Attempting to utilize the red, green, and blue functions in LESS CSS to extract RGB values and subsequently inject them into a gradient

Is there a way to extract individual r, g, b values from a hex code and then use them in a background gradient without dealing with rgb values separately? I attempted to utilize the red(@color), green(@color), blue(@color) functions but encountered an error stating: error evaluating function rgb: color functions require numerical parameters

@color: #E28141;

.test-gradient (@color, @topalpha, @topstop, @bottomalpha, @bottomstop) {
@red1:red(@color);
@green1:green(@color);
@blue1:blue(@color);

background: linear-gradient(to bottom, rgba(@red1,@green1,@blue1,@topalpha) @topstop,rgba(@red1,@green1,@blue1,@bottomalpha) @bottomstop);
}

Martin confirmed that the code above actually functions as intended, so here is the actual mixin I am trying to execute:

.test-gradient (@color, @topalpha, @topstop, @bottomalpha, @bottomstop) {
@red1:red(@color); @green1:green(@color); @blue1:blue(@color);
background: rgb(@red1,@green1,@blue1);
background: -moz-linear-gradient(top, rgba(@red1,@green1,@blue1,@topalpha) @topstop, rgba(@red1,@green1,@blue1,@bottomalpha) @bottomstop);
background: -webkit-gradient(linear, left top, left bottom, color-stop(@topstop,rgba(@red1,@green1,@blue1,@topalpha)), color-stop(@bottomstop,rgba(@red1,@green1,@blue1,@bottomalpha)));
background: -webkit-linear-gradient(top, rgba(@red1,@green1,@blue1,@topalpha) @topstop,rgba(@@red1,@green1,@blue1,@bottomalpha) @bottomstop);
background: -o-linear-gradient(top, rgba(@red1,@green1,@blue1,@topalpha) @topstop,rgba(@red1,@green1,@blue1,@bottomalpha) @bottomstop);
background: -ms-linear-gradient(top, rgba(@red1,@green1,@blue1,@topalpha) @topstop,rgba(@red1,@green1,@blue1,@bottomalpha) @bottomstop);
background: linear-gradient(to bottom, rgba(@red1,@green1,@blue1,@topalpha) @topstop,rgba(@red1,@green1,@blue1,@bottomalpha) @bottomstop);
}
cssfunctionless

Answer №1

It appears to be functioning correctly for me,

I experimented with this:

.test-gradient(@color, .5, 10%, .9, 90%);

and here is the result I obtained:

background: linear-gradient(to bottom, rgba(226, 129, 65, 0.5) 10%, rgba(226, 129, 65, 0.9) 90%);

Therefore, I am uncertain about what values you were inputting into the variables. Perhaps you could provide an example that triggers the error.

Edit:

You just had a few errors (used green instead of @green once and @@red in another instance). This should now function correctly:

.test-gradient (@color, @topalpha, @topstop, @bottomalpha, @bottomstop) {
@red1:red(@color);
@green1:green(@color);
@blue1:blue(@color);
background: rgb(@red1, @green1, @blue1);
background: -moz-linear-gradient(top, rgba(@red1,@green1,@blue1,@topalpha) @topstop, rgba(@red1,@green1,@blue1,@bottomalpha) @bottomstop);
background: -webkit-gradient(linear, left top, left bottom, color-stop(@topstop,rgba(@red1,@green1,@blue1,@topalpha)), color-stop(@bottomstop,rgba(@red1,@green1,@blue1,@bottomalpha)));
background: -webkit-linear-gradient(top, rgba(@red1,@green1,@blue1,@topalpha) @topstop,rgba(@red1,@green1,@blue1,@bottomalpha) @bottomstop);
background: -o-linear-gradient(top, rgba(@red1,@green1,@blue1,@topalpha) @topstop,rgba(@red1,@green1,@blue1,@bottomalpha) @bottomstop);
background: -ms-linear-gradient(top, rgba(@red1,@green1,@blue1,@topalpha) @topstop,rgba(@red1,@green1,@blue1,@bottomalpha) @bottomstop);
background: linear-gradient(to bottom, rgba(@red1,@green1,@blue1,@topalpha) @topstop,rgba(@red1,@green1,@blue1,@bottomalpha) @bottomstop);
}

Note:

The functions tint, shade, multiply, screen, overlay, hardlight, difference, exclusion, average, negation, softlight, red, green, blue, and contrast were only introduced in LESS 1.3.1.

In previous versions of LESS, for example:

red(#E28141)

would output

red(#E28141)

instead of the number

226

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