SCSS Color Loop

I'm learning SASS/SCSS and I'm stuck in generating a color loop that outputs 5 color classes with the addition of classes darkest-lightest.
So for example I can use

class="color1 darkest"

in the end.

This is what I got so far.

<pre>

$color1: #F4858E!default;

$color1-darkest: darken( $color1, 20% )!default;

$color1-darker: darken( $color1, 10% )!default;

$color1-lighter: lighten( $color1, 10% )!default;

$color1-lightest: lighten( $color1, 20% )!default;



$color2: #BFE2CA!default;

$color2-darkest: darken( $color2, 20% )!default;

$color2-darker: darken( $color2, 10% )!default;

$color2-lighter: lighten( $color2, 10% )!default;

$color2-lightest: lighten( $color2, 20% )!default;



$color3: #A6DAEF!default;

$color3-darkest: darken( $color3, 20% )!default;

$color3-darker: darken( $color3, 10% )!default;

$color3-lighter: lighten( $color3, 10% )!default;

$color3-lightest: lighten( $color3, 20% )!default;



$color4: #FED88F!default;

$color4-darkest: darken( $color4, 20% )!default;

$color4-darker: darken( $color4, 10% )!default;

$color4-lighter: lighten( $color4, 10% )!default;

$color4-lightest: lighten( $color4, 20% )!default;



$color5: #D1B6F7!default;

$color5-darkest: darken( $color5, 20% )!default;

$color5-darker: darken( $color5, 10% )!default;

$color5-lighter: lighten( $color5, 10% )!default;

$color5-lightest: lighten( $color5, 20% )!default;





$colors: $color1, $color2, $color3, $color4, $color5;



@for $i from 1 through length($colors) {

    .color#{$i} {

        color: ($colors, $i);

        &.darkest {

            color: ($colors, $i)-darkest;

        }

        &.darker {

           color: ($colors, $i)-darker;

        }

        &.lighter {

           color: ($colors, $i)-lighter;

        }

        &.lightest {

           color: ($colors, $i)-lightest;

        }



    }

}

</pre>

The output is pretty wrong at the moment...
Thanks for your help in advance!

asked Nov 26 '18 at 10:14

Mike

Do you need, for example, the $color3-lightest variable also in another point of your code or only in that loop? I mean, do you have to create all those variables or did you just do it for your list&loops?

– ReSedano
Nov 26 '18 at 14:18

i also would remove all variables with $color-something and create the colors on the fly in your loop

– Dirk
Nov 26 '18 at 15:23

@ReSedano I need both. For use in SASS (for predefined Elements) as well as classes (for flexible elements in HTML).

– Mike
Nov 28 '18 at 10:49

@Dirk the question is: how? :)

– Mike
Nov 28 '18 at 10:49

add a comment |

I'm learning SASS/SCSS and I'm stuck in generating a color loop that outputs 5 color classes with the addition of classes darkest-lightest.
So for example I can use

class="color1 darkest"

in the end.

This is what I got so far.

<pre>

$color1: #F4858E!default;

$color1-darkest: darken( $color1, 20% )!default;

$color1-darker: darken( $color1, 10% )!default;

$color1-lighter: lighten( $color1, 10% )!default;

$color1-lightest: lighten( $color1, 20% )!default;



$color2: #BFE2CA!default;

$color2-darkest: darken( $color2, 20% )!default;

$color2-darker: darken( $color2, 10% )!default;

$color2-lighter: lighten( $color2, 10% )!default;

$color2-lightest: lighten( $color2, 20% )!default;



$color3: #A6DAEF!default;

$color3-darkest: darken( $color3, 20% )!default;

$color3-darker: darken( $color3, 10% )!default;

$color3-lighter: lighten( $color3, 10% )!default;

$color3-lightest: lighten( $color3, 20% )!default;



$color4: #FED88F!default;

$color4-darkest: darken( $color4, 20% )!default;

$color4-darker: darken( $color4, 10% )!default;

$color4-lighter: lighten( $color4, 10% )!default;

$color4-lightest: lighten( $color4, 20% )!default;



$color5: #D1B6F7!default;

$color5-darkest: darken( $color5, 20% )!default;

$color5-darker: darken( $color5, 10% )!default;

$color5-lighter: lighten( $color5, 10% )!default;

$color5-lightest: lighten( $color5, 20% )!default;





$colors: $color1, $color2, $color3, $color4, $color5;



@for $i from 1 through length($colors) {

    .color#{$i} {

        color: ($colors, $i);

        &.darkest {

            color: ($colors, $i)-darkest;

        }

        &.darker {

           color: ($colors, $i)-darker;

        }

        &.lighter {

           color: ($colors, $i)-lighter;

        }

        &.lightest {

           color: ($colors, $i)-lightest;

        }



    }

}

</pre>

The output is pretty wrong at the moment...
Thanks for your help in advance!

asked Nov 26 '18 at 10:14

Mike

Do you need, for example, the $color3-lightest variable also in another point of your code or only in that loop? I mean, do you have to create all those variables or did you just do it for your list&loops?

