Use hexadecimal color as string in darken function
up vote
2
down vote
favorite
In a less style, I have a variable that is a string.
@colorString: 'DADADA';
I can turn it into a color:
@color: ~'#@{colorString}';
I can use @color to set some value from a style:
div { color: @color }
but I cannot use it with the darken() function (or any other built-in function that manages colors).
Example:
background: linear-gradient(to bottom right,darken( @color , 20%), @color);
The compiler outputs
error evaluating function darken: color.toHSL is not a function
it seems that the @color is not a color (#DADADA) but rather is a string ('#DADADA') and the function can't parse it.
Any idea to solve this, without changing the @colorString (it mus be a string)?
css less
asked Nov 22 at 17:03
Christian Benseler
4,78652243
add a comment |
up vote
2
down vote
favorite
In a less style, I have a variable that is a string.
@colorString: 'DADADA';
I can turn it into a color:
@color: ~'#@{colorString}';
I can use @color to set some value from a style:
div { color: @color }
but I cannot use it with the darken() function (or any other built-in function that manages colors).
Example:
background: linear-gradient(to bottom right,darken( @color , 20%), @color);
The compiler outputs
error evaluating function darken: color.toHSL is not a function
it seems that the @color is not a color (#DADADA) but rather is a string ('#DADADA') and the function can't parse it.
Any idea to solve this, without changing the @colorString (it mus be a string)?
css less
asked Nov 22 at 17:03
Christian Benseler
4,78652243
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In a less style, I have a variable that is a string.
@colorString: 'DADADA';
I can turn it into a color:
@color: ~'#@{colorString}';
I can use @color to set some value from a style:
div { color: @color }
but I cannot use it with the darken() function (or any other built-in function that manages colors).
Example:
background: linear-gradient(to bottom right,darken( @color , 20%), @color);
The compiler outputs
error evaluating function darken: color.toHSL is not a function
it seems that the @color is not a color (#DADADA) but rather is a string ('#DADADA') and the function can't parse it.
Any idea to solve this, without changing the @colorString (it mus be a string)?
css less
asked Nov 22 at 17:03
Christian Benseler
4,78652243
In a less style, I have a variable that is a string.
@colorString: 'DADADA';
I can turn it into a color:
@color: ~'#@{colorString}';
I can use @color to set some value from a style:
div { color: @color }
but I cannot use it with the darken() function (or any other built-in function that manages colors).
Example:
background: linear-gradient(to bottom right,darken( @color , 20%), @color);
The compiler outputs
error evaluating function darken: color.toHSL is not a function
it seems that the @color is not a color (#DADADA) but rather is a string ('#DADADA') and the function can't parse it.
Any idea to solve this, without changing the @colorString (it mus be a string)?
css less
css less
asked Nov 22 at 17:03
Christian Benseler
4,78652243
asked Nov 22 at 17:03
Christian Benseler
4,78652243
asked Nov 22 at 17:03
Christian Benseler
4,78652243
asked Nov 22 at 17:03
Christian Benseler
4,78652243
asked Nov 22 at 17:03
Christian Benseler
4,78652243
4,78652243
add a comment |
add a comment |
1 Answer
1
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oldest
votes
up vote
2
down vote
accepted
You need to parse the the string to color using color().
@colorString: 'DADADA';
@color: color('#@{colorString}');
background: linear-gradient(to bottom right, darken(@color, 20%), @color);
Docs: http://lesscss.org/functions/#misc-functions-color
answered Nov 22 at 17:27
The.Bear
3,15811124
1
I think this is the problem. I tried in my application, which uses 3.7.1 and the color() worked! Thanks!
– Christian Benseler
Nov 22 at 17:44
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You need to parse the the string to color using color().
@colorString: 'DADADA';
@color: color('#@{colorString}');
background: linear-gradient(to bottom right, darken(@color, 20%), @color);
Docs: http://lesscss.org/functions/#misc-functions-color
answered Nov 22 at 17:27
The.Bear
3,15811124
1
I think this is the problem. I tried in my application, which uses 3.7.1 and the color() worked! Thanks!
– Christian Benseler
Nov 22 at 17:44
add a comment |
up vote
2
down vote
accepted
You need to parse the the string to color using color().
@colorString: 'DADADA';
@color: color('#@{colorString}');
background: linear-gradient(to bottom right, darken(@color, 20%), @color);
Docs: http://lesscss.org/functions/#misc-functions-color
answered Nov 22 at 17:27
The.Bear
3,15811124
1
I think this is the problem. I tried in my application, which uses 3.7.1 and the color() worked! Thanks!
– Christian Benseler
Nov 22 at 17:44
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You need to parse the the string to color using color().
@colorString: 'DADADA';
@color: color('#@{colorString}');
background: linear-gradient(to bottom right, darken(@color, 20%), @color);
Docs: http://lesscss.org/functions/#misc-functions-color
answered Nov 22 at 17:27
The.Bear
3,15811124
You need to parse the the string to color using color().
@colorString: 'DADADA';
@color: color('#@{colorString}');
background: linear-gradient(to bottom right, darken(@color, 20%), @color);
Docs: http://lesscss.org/functions/#misc-functions-color
answered Nov 22 at 17:27
The.Bear
3,15811124
answered Nov 22 at 17:27
The.Bear
3,15811124
answered Nov 22 at 17:27
The.Bear
3,15811124
answered Nov 22 at 17:27
The.Bear
3,15811124
3,15811124
1
I think this is the problem. I tried in my application, which uses 3.7.1 and the color() worked! Thanks!
– Christian Benseler
Nov 22 at 17:44
add a comment |
1
I think this is the problem. I tried in my application, which uses 3.7.1 and the color() worked! Thanks!
– Christian Benseler
Nov 22 at 17:44
1
1
I think this is the problem. I tried in my application, which uses 3.7.1 and the color() worked! Thanks!
– Christian Benseler
Nov 22 at 17:44
I think this is the problem. I tried in my application, which uses 3.7.1 and the color() worked! Thanks!
– Christian