Question

我将xp存储在json文件中以进行练习。如何显示前10位用户？我第一次本来希望可以进行排序，但我不知道这样会更有效率。

let xpAdd = Math.floor(Math.random() * 9) + 8;

  if(!xp[message.author.id]) {
    xp[message.author.id] = {
      xp:0,
      level:1
    };
  }

  let curxp = xp[message.author.id].xp;
  let curlvl = xp[message.author.id].level;
  let nxtLevel = xp[message.author.id].level * 300;

  xp[message.author.id].xp = curxp + xpAdd;

 fs.writeFile("./xp.json", JSON.stringify(xp), (error) => {
    if(error) console.log(error);
  });

这是我存储的代码

并以此显示xp的水平

if(!xp[message.author.id]) {
        xp[message.author.id] = {
            xp: 0,
            level:1
        };
    }

    let curxp = xp[message.author.id].xp;
    let curlvl = xp[message.author.id].level;
    let nxtLevelXp = curlvl * 300;
    let difference = nxtLevelXp - curxp;

Answer 1

我建议您将对象转换为数组，以便可以按自己的喜好对其进行格式化和排序。

// sample object
// I'm going to show the top three in this example for the interest of space
const xp = {
  "ID #1": {
     level: 3,
     xp: 300,
   },
   "ID #2": {
     level: 4,
     xp: 400,
   },
   "ID #3": {
     level: 2,
     xp: 200,
   },
   "ID #4": {
     level: 1,
     xp: 100,
   },
};

// sort entries in order of exp (descending), then single out the top three
let entries = Object.entries(xp).sort(([, a], [,  b]) => b.xp > a.xp ? 1 : -1).slice(0, 3);

// map entries to the prefered format using the data in their objects
// (in your actual command, you should use `client.users.cache.get(id).tag`,
// but since I don't have access to the client object here, I'm just writing `id`
entries = entries.map(([id, { level, xp }], idx) => `${idx + 1} -  ${id} (level ${level}; ${xp} xp)`);

// join each user by a line break
console.log(entries.join('\n'));

Answer 2

  if(!xpd[message.author.id]) {
        xpd[message.author.id] = {
            xp: 0,
            level:1
        };
    }
    const xp = {
        "xpd[message.author.id]":{
           level: xpd[message.author.id].level,
           xp: xpd[message.author.id].xp,
        },
        "xpd[message.author.id]": {
            level: xpd[message.author.id].level,
            xp: xpd[message.author.id].xp,
        },
        "xpd[message.author.id]": {
            level: xpd[message.author.id].level,
            xp: xpd[message.author.id].xp,
        },
        "xpd[message.author.id]": {
            level: xpd[message.author.id].level,
            xp: xpd[message.author.id].xp,
        },
      };
      
      let entries = Object.entries(xp).sort(([, a], [,  b]) => b.xp > a.xp ? 1 : -1).slice(0, 3);
      

      entries = entries.map(([id, {level, xp }], idx) => `${idx + 1} -  ${id} (${level}; ${xp} xp)`);
      
    console.log(entries.join('\n'));

我尝试了您的解决方案，但只显示了一个用户，我应该添加什么？

排行榜不和谐js

2 个答案: