Question

我正在尝试在3个不同的图像之间来回切换按钮，但是按钮并没有在上一个图像中回来，只是前进到下一个图像，有人可以帮我吗？

    if message.content.startswith('!image'):

        msg1 = await Bot.send_file(message.channel, 'image1.png')

        toReact = ('⏩')
        for reaction in toReact:
            await Bot.add_reaction(msg1, reaction)
        def checkReaction(reaction, user):
            e = str(reaction.emoji)
            return e.startswith('⏩')
        res = await Bot.wait_for_reaction(message=msg1, user=message.author, timeout=30, check=checkReaction)

        if res is None:
            await Bot.delete_message(msg1)

        elif '⏩' in str(res.reaction.emoji):
            await Bot.delete_message(msg1)
            msg2 = await Bot.send_file(message.channel, 'image2.png')

            toReact = ['⏪', '⏩']
            for reaction in toReact:
                await Bot.add_reaction(msg2, reaction)
            def checkReaction2(reaction, user):
                e = str(reaction.emoji)
                return e.startswith('⏪','⏩')
            res2 = await Bot.wait_for_reaction(message=msg2, user=message.author, timeout=30, check=checkReaction2)

            if res2 is None:
                await Bot.delete_message(msg2)

            elif '⏩' in str(res.reaction.emoji):
                await Bot.delete_message(mensagem2)
                await Bot.send_file(message.channel, 'image3.png')

Answer 1

这就是我想出的。

基本上，我们有一个循环，可根据所需反应检查每个反应，然后删除旧消息并发送新消息（如果我们看到了所寻找的反应之一）。

from discord.ext import commands

bot = commands.Bot(command_prefix='!')

left = '⏪'
right = '⏩'

messages = ("1", "2", "3")

def predicate(message, l, r):
    def check(reaction, user):
        if reaction.message.id != message.id or user == bot.user:
            return False
        if l and reaction.emoji == left:
            return True
        if r and reaction.emoji == right:
            return True
        return False

    return check


@bot.command(pass_context=True)
async def series(ctx):
    index = 0
    while True:
        msg = await bot.say(messages[index])
        l = index != 0
        r = index != len(messages) - 1
        if l:
            await bot.add_reaction(msg, left) 
        if r:
            await bot.add_reaction(msg, right)
        # bot.wait_for_reaction
        react, user = await bot.wait_for_reaction(check=predicate(msg, l, r))
        if react.emoji == left:
            index -= 1
        elif react.emoji == right:
            index += 1
        await bot.delete_message(msg)

bot.run("TOKEN")

有人要求提供一种用于编辑消息的版本，而不是发送新消息，我还已将其更新为最新版本：

@bot.command(pass_context=True)
async def series(ctx):
    index = 0
    msg = None
    action = ctx.send
    while True:
        res = await action(content=messages[index])
        if res is not None:
            msg = res
        l = index != 0
        r = index != len(messages) - 1
        if l:
            await msg.add_reaction(left) 
        if r:
            await msg.add_reaction(right)
        react, user = await bot.wait_for('reaction_add', check=predicate(msg, l, r))
        if react.emoji == left:
            index -= 1
        elif react.emoji == right:
            index += 1
        action = msg.edit

反应分页按钮前进和后退python

1 个答案: