Use @if and @else to Add Logic To Your Styles isn't working

HTML-CSS
1

Tell us what’s happening:
I am not able to complete the following test when I run the code: “Your mixin should have an @if statement to check if $val is light, and to set the border to 1px solid black.” Can anyone help me understand why the if statement isn’t working? Thanks.

Your code so far


<style type='text/sass'>
  @mixin border-stroke($val) {
    @if $val == light {
      border: lpx solid black; 
    }
    @else if $val == medium {
      border: 3px solid black;
    }
    @else if $val == heavy {
      border: 6px solid black;
    }
    @else {
      border: none;
    }
  
  #box {
    width: 150px;
    height: 150px;
    background-color: red;
    @include border-stroke(medium);
  }  
</style>

<div id="box"></div>

Your browser information:

User Agent is: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/67.0.3396.87 Safari/537.36.

Link to the challenge:
https://learn.freecodecamp.org/front-end-libraries/sass/use-if-and-else-to-add-logic-to-your-styles

3

The issue persists when I match up the braces.

6

for some reason I am unable to pass this challenge. I am posting a copy of my code, but I think everything is the way it has been requested.

@mixin border-stroke($val) { @if $val == light { border: 1px solid black; }
@else if $val == medium { border: 3px solid black; }
@else if @val == heavy { border: 6px solid black; }
@else { border: none; }

#box {
width: 150px;
height: 150px;
background-color: red;
@include border-stroke(medium);
}

7

Dang it.. I reviewed my coding again, and noticed that I typed @val instead of $val.. Solved my own problem..