– ReSedano
Nov 26 '18 at 14:18

i also would remove all variables with $color-something and create the colors on the fly in your loop

– Dirk
Nov 26 '18 at 15:23

@ReSedano I need both. For use in SASS (for predefined Elements) as well as classes (for flexible elements in HTML).

– Mike
Nov 28 '18 at 10:49

@Dirk the question is: how? :)

– Mike
Nov 28 '18 at 10:49

add a comment |

I'm learning SASS/SCSS and I'm stuck in generating a color loop that outputs 5 color classes with the addition of classes darkest-lightest.
So for example I can use

class="color1 darkest"

in the end.

This is what I got so far.

<pre>

$color1: #F4858E!default;

$color1-darkest: darken( $color1, 20% )!default;

$color1-darker: darken( $color1, 10% )!default;

$color1-lighter: lighten( $color1, 10% )!default;

$color1-lightest: lighten( $color1, 20% )!default;



$color2: #BFE2CA!default;

$color2-darkest: darken( $color2, 20% )!default;

$color2-darker: darken( $color2, 10% )!default;

$color2-lighter: lighten( $color2, 10% )!default;

$color2-lightest: lighten( $color2, 20% )!default;



$color3: #A6DAEF!default;

$color3-darkest: darken( $color3, 20% )!default;

$color3-darker: darken( $color3, 10% )!default;

$color3-lighter: lighten( $color3, 10% )!default;

$color3-lightest: lighten( $color3, 20% )!default;



$color4: #FED88F!default;

$color4-darkest: darken( $color4, 20% )!default;

$color4-darker: darken( $color4, 10% )!default;

$color4-lighter: lighten( $color4, 10% )!default;

$color4-lightest: lighten( $color4, 20% )!default;



$color5: #D1B6F7!default;

$color5-darkest: darken( $color5, 20% )!default;

$color5-darker: darken( $color5, 10% )!default;

$color5-lighter: lighten( $color5, 10% )!default;

$color5-lightest: lighten( $color5, 20% )!default;





$colors: $color1, $color2, $color3, $color4, $color5;



@for $i from 1 through length($colors) {

    .color#{$i} {

        color: ($colors, $i);

        &.darkest {

            color: ($colors, $i)-darkest;

        }

        &.darker {

           color: ($colors, $i)-darker;

        }

        &.lighter {

           color: ($colors, $i)-lighter;

        }

        &.lightest {

           color: ($colors, $i)-lightest;

        }



    }

}

</pre>

The output is pretty wrong at the moment...
Thanks for your help in advance!

asked Nov 26 '18 at 10:14

Mike

I'm learning SASS/SCSS and I'm stuck in generating a color loop that outputs 5 color classes with the addition of classes darkest-lightest.
So for example I can use

class="color1 darkest"

in the end.

This is what I got so far.

<pre>

$color1: #F4858E!default;

$color1-darkest: darken( $color1, 20% )!default;

$color1-darker: darken( $color1, 10% )!default;

$color1-lighter: lighten( $color1, 10% )!default;

$color1-lightest: lighten( $color1, 20% )!default;



$color2: #BFE2CA!default;

$color2-darkest: darken( $color2, 20% )!default;

$color2-darker: darken( $color2, 10% )!default;

$color2-lighter: lighten( $color2, 10% )!default;

$color2-lightest: lighten( $color2, 20% )!default;



$color3: #A6DAEF!default;

$color3-darkest: darken( $color3, 20% )!default;

$color3-darker: darken( $color3, 10% )!default;

$color3-lighter: lighten( $color3, 10% )!default;

$color3-lightest: lighten( $color3, 20% )!default;



$color4: #FED88F!default;

$color4-darkest: darken( $color4, 20% )!default;

$color4-darker: darken( $color4, 10% )!default;

$color4-lighter: lighten( $color4, 10% )!default;

$color4-lightest: lighten( $color4, 20% )!default;



$color5: #D1B6F7!default;

$color5-darkest: darken( $color5, 20% )!default;

$color5-darker: darken( $color5, 10% )!default;

$color5-lighter: lighten( $color5, 10% )!default;

$color5-lightest: lighten( $color5, 20% )!default;





$colors: $color1, $color2, $color3, $color4, $color5;



@for $i from 1 through length($colors) {

    .color#{$i} {

        color: ($colors, $i);

        &.darkest {

            color: ($colors, $i)-darkest;

        }

        &.darker {

           color: ($colors, $i)-darker;

        }

        &.lighter {

           color: ($colors, $i)-lighter;

        }

        &.lightest {

           color: ($colors, $i)-lightest;

        }



    }

}

</pre>

The output is pretty wrong at the moment...
Thanks for your help in advance!

loops colors sass

asked Nov 26 '18 at 10:14

Mike

asked Nov 26 '18 at 10:14

Mike

asked Nov 26 '18 at 10:14

Mike

asked Nov 26 '18 at 10:14

Mike

asked Nov 26 '18 at 10:14

Mike

Do you need, for example, the $color3-lightest variable also in another point of your code or only in that loop? I mean, do you have to create all those variables or did you just do it for your list&loops?

– ReSedano
Nov 26 '18 at 14:18

i also would remove all variables with $color-something and create the colors on the fly in your loop

– Dirk
Nov 26 '18 at 15:23

@ReSedano I need both. For use in SASS (for predefined Elements) as well as classes (for flexible elements in HTML).

– Mike
Nov 28 '18 at 10:49

@Dirk the question is: how? :)

– Mike
Nov 28 '18 at 10:49

add a comment |

Do you need, for example, the $color3-lightest variable also in another point of your code or only in that loop? I mean, do you have to create all those variables or did you just do it for your list&loops?

– ReSedano
Nov 26 '18 at 14:18

i also would remove all variables with $color-something and create the colors on the fly in your loop

– Dirk
Nov 26 '18 at 15:23

@ReSedano I need both. For use in SASS (for predefined Elements) as well as classes (for flexible elements in HTML).

– Mike
Nov 28 '18 at 10:49

@Dirk the question is: how? :)

– Mike
Nov 28 '18 at 10:49

Do you need, for example, the $color3-lightest variable also in another point of your code or only in that loop? I mean, do you have to create all those variables or did you just do it for your list&loops?

– ReSedano
Nov 26 '18 at 14:18

i also would remove all variables with $color-something and create the colors on the fly in your loop

– Dirk
Nov 26 '18 at 15:23

@ReSedano I need both. For use in SASS (for predefined Elements) as well as classes (for flexible elements in HTML).

– Mike
Nov 28 '18 at 10:49

@Dirk the question is: how? :)

– Mike
Nov 28 '18 at 10:49

add a comment |

1 Answer
1

active

oldest

votes

just like this

// scss

$color1: #F4858E;

$color2: #BFE2CA;

$color3: #A6DAEF;

$color4: #FED88F;

$color5: #D1B6F7;



$colors: $color1, $color2, $color3, $color4, $color5;



@for $i from 1 through length($colors) {

    .color#{$i} {

        color: nth($colors, $i);

        &.darkest {

          color: darken(nth($colors, $i), 20%);

        }

        &.darker {

          color: darken(nth($colors, $i), 10%);

        }

        &.lighter {

          color: lighten(nth($colors, $i), 10%);

        }

        &.lightest {

          color: lighten(nth($colors, $i), 20%);

        }

    }

}

gives me that output

//css

.color1 {

  color: #F4858E;

}

.color1.darkest {

  color: #ec2737;

}

.color1.darker {

  color: #f05663;

}

.color1.lighter {

  color: #f8b4b9;

}

.color1.lightest {

  color: #fce3e5;

}



.color2 {

  color: #BFE2CA;

}

.color2.darkest {

  color: #79c290;

}

.color2.darker {

  color: #9cd2ad;

}

.color2.lighter {

  color: #e2f2e7;

}

.color2.lightest {

  color: white;

}



.color3 {

  color: #A6DAEF;

}

.color3.darkest {

  color: #50b6df;

}

.color3.darker {

  color: #7bc8e7;

}

.color3.lighter {

  color: #d1ecf7;

}

.color3.lightest {

  color: #fcfeff;

}



.color4 {

  color: #FED88F;

}

.color4.darkest {

  color: #fdb52a;

}

.color4.darker {

  color: #fec65c;

}

.color4.lighter {

  color: #feeac2;

}

.color4.lightest {

  color: #fffbf4;

}



.color5 {

  color: #D1B6F7;

}

.color5.darkest {

  color: #975aed;

}

.color5.darker {

  color: #b488f2;

}

.color5.lighter {

  color: #eee4fc;

}

.color5.lightest {

  color: white;

}

is that what you want ?

answered Nov 28 '18 at 13:52

Dirk

721413

Sorry for the late reply. Yes, this works. Thank you very much!

– Mike
Jan 28 at 13:58

add a comment |

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1 Answer
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1 Answer
1

active

oldest

votes

just like this

// scss

$color1: #F4858E;

$color2: #BFE2CA;

$color3: #A6DAEF;

$color4: #FED88F;

$color5: #D1B6F7;



$colors: $color1, $color2, $color3, $color4, $color5;



@for $i from 1 through length($colors) {

    .color#{$i} {

        color: nth($colors, $i);

        &.darkest {

          color: darken(nth($colors, $i), 20%);

        }

        &.darker {

          color: darken(nth($colors, $i), 10%);

        }

        &.lighter {

          color: lighten(nth($colors, $i), 10%);

        }

        &.lightest {

          color: lighten(nth($colors, $i), 20%);

        }

    }

}

gives me that output

//css

.color1 {

  color: #F4858E;

}

.color1.darkest {

  color: #ec2737;

}

.color1.darker {

  color: #f05663;

}

.color1.lighter {

  color: #f8b4b9;

}

.color1.lightest {

  color: #fce3e5;

}



.color2 {

  color: #BFE2CA;

}

.color2.darkest {

  color: #79c290;

}

.color2.darker {

  color: #9cd2ad;

}

.color2.lighter {

  color: #e2f2e7;

}

.color2.lightest {

  color: white;

}



.color3 {

  color: #A6DAEF;

}

.color3.darkest {

  color: #50b6df;

}

.color3.darker {

  color: #7bc8e7;

}

.color3.lighter {

  color: #d1ecf7;

}

.color3.lightest {

  color: #fcfeff;

}



.color4 {

  color: #FED88F;

}

.color4.darkest {

  color: #fdb52a;

}

.color4.darker {

  color: #fec65c;

}

.color4.lighter {

  color: #feeac2;

}

.color4.lightest {

  color: #fffbf4;

}



.color5 {

  color: #D1B6F7;

}

.color5.darkest {

  color: #975aed;

}

.color5.darker {

  color: #b488f2;

}

.color5.lighter {

  color: #eee4fc;

}

.color5.lightest {

  color: white;

}

is that what you want ?

answered Nov 28 '18 at 13:52

Dirk

721413

Sorry for the late reply. Yes, this works. Thank you very much!

– Mike
Jan 28 at 13:58

add a comment |

just like this

// scss

$color1: #F4858E;

$color2: #BFE2CA;

$color3: #A6DAEF;

$color4: #FED88F;

$color5: #D1B6F7;



$colors: $color1, $color2, $color3, $color4, $color5;



@for $i from 1 through length($colors) {

    .color#{$i} {

        color: nth($colors, $i);

        &.darkest {

          color: darken(nth($colors, $i), 20%);

        }

        &.darker {

          color: darken(nth($colors, $i), 10%);

        }

        &.lighter {

          color: lighten(nth($colors, $i), 10%);

        }

        &.lightest {

          color: lighten(nth($colors, $i), 20%);

        }

    }

}

gives me that output

//css

.color1 {

  color: #F4858E;

}

.color1.darkest {

  color: #ec2737;

}

.color1.darker {

  color: #f05663;

}

.color1.lighter {

  color: #f8b4b9;

}

.color1.lightest {

  color: #fce3e5;

}



.color2 {

  color: #BFE2CA;

}

.color2.darkest {

  color: #79c290;

}

.color2.darker {

  color: #9cd2ad;

}

.color2.lighter {

  color: #e2f2e7;

}

.color2.lightest {

  color: white;

}



.color3 {

  color: #A6DAEF;

}

.color3.darkest {

  color: #50b6df;

}

.color3.darker {

  color: #7bc8e7;

}

.color3.lighter {

  color: #d1ecf7;

}

.color3.lightest {

  color: #fcfeff;

}



.color4 {

  color: #FED88F;

}

.color4.darkest {

  color: #fdb52a;

}

.color4.darker {

  color: #fec65c;

}

.color4.lighter {

  color: #feeac2;

}

.color4.lightest {

  color: #fffbf4;

}



.color5 {

  color: #D1B6F7;

}

.color5.darkest {

  color: #975aed;

}

.color5.darker {

  color: #b488f2;

}

.color5.lighter {

  color: #eee4fc;

}

.color5.lightest {

  color: white;

}

is that what you want ?

answered Nov 28 '18 at 13:52

Dirk

721413

Sorry for the late reply. Yes, this works. Thank you very much!

