CODEWITHSUNDEEP
csssassmixins
user2339729
user2339729

Reputation: 59

Sass variable inside mixing include

I'm trying to create an each loop that creates a class name and calls a mixin with that name as the parameter. The idea is that I can make a list:

$list: (item1, item2, item3, ect);

Then run an each loop to create a class using that name and call a mixin with that name as the parameter.

@each $name in $list {
    .#{$name} {
        @include mixin($name);
    }
}

That parameter in the mixin calls another variable of the same name to set the attributes I need within the mixin. This seems like something I should be able to do but can't seem to figure out how to make the mixin parameter as a variable instead of just a string of text. Do I have to use interpolation in a specific way to be able to add the $ in front of the list name? Doing this doesn't seem to work: 

@include mixin($#{$name});

What I need it to do is use the values of item1, item2, item3, ect to create the class name but then send them as variables $item1, $item2, $item3 through the mixin and I'm not sure what else to try.

Upvotes: 0

Views: 2709

Answers (1)

Nathanael McDaniel
Nathanael McDaniel

Reputation: 333

Using just the variable name $name as you did the first time will work.

Here is an example I made using your code.

Upvotes: 1

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