如何让我的不和谐机器人一次打印整个播放列表？答案

【问题标题】：How do I make my discord bot print an entire playlist at once?如何让我的不和谐机器人一次打印整个播放列表？
【发布时间】：2021-07-06 17:06:57
【问题描述】：

使用 discord.py 构建一个不和谐机器人，我集成了 Spotify API 以打印播放列表中的歌曲列表。

我现在的代码一次打印每首歌 1 个句子（总共 50 个），这非常低效并且意味着很多 ping。关于如何将其全部打印成 1 个大块或小于 50 个的任何想法？

@client.command(name="ukmusic")
async def playlist(ctx):
    #Authortization for API usage
    headers = {
        "Authorization": "Bearer {}".format(oauth)
        }
    #These create the link that the program fetches (The playlist)
    endpoint = "https://api.spotify.com/v1/playlists/37i9dQZEVXbLnolsZ8PSNw"
    data = urlencode({"market": "GB"})
    lookup_url = f"{endpoint}?{data}"
    #This prints what the link looks like and the status code (200 if it works correctly)
    print (lookup_url)
    r = requests.get(lookup_url, headers = headers)
    print (r.status_code)

    #This prints out the playlist 
    await ctx.send("Here is what the good people of Britiania are listening to on Spotify")
    em = discord.Embed(title = "Song - Artist - Album\n")

    for item in r.json()['tracks']['items']:
        await ctx.send(
            item['track']['name'] + ' - ' +
            item['track']['artists'][0]['name'] + ' - ' +
            item['track']['album']['name']
        )

【问题讨论】：

如果您有一个需要令牌的请求，最好包含该数据的样本。

标签： python discord.py spotipy

【解决方案1】：

最简单的方法可能是制作一个临时字符串，您可以使用它来“构建”您的曲目列表：

@client.command(name="ukmusic")
async def playlist(ctx):
    #Authortization for API usage
    headers = {
        "Authorization": "Bearer {}".format(oauth)
        }
    #These create the link that the program fetches (The playlist)
    endpoint = "https://api.spotify.com/v1/playlists/37i9dQZEVXbLnolsZ8PSNw"
    data = urlencode({"market": "GB"})
    lookup_url = f"{endpoint}?{data}"
    #This prints what the link looks like and the status code (200 if it works correctly)
    print (lookup_url)
    r = requests.get(lookup_url, headers = headers)
    print (r.status_code)

    #This prints out the playlist 
    await ctx.send("Here is what the good people of Britiania are listening to on Spotify")
    em = discord.Embed(title = "Song - Artist - Album\n")

    allTracks = ""

    for item in r.json()['tracks']['items']:
        allTracks += item['track']['name'] + ' - ' + item['track']['artists'][0]['name'] + ' - ' + item['track']['album']['name'] + '\n'

    await ctx.send(allTracks)

【讨论】：

【解决方案2】：

制作一个保存所有数据的列表，然后使用.join() 发送。

我建议使用f string 以更简单的方式制作字符串。

您所说的其他方法是将数据保存到单独的变量中以嵌入 3 列

@client.command(name="ukmusic")
async def playlist(ctx):
    #Authortization for API usage
    headers = {
        "Authorization": "Bearer {}".format(oauth)
        }
    #These create the link that the program fetches (The playlist)
    endpoint = "https://api.spotify.com/v1/playlists/37i9dQZEVXbLnolsZ8PSNw"
    data = urlencode({"market": "GB"})
    lookup_url = f"{endpoint}?{data}"
    #This prints what the link looks like and the status code (200 if it works correctly)
    print (lookup_url)
    r = requests.get(lookup_url, headers = headers)
    print (r.status_code)

    #This prints out the playlist 
    em = discord.Embed(title = "Here is what the good people of Britiania are listening to on Spotify")

    tracks_names = []
    tracks_artists = []
    tracks_album = []

    # save the data into lists (could be better using a dict with a nested list
    for item in r.json()['tracks']['items']:
        tracks_names.append(item['track']['name'])
        tracks_artists.append(item['track']['artists'][0]['name'])
        tracks_album.append(item['track']['album']['name'])
        
    # add the 3 columns 
    em.add_field(name="Song", value='\n'.join(tracks_names), inline=True)
    em.add_field(name="Artist", value='\n'.join(tracks_artists), inline=True)
    em.add_field(name="Album", value='\n'.join(tracks_album), inline=True)

    await ctx.send(embed=em)

【讨论】：