Question

即使You Win!更高，代码也会继续打印computer_card。

import random

computer_card = random.choice([2,3,4,5,6,7,8,9,10,11,12,13,14])
player_card = raw_input('Please pick a card \n')

if player_card == 'Jack' or player_card == 'jack':
    player_card = 11

if player_card == 'Queen' or player_card == 'queen':
    player_card = 12

if player_card == 'King' or player_card == 'king':
    player_card = 13

if player_card == 'Ace' or player_card == 'ace':
    player_card = 14

if  computer_card == player_card:
    print 'Computer Choice: ', computer_card
    print 'Player Choice: ', player_card
    print 'Tie!'

elif computer_card > player_card:
    print 'Computer Choice: ', computer_card
    print 'Player Choice: ', player_card
    print 'Computer Wins!'

elif computer_card < player_card:
    print 'Computer Choice: ', computer_card
    print 'Player Choice: ', player_card
    print 'You Win!'

else:
    print 'I am not sure what is really going on if you got to this line.'

Answer 1

问题：

raw_string（）读取的输入转换为字符串。这一行，

Pre-main time: 0.52 seconds App did launch to javascript did load time: 2.12 seconds JS render time: 0.74 seconds Total time: 3.34 seconds

读取输入，将其转换为字符串并将其放入player_card。并player_card=raw_input()

一种可能的解决办法：

如果输入不是['jack'，'queen'，'king'或'ace']，则更改比较流并将输入转换为int

'1' > 9

进一步改进：虽然此解决方案应解决当前问题，但代码可以大大改进。一些改进建议，

确保您的代码可以处理来自用户的无效输入[永不信任用户输入]
使用dict或list将用户输入转换为数字值而不是逐个处理案例。

Answer 2

raw_input将字符串放入变量player_card。当玩家选择King，Queen，Ace，Jack时，player_card语句将if的值更改为int。但如果player_card不是这些，player_card中的值仍为字符串。 Like @Fallen said，可以通过将player_card转换为int。

来解决此问题

计划未按预期结束

2 个答案: