类型错误：<member>.ban 不是一个函数

Question

我试图用discord.js v13制作一个ban斜杠命令，但它给了我一个错误！怎么了？

const { MessageEmbed } = require("discord.js");
module.exports = {
data: new SlashCommandBuilder()
.setName('ban')
.setDescription('Ban someone!')
.addUserOption(option => option.setName('user').setDescription('Select a user').setRequired(true))
.addStringOption(option => option.setName('reason').setDescription('Enter a reason').setRequired(true))
,
async execute(interaction) {
if (!interaction.member.permissions.has("BAN_MEMBERS")) return interaction.reply({ content: "Jij kan dit niet doen", ephemeral: true });
const banUser = interaction.options.getUser('user');
const reason = interaction.options.getString('reason');
let member = interaction.options._hoistedOptions[0]
console.log(member);
console.log(banUser);
console.log(interaction.options);

member.ban({ reason: reason }).catch(err => {
if (err) return interaction.reply({ content: `Er is een fout opgetreden tijdens het uitvoeren van dit commando!nMogelijke redenen: n- De gebruiker zit niet in de server;n- De gebruiker heeft een te hoge rang;n- Ik heb niet genoeg permissies om dit commando uit te voeren.`, ephemeral: true })
console.log(err)
});
var logEmbed = new MessageEmbed()
.setColor("BLURPLE")
.setFooter({ text: interaction.member.displayName, iconURL: interaction.user.displayAvatarURL })
.setTimestamp()
.setDescription(`**Gebanned:**
${member.user.username} (${member.id})
**Ban door:** ${interaction.author}
**Reden:** ${reason}`);
const logChannel = interaction.client.channels.cache.get("934107862072950814");
if (!logChannel) return interaction.reply({ content: "Ik kan niet bij het logs kanaal, meld dit bij de bot developer!", ephemeral: true });
logChannel.send({ embeds: [logEmbed] });
},
};

我的错误：

类型错误：member.ban不是函数

有人知道如何修复吗？

Answer 1

使用interaction.options.getMember()获取GuildMember对象

let member = interaction.options.getMember('user')

Answer 2

您最好使用它，因为您不需要获取用户：

interaction.guild.bans.create(banUser.id, {reason: "reason here"}).then(u => {
//u is the user so you can get his nickname or id
//your code here
});