2

我希望我的排行榜根据有多少用户分成多个页面。例如，如果有 5 个用户，则只剩下一个页面，如果有 23 个用户，则将它们分成 3 个页面，其中前 2 个显示 10 个用户，最后一个显示最后 3 个用户。我已经尝试了很多方法，但我已经到了我不知道该怎么做的地步。你能解决吗？只要可行，任何方法都可以接受。

@client.command()
async def leaderboard(ctx):
    cursor = levelsystem_db.cursor()
    cursor.execute("SELECT users.user_level, users.user_xp, users.client_id\nFROM users\nORDER BY users.user_level DESC , users.user_xp DESC")
    leaderboard = cursor.fetchall()
    embed = discord.Embed(
        title=f"Leaderboard del {ctx.guild.name}",
        colour=0x003399
    )
    for i, pos in enumerate(leaderboard, start=1):
        lvl, xp, member_id = pos
        cursor.execute(f"SELECT xp_level FROM levels WHERE level = {lvl+1}")
        xp_level = cursor.fetchall()
        xp_level = xp_level[0][0]
        member = await client.fetch_user(member_id)
        sorted_users = ## get all the members and sort them...
        pages = grouper(sorted_users, 10)
        print(pages)
        embed.add_field(
            name=f"{i}. {member.display_name}",
            value=f"Livello attuale: {lvl}\n\t\tXP: {xp}/{xp_level}\n",
            inline=False
        )
    await ctx.send(embed=embed)

def grouper(iterable, n, fillvalue=None):
    args = [iter(iterable)] * n
    return zip_longest(*args, fillvalue=fillvalue)
4

3 回答 3

1

您可以使用反应对嵌入进行分页：

import asyncio

@client.command()
async def leaderboard(ctx):
    cursor = levelsystem_db.cursor()
    cursor.execute("SELECT user_level, user_xp, client_id FROM users ORDER BY user_level DESC, user_xp DESC")
    users = cursor.fetchall()
    pages = []
    page_count = -(-len(users)//10)) #will return 3 if users' length is 23
    for i in range(page_count):
        embed = discord.Embed(
            title = f"Leaderboard del {ctx.guild.name}",
            colour = 0x003399
        )
        embed.set_footer(
            text = f"Page {i + 1}/{page_count}"
        )
        for rank, data in enumerate(users[i * 10:i * 10 + 10], start=i * 10 + 1):
            level = data[0]
            xp = data[1]
            user = await client.fetch_user(data[2])
    
            embed.add_field(
                name = f"#{rank} {user}",
                value = f"Level: {level}\nExperience: {xp}"
                inline = False
            )
        pages.append(embed)

    index = 0
    message = await ctx.send(embed=pages[0])
    emojis = ["◀️", "⏹", "▶️"]
    for emoji in emojis:
        await message.add_reaction(emoji)
    while not client.is_closed():
        try:
            react, user = await client.wait_for(
                "reaction_add",
                timeout = 60.0,
                check = lambda r, u: r.emoji in emojis and u.id == ctx.author.id and r.message.id == message.id
            )
            if react.emoji == emojis[0] and index > 0:
                index -= 1
            elif react.emoji == emojis[1]:
                await message.delete()
                break
            elif react.emoji == emojis[2] and index < len(pages) - 1:
                index += 1

            await message.edit(embed=pages[index])
        except asyncio.TimeoutError:
            await message.delete()
            break
于 2021-05-08T19:27:00.397 回答
0

根据 Discord 的关于嵌入限制的文档，可在此处找到，您最多可以有 25 个字段对象，并且嵌入最多不应超过 6000 个字符。

考虑到这些信息，您可以创建一个跟踪字符数和字段数的函数。一旦超过这些限制之一，就应该创建一个单独的“页面”以使嵌入仍然保持在这些限制之下。

您将需要存储消息的当前状态并拥有一个reactor_add 事件监听器。在您的反应添加事件侦听器中，您需要再次检查消息并检查它位于哪个页面并编辑消息以进行新嵌入。

我希望这有帮助。如果您遇到困难，我会检查评论并帮助您！

于 2021-03-24T15:11:31.333 回答
0

如果这就是你的意思，但试试这个：

import math


@client.command()
async def test(ctx):
    users = [
        "user1",
        "user2",
        "user3",
        "user4",
        "user5",
        "user6",
        "user7",
        "user8",
        "user9",
        "user10",
        "user11",
        "user12",
        "user13",
        "user14",
        "user15",
        "user16",
        "user17",
        "user18",
        "user19",
        "user20",
        "user21",
        "user22",
        "user23",
    ]
    pages = {}
    numOfPages = math.ceil(len(users) / 10)
    j = 0
    for i in range(numOfPages):
        pages[f"page{i}"] = discord.Embed(title=f"Page {i+1}")
        displayUsers = ""
        while j < 10 + i * 10:
            try:
                displayUsers += users[j] + "\n"
            except:
                break
            j += 1
        pages[f"page{i}"].add_field(name="Users:", value=displayUsers)
        await ctx.send(embed=pages[f"page{i}"])

discord.Embed它使用 for 循环根据用户计数创建对象，然后依次发送它们。

于 2021-05-08T17:39:40.753 回答