Question

所以我正在尝试编写一个很好的战争游戏，当我尝试将一个对象（一张牌）从一个列表（手）移动到另一个列表时，我遇到了错误。还有其他一些关于此事的帖子，但我无法拼凑出怎么做...... 这是代码：

import random
cardvalues = {"ace" : 13 , "2" : 2 , "3" : 3 , "4" : 4 , "5" : 5 , "6" : 6, "7" : 7 , "8" : 8 , "9" : 9, "10" : 10 , "jack" : 11 , "queen" : 12 , "king" : 13}
suits = {"clubs" : 1, "diamonds" : 2 , "spades" : 3 , "hearts" : 4}
deck = []
currentDeck = []

class card:
    def __init__ (self, value, suit):
        if value not in cardvalues:
            raise RuntimeError("must input valid card value")...

### [此处的代码已被删除，因为它不会导致问题]

def battle():            
    if hand1[0] >= hand2[0]:
        hand1.append[hand2.pop(0)] #The error happens in this function in the .append
        print("player 1 won the battle")
    elif hand1[0] < hand2[0]:
        hand2.append[hand1.pop(0)]
        print("player 2 won the battle")
    else:
        raise RuntimeError("something wrong")


while len(hand1) > 0:
    while len(hand2) > 0:
        battle()
if len(hand1) == 0:
    print("PLAYER 2 WON THE WAR")
elif len(hand2) == 0:
    print("PLAYER 1 WON THE WAR")
else:
    print("no one won?")

谢谢！

我不仅需要帮助找到我的错误，但这是什么意思？

Answer 1

您在尝试拨打0 <= n < s.size()时使用function keyword_replace($where) { $keywords_to_replace = get_option('keywords_to_replace'); $keyword_links = get_option('keyword_links'); $KWs = explode("\n", $keywords_to_replace); $URLs = explode("\n", $keyword_links); $pattern = array(); $replacement = array(); for($i=0; $i< count($KWs); $i++) { $pattern2 = '/<a[^>]*>(.*?)'.$KWs[$i].'(.*?)</a>/'; if(preg_match($pattern2, $where)) { continue; } else { $pattern[$i] = '\'(?!((<.*?)|(<a.*?)))(\b'. $KWs[$i] . '\b)(?!(([^<>]*?)>)|([^>]*?</a>))\'si'; $replacement[$i] .= '<a href="'.$URLs[$i].'" target="_blank">'.$KWs[$i].'</a>'; } } return preg_replace($pattern, $replacement, $where, -1); } add_filter('the_content','keyword_replace');代替[]。

'builtin_function_or_method'对象不是可下载错误'是什么意思？

1 个答案: