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How to get "snowflake" value for user_id?

I'm trying to write a command that messages a specified user. The user can type something like /warn @user insert warning here. Here is the code:

    const user = await client.users
      .fetch(interaction.options.getString("user"))
      .catch(console.error);

    const embed = new MessageEmbed()
      .setColor("#FCE100")
      .setTitle(`⚠️ Warning!`)
      .setDescription(`${interaction.options.getString("warning")}`);
    await user.send({ embeds: embed }).catch(() => {
      interaction.channel.send("Error: user not found");
    });

Here's the error I'm getting:

user_id: Value "<@!9872345978#####>" is not snowflake.

How do I get the correct "snowflake" value to actually be able to DM the user?

about 4 years ago · Juan Pablo Isaza
1 answers
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0

You have to use getUser instead of getString and also change your user option type to "USER" to make it work!

about 4 years ago · Juan Pablo Isaza Report
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