程序似乎忽略了整个if语句，因为它打印了if语句之外的所有打印操作，但没有在声明中。任何帮助都会非常感激，谢谢你。

Answer 1

您需要将numberofplayers投放到int，否则您需要进行比较int和string：

numberofplayers = int(input("# of players"))

Python 3中的

input()返回一个字符串对象，与以前的Python2 input()不同。

这是正在发生的事情：

>>> a = input()
2
>>> a == 2
False
>>> a == '2'
True

Answer 2

虽然保罗的答案解决了你眼前的问题，但还有其他一些事情需要指出。

由于您的功能至少会创建两个玩家，因此请正确命名：create_players。它可以避免以后混淆。

不要依赖函数范围之外的全局变量或值。您应该将它们明确地传递给您的函数。

def create_players(numberofplayers):
    ...

numberofplayers = int(input('# of players'))
create_players(numberofplayers)

这些行没有用处：
```
name = str()

numberofplayers = int(0)
```
您不需要将变量定义为类型以便使用它们，并且您立即用实际值替换它们。

作为单个if-elif循环，您的分支for会更好：

names = []
for player in range(1, numberofplayers+1):
    names.append( input("Please enter Player %d's name: " % player) )
print('Welcome to the game ' + ' and '.join([', '.join(names[:-1]), names[-1]]) + '!')

如果语句没有在Python中执行模块内部

2 个答案: