1

所以我试图在我的 discord.js 机器人中使用 quick.db 为我的 xp 系统创建一个排行榜命令。该命令的代码：

const db = require("quick.db");
        const Discord = require("discord.js");

        let firstplace = 0;
        let firstplaceperson = "";
        let secondplace = 0;
        let secondplaceperson = "";
        let thirdplace = 0;
        let thirdplaceperson = "";


        message.guild.members.forEach(member => {
            console.log(member.displayName)
            let xp = db.get(`xp_${message.guild.id}_${member.id}`);
            if (xp === null) db.set(`xp_${message.guild.id}_${member.id}`, 0);
            xp = db.get(`xp_${message.guild.id}_${member.id}`);

            if (xp >= firstplace) {
                firstplace = xp;
                firstplaceperson = member.displayName;
            }

            if (xp <= firstplace && xp > secondplace && firstplaceperson != member.displayName) {
                secondplace = xp;
                secondplaceperson = member.displayName;
            }

            if (xp <= secondplace && xp > thirdplace && firstplaceperson != member.displayName && secondplaceperson != member.displayName) {
                thirdplace = xp;
                thirdplaceperson = member.displayName;
            }
        });




        console.log(firstplace);
        console.log(firstplaceperson);

        console.log(secondplace);
        console.log(secondplaceperson);

        console.log(thirdplace);
        console.log(thirdplaceperson);

        let embed = {
            color: 000000,
            timestamp: new Date(),
            fields: [
                {
                    name: message.guild.language.first_place,
                    value: firstplaceperson + " " + firstplace
                },
                {
                    name: message.guild.language.second_place,
                    value: secondplaceperson + " " + secondplace
                },
                {
                    name: message.guild.language.third_place,
                    value: thirdplaceperson + " " + thirdplace
                }
            ]
        }

        if (thirdplace === 0) {
            embed.fields = [
                {
                    name: message.guild.language.first_place,
                    value: firstplaceperson + " " + firstplace
                },
                {
                    name: message.guild.language.second_place,
                    value: secondplaceperson + " " + secondplace
                },
                {
                    name: message.guild.language.third_place,
                    value: message.guild.language.invalid_xp
                }
            ]
        }


        if (secondplace === 0) {
            embed.fields = [
                {
                    name: message.guild.language.first_place,
                    value: firstplaceperson + " " + firstplace
                },
                {
                    name: message.guild.language.second_place,
                    value: message.guild.language.invalid_xp
                },
                {
                    name: message.guild.language.third_place,
                    value: message.guild.language.invalid_xp
                }
            ]
        }

        message.channel.send({ embed: embed })

    }

我添加了 console.log 部分以查看问题是否出在我试图获得前三名成员的代码的第一部分。问题是，即使在不和谐服务器中有其他人的 xp 大于 0，两者都secondplace保持thirdplace为 0。forEach 函数也会遍历公会中的每个成员，所以这不是问题。

正如预期的那样，因为secondplaceandthirdplace值保持为 0，嵌入将在最后两个字段中以 message.guild.language.invalid_xp 值结束。而第一个字段正确显示第一名。

4

2 回答 2

0

注意：如果您使用 V12 或更高版本，则应为

message.guild.members.cache.forEach()

于 2021-07-29T13:57:45.917 回答
0

我认为您用于对成员进行分类的“系统”有点糟糕。尝试对排行榜使用更简单的东西，例如数组。

let leaderboard = new Array()

function update(user,xp) {
  if (!leaderboard[0]) return leaderboard.push([user,xp])
  if (xp >= leaderboard[0][1]) return leaderboard.splice(0,0,[user,xp]);
  if (!leaderboard[1]) return leaderboard.push([user,xp])
  if (xp >= leaderboard[1][1]) return leaderboard.splice(1,0,[user,xp]);
  if (!leaderboard[2]) return leaderboard.push([user,xp])
  if (xp >= leaderboard[2][1]) return leaderboard.splice(2,0,[user,xp]);
}

您现在可以在 forEach 循环的每次迭代中调用该函数。它看起来像这样：

message.guild.members.forEach(member => {
  console.log(member.displayName)
  let xp = db.get(`xp_${message.guild.id}_${member.id}`);
  if (xp === null) db.set(`xp_${message.guild.id}_${member.id}`, 0);
  xp = db.get(`xp_${message.guild.id}_${member.id}`);
  
  ((user,xp) => {
    if (!leaderboard[0]) return leaderboard.push([user,xp])
    if (xp >= leaderboard[0][1]) return leaderboard.splice(0,0,[user,xp]);
    if (!leaderboard[1]) return leaderboard.push([user,xp])
    if (xp >= leaderboard[1][1]) return leaderboard.splice(1,0,[user,xp]);
    if (!leaderboard[2]) return leaderboard.push([user,xp])
    if (xp >= leaderboard[2][1]) return leaderboard.splice(2,0,[user,xp]);
  })()

（使用匿名函数，它看起来更酷（：你不必以这种方式声明函数）

于 2020-07-17T17:50:39.783 回答