Question

Is there way to set @include mixin(); to variable?
I tried this

@mixin bg-gradient($fallback, $type, $positionX, $positionY, $from, $from-percent, $to, $to-percent){
    background: $fallback;
    background: -webkit-#{$type}-gradient($positionX $positionY, $from $from-percent, $to $to-percent);
    background:    -moz-#{$type}-gradient($positionX $positionY, $from $from-percent, $to $to-percent);
    background:         #{$type}-gradient($positionX $positionY, $from $from-percent, $to $to-percent);
}

$navBg: @include bg-gradient(#eee, radial, top, center, #999, 0%, #555, 100%);
body { $navBg; } // it gave me an error

Answer 1

I’m not aware of any way to do that specifically, but if you’re trying to just dry your settings on that particular type of gradient, you could write a wrapper mixin for it:

@mixin navBg() {
    @include bg-gradient(#eee, radial, top, center, #999, 0%, #555, 100%);
}
body { @include navBg; }

Edit:

Here‘s a list of data types supported by SASS variables. Neither mixin calls, nor the result of them (entire CSS rules), are included. I also tried treating the include as a string and interpolating it, but that only works for end-result CSS, not further directives.