– Mike
Jan 28 at 13:58

add a comment |

just like this

// scss

$color1: #F4858E;

$color2: #BFE2CA;

$color3: #A6DAEF;

$color4: #FED88F;

$color5: #D1B6F7;



$colors: $color1, $color2, $color3, $color4, $color5;



@for $i from 1 through length($colors) {

    .color#{$i} {

        color: nth($colors, $i);

        &.darkest {

          color: darken(nth($colors, $i), 20%);

        }

        &.darker {

          color: darken(nth($colors, $i), 10%);

        }

        &.lighter {

          color: lighten(nth($colors, $i), 10%);

        }

        &.lightest {

          color: lighten(nth($colors, $i), 20%);

        }

    }

}

gives me that output

//css

.color1 {

  color: #F4858E;

}

.color1.darkest {

  color: #ec2737;

}

.color1.darker {

  color: #f05663;

}

.color1.lighter {

  color: #f8b4b9;

}

.color1.lightest {

  color: #fce3e5;

}



.color2 {

  color: #BFE2CA;

}

.color2.darkest {

  color: #79c290;

}

.color2.darker {

  color: #9cd2ad;

}

.color2.lighter {

  color: #e2f2e7;

}

.color2.lightest {

  color: white;

}



.color3 {

  color: #A6DAEF;

}

.color3.darkest {

  color: #50b6df;

}

.color3.darker {

  color: #7bc8e7;

}

.color3.lighter {

  color: #d1ecf7;

}

.color3.lightest {

  color: #fcfeff;

}



.color4 {

  color: #FED88F;

}

.color4.darkest {

  color: #fdb52a;

}

.color4.darker {

  color: #fec65c;

}

.color4.lighter {

  color: #feeac2;

}

.color4.lightest {

  color: #fffbf4;

}



.color5 {

  color: #D1B6F7;

}

.color5.darkest {

  color: #975aed;

}

.color5.darker {

  color: #b488f2;

}

.color5.lighter {

  color: #eee4fc;

}

.color5.lightest {

  color: white;

}

is that what you want ?

answered Nov 28 '18 at 13:52

Dirk

721413

just like this

// scss

$color1: #F4858E;

$color2: #BFE2CA;

$color3: #A6DAEF;

$color4: #FED88F;

$color5: #D1B6F7;



$colors: $color1, $color2, $color3, $color4, $color5;



@for $i from 1 through length($colors) {

    .color#{$i} {

        color: nth($colors, $i);

        &.darkest {

          color: darken(nth($colors, $i), 20%);

        }

        &.darker {

          color: darken(nth($colors, $i), 10%);

        }

        &.lighter {

          color: lighten(nth($colors, $i), 10%);

        }

        &.lightest {

          color: lighten(nth($colors, $i), 20%);

        }

    }

}

gives me that output

//css

.color1 {

  color: #F4858E;

}

.color1.darkest {

  color: #ec2737;

}

.color1.darker {

  color: #f05663;

}

.color1.lighter {

  color: #f8b4b9;

}

.color1.lightest {

  color: #fce3e5;

}



.color2 {

  color: #BFE2CA;

}

.color2.darkest {

  color: #79c290;

}

.color2.darker {

  color: #9cd2ad;

}

.color2.lighter {

  color: #e2f2e7;

}

.color2.lightest {

  color: white;

}



.color3 {

  color: #A6DAEF;

}

.color3.darkest {

  color: #50b6df;

}

.color3.darker {

  color: #7bc8e7;

}

.color3.lighter {

  color: #d1ecf7;

}

.color3.lightest {

  color: #fcfeff;

}



.color4 {

  color: #FED88F;

}

.color4.darkest {

  color: #fdb52a;

}

.color4.darker {

  color: #fec65c;

}

.color4.lighter {

  color: #feeac2;

}

.color4.lightest {

  color: #fffbf4;

}



.color5 {

  color: #D1B6F7;

}

.color5.darkest {

  color: #975aed;

}

.color5.darker {

  color: #b488f2;

}

.color5.lighter {

  color: #eee4fc;

}

.color5.lightest {

  color: white;

}

is that what you want ?

answered Nov 28 '18 at 13:52

Dirk

721413

answered Nov 28 '18 at 13:52

Dirk

721413

answered Nov 28 '18 at 13:52

Dirk

721413

answered Nov 28 '18 at 13:52

Dirk

721413

Sorry for the late reply. Yes, this works. Thank you very much!

– Mike
Jan 28 at 13:58

add a comment |

Sorry for the late reply. Yes, this works. Thank you very much!

– Mike
Jan 28 at 13:58

Sorry for the late reply. Yes, this works. Thank you very much!

– Mike
Jan 28 at 13:58

add a comment |

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