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Python 3.x 关于匹配的Hangman Python 3_Python 3.x_Matching - Fatal编程技术网

Python 3.x 关于匹配的Hangman Python 3

Python 3.x 关于匹配的Hangman Python 3,python-3.x,matching,Python 3.x,Matching,我在编写一个简单的刽子手游戏时遇到了问题。例如,神秘 单词是“btta”,用户可以通过输入每个字母猜单词 反过来。例如,输入字母“t”,它将在索引1中填充“t” 与“batt”索引相关的已声明列表框。在填写index1之后 框(['','','','''.'])它不会在索引2中填充t。我怎样才能解决它 riddle='btta' //Mystery word that user needs to guess length=len(riddle) listriddle=list(r

我在编写一个简单的刽子手游戏时遇到了问题。例如,神秘 单词是“btta”,用户可以通过输入每个字母猜单词 反过来。例如,输入字母“t”,它将在索引1中填充“t” 与“batt”索引相关的已声明列表框。在填写index1之后 框(['','','','''.'])它不会在索引2中填充t。我怎样才能解决它

riddle='btta'         //Mystery word that user needs to guess
length=len(riddle)
listriddle=list(riddle)
**box**=['_','_','_','_']  // a list where guessed letters are stored to form the mystery word
wrongcount=0

while(wrongcount<7):
     guess=str(input("Enter a letter: "))
     g=0
     while(g<length):
        if((listriddle[g]==guess) & (box[g]=='_')):
              zebra=listriddle.index(guess)
              box[zebra]=guess
              print(box)
              g+=1
        else:
             g+=1
    print('Try Again')
    wrongcount+=1
ridle='btta'//用户需要猜测的神秘单词
长度=长度(谜语)
listriddle=列表(谜语)
**方框**=['''''''''''''''''''.'''.'']//存储猜测字母以形成神秘单词的列表
错误计数=0

while(errorcount)虽然这段代码可以回答这个问题,但通常最好解释一下为什么以及这段代码在这种情况下有帮助。请参阅
    if ((listriddle[g] == guess) & (box[g] == '_')):
        box[g] = guess
        print(box)
        g += 1
    else:
        g += 1