Command !so custom

Asked Jun 25, 2021·2 replies·Last activity 5 years ago·Open in Discord ↗

Velvet FoxOriginal postJun 25, 2021

<?php
$so = $_GET["option"];
> $default = 'Go check out $(touser) at twitch.tv/$(touser) and go show them some love ';

Hello! I want to do a shoutout command or custom !so already works with the 5 friends I have in the array, but I also have the variable “default” for those who are not in the array and give them another message, only that the $(touser) does not work…

Example:

!so gusbars => Go check out my great friend who is part of our crew and go show them some love in ....

!so otheruser1 => Go check out $(touser) at $(touser) and go show them some love

help:( I really appreciate the help

2 replies

Bashful GoatJun 25, 2021

You're already passing in the name of the user with <url>?options=<username> or more likely <url>?options=$(touser), so you don't need to return $(touser) in your response. Add some conditionals for the names your looking for and format your default response with the actual data like so:

php
$default = 'Go check out '  . $so . ' at twitch.tv/' . $so . ' and go show them some love ';

Format it how you like as I don't use PHP much these days.

Velvet FoxJun 25, 2021

Oh perfect!!! thank you very much for the help, I did not know very well haha 😅