Question

我正在尝试创建一个用于猜歌曲的游戏，但是即使我更改了范围，它也不允许我播放超过一首歌曲，我也想知道如何添加代码来猜测乐曲。

import random

for x in range(0,1):
    randNum = int(random.randint(0,1))

    song = open("Songs.txt", "r")
    songname = str(song.readlines()[randNum])
    print(songname[0])
    song.close()

    artist = open("Artists.txt", "r")
    artistname = artist.readlines()[randNum]
    print(artistname[0])
    artist.close()
    y = 0

    songGuess = input("What is the song called?")
    while(y<=2):
        if songGuess == songname:
            print("Answer correct!")
            break
        else:
            y = y + 1
            songguess = input("Incorrect! Try again:")

        if y == 2:
            print("GAME OVER")
            break

Answer 1

您需要将random.randint的范围更改为random.randint(0,len(song.readlines())-1)，以便从所有列出的歌曲中选择随机索引，并对歌手做同样的事情。
更好的方法是使用random.choice从列表中选择一个随机元素。
您对范围range(0,1)的使用将使您的循环仅运行一次，并相应地更新范围
您可以使用with关键字自动关闭文件，而不必显式关闭文件。

因此，根据上述更改，固定代码可能看起来像

import random

num_attempts = 10
#Run loop for num_attempts times
for x in range(0, num_attempts):

    songname = ''
    artistname = ''
    #Use with to open file
    with open('Songs.txt') as song:
        #Read all songs into a list
        songs = song.readlines()
        #Choose a random song
        songname = random.choice(songs)
        print(songname)

    with open('Artists.txt') as artist:
        # Read all artists into a list
        artists = artist.readlines()
        # Choose a random artist
        artistname = random.choice(artists)
        print(artistname)

    y = 0

    #Play your game
    songGuess = input("What is the song called?")
    while(y<=2):
        if songGuess == songname:
            print("Answer correct!")
            break
        else:
            y = y + 1
            songguess = input("Incorrect! Try again:")

        if y == 2:
            print("GAME OVER")
            break

如何修复“猜歌游戏” python

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