如何使我的discord bot一次打印整个播放列表？

Question

使用discord.py构建一个discord机器人，我和Spotify API集成，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一句话（总共50句），这是非常低效的，意味着大量的ping。有没有办法把它全部打印成一大块或少于50块的

@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']
        )

Answer 1

最简单的方法可能是制作一个临时字符串，用于“构建”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)

Answer 2

制作一个保存所有数据的列表，然后使用.join()发送它

我建议使用f string来简化字符串

另一种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)