setNickname is not including spaces in discord.js v13

Issue

When I set someone’s nickname for example Stack Overflow", It does not set It correctly. It sets my nickname to "stackoverflow", Without spaces.

My code:

const { MessageEmbed } = require('discord.js');
module.exports = {
    name: 'changenickname',
    aliases: ['changenick'],
    permissions: ["MANAGE_NICKNAMES"],
    cooldown: 5,
    description: "change your target's nickname",
    usage: "d!changenickname <@user> <nickname>",
    async execute(message, args, cmd, client, Discord, profileData){
        const target = message.mentions.members.first();
        const nickname = args.slice(1).join('');
        if(!target) return message.channel.send('Please specify a target to change nickname');
        if(!nickname) return message.channel.send('Please specify a nickname to change');
        if (member.roles.highest.position >= message.guild.me.roles.highest.position) return message.reply('I cannot moderate this user as their highest role is higher than mine or I have the same highest role position as them.')


        target.setNickname(nickname);
        const done = new MessageEmbed()
        .setColor("GREEN")
        .setTitle("✅ Done")
        .setDescription(`Changed <@${target.id}> nickname to ${nickname}`)
        .setTimestamp()
        message.channel.send({embeds: [done]})
    }
}

Note: The command sets the nickname without any error, but it must set the nicknames with spaces(if there were in args).

I am using discord.js v13 and node.js v16

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Solution

At line 11 you have this:

const nickname = args.slice(1).join('');

In your case args.slice(1) returns the array ['The', 'God']. The join() method takes every element of the array and puts it together in one string with the given string between every element. The string you gave to the function is '', so it will put the elements together without anything between it. So from the array ['The', 'God'] it makes the string TheGod instead of The God.

So what you would need is .join(' '), not .join('').

I hope that helps.

Answered By – Iliannnn

Answer Checked By – Cary Denson (AngularFixing Admin)

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