Setting variable to @mixin in Sass?

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

Rick · Accepted Answer · 2012-04-05T06:06:22.227

26

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.

edited Apr 05 '12 at 06:06

answered Apr 05 '12 at 05:46

Rick

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2

Thnx for the solution and for additional information! I really appreciate it – FoxKllD Apr 05 '12 at 18:55

Bruce Smith · Answer 2 · 2015-02-12T15:18:59.450

9

If you are trying to set a returned value to a SASS variable, you need to use @function, not @mixin. This hung me up for a little while and was not aware of @function. For example...

@function font($font-size, $line-height: 1.4, $font-family: Arial) {

    @if $line-height == 1.4 {
        $line-height: round($line-height*$font-size);
    }

    @return #{$font-size}/#{$line-height} $font-family;
}

$font-medium: font(20px);

BTW, although this doesn't address what this user is looking for exactly, this is about the only thing that pops up on an internet search about setting a var to a mixin so I wanted to share this here for others to know what to do if this situation pops up.

edited Feb 12 '15 at 15:18

answered Feb 11 '15 at 16:03

Bruce Smith

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That won't help with the OP's use case since they're not expecting a single return value. – cimmanon Feb 11 '15 at 16:56
Why would you create a `function` with hard-coded values? Kinda defeats the point eh? – Edward J Beckett Aug 27 '15 at 03:00

score 4 · Answer 3 · edited Sep 08 '13 at 19:04

@mixin gradient ($place, $bcolor1, $bcolor2){`<br>`
  background-image: linear-gradient($place, $bcolor1, $bcolor2)`<br>`
  background-image: -o-linear-gradient($place, $bcolor1, $bcolor2)`<br>`
  background-image: -moz-linear-gradient($place, $bcolor1, $bcolor2)`<br>`
  background-image: -webkit-linear-gradient($place, $bcolor1, $bcolor2)`<br>`
  background-image: -ms-linear-gradient($place, $bcolor1, $bcolor2)`<br>`
}

body{
  @include gradient(bottom, #F90000 18%, #FF4C4C 66%)
}

Setting variable to @mixin in Sass?

3 Answers3