迷宫游戏 - 输入预计最多1个参数，得到3

Question

import random 
from random import randint 


print("Welcome to Brandon's Maze Game", 
    "You have to get 10 shields or find the exit to win", 
    "Good Luck :D") 


counter = 0 
shields = 3 
fate = randint(1,2) 
direction = input("You have come to a stop, do you want to turn Left(L) or  Right(R)? ") 

if direction == "L": 
    if fate == 1: 
     shields += 3 
     counter += 1 
     direction = input("You came across a chest, you now have ",shields, "! What way do you want to go next? Left(L) or Right(R)? ") 
    if fate == 2: 
     shields -= 1 
     counter += 1 
     direction = input("Uh oh, you got attacked and lost a shield, you now have ",shields," shields. Do you want to go Left(L) or Right(R)? ") 


if direction == "R": 
    if fate == 1: 
     shields += 3 
     counter += 1 
     direction = input("You came across a chest, you now have ",shields, "! What way do you want to go next? Left(L) or Right(R)? ") 
    if fate == 2: 
     shields -= 1 
     counter += 1 
     direction = input("Uh oh, you got attacked and lost a shield, you now have ",shields," shields. Do you want to go Left(L) or Right(R)? ") 

if counter == 10: 
     print("Congratulations, you made it to the end with ",shields," shields.") 

我正在做一个迷宫游戏，用户可以选择左（“L”）或右（“R”），然后该程序使选择是否让玩家找到胸部或受到攻击。当用户发现聊天时他们获得+3盾牌，如果他们受到攻击，他们将失去1盾牌。 当我输入“L”或“R”它说：在线19 TypeError：输入预计最多1个参数，得到3. 不知道发生了什么，因为我只输入1值吗？， 任何帮助表示。

Answer 1

你是给出input三个参数（字符串，整数，字符串）。下面是用它打印数量的参数的函数演示：

>>> shields = 42 
>>> def foo(*args): 
...  print(len(args)) 
... 
>>> foo('fiz', shields, 'buzz') 
3 

你需要的是给一个字符串input：

>>> foo('fiz {} buzz'.format(shields)) 
1 

Answer 2

input()不像print()。提示符是单个字符串，因此将字符串与+（其中有逗号）进行连接。请注意，您必须将非字符串（如int）转换为字符串。

print("Congratulations, you made it to the end with " + str(shields) + " shields.") 

或字符串格式化：

print("Congratulations, you made it to the end with {:d} shields.".format(shields)) 

，或者使用文本字符串插值如果您对Python的3.6

print(f"Congratulations, you made it to the end with {shields} shields.") 

Answer 3

所以问题是，你在输入了比对参数的更多功能不像印刷品，你不能这样做。使用加号可以很容易地解决这个问题，以便将字符串连接起来而不是逗号，但要做到这一点，您必须通过将str（屏蔽）转换为int来将其转换为字符串。这是我写的Python 2.7，所以你可能不得不改变一些东西，但它应该都在那里。我还为if语句添加了.upper（），以便它可以同时使用大写和小写输入。 ps for python 3你会做input（）而不是raw_input（）。对不起，在2.7写我不跟3.好的，如果你有任何问题随时问

import random 
from random import randint 


print("Welcome to Brandon's Maze Game", 
    "You have to get 10 shields or find the exit to win", 
    "Good Luck :D") 


counter = 0 
shields = 3 
fate = randint(1,2) 
direction = raw_input("You have come to a stop, do you want to turn Left(L) or Right(R)? ") 

if direction.upper() == "L": 
    if fate == 1: 
     shields += 3 
     counter += 1 
     direction = raw_input("You came across a chest, you now have "+str(shields)+ "! What way do you want to go next? Left(L) or Right(R)? ") 
    if fate == 2: 
     shields -= 1 
     counter += 1 
     direction = raw_input("Uh oh, you got attacked and lost a shield, you now have "+str(shields)+" shields. Do you want to go Left(L) or Right(R)? ") 


if direction.upper() == "R": 
    if fate == 1: 
     shields += 3 
     counter += 1 
     direction = raw_input("You came across a chest, you now have "+str(shields)+"! What way do you want to go next? Left(L) or Right(R)? ") 
    if fate == 2: 
     shields -= 1 
     counter += 1 
     direction = raw_input("Uh oh, you got attacked and lost a shield, you now have "+str(shields)+" shields. Do you want to go Left(L) or Right(R)? ") 

if counter == 10: 
     print("Congratulations, you made it to the end with "+str(shields)+" shields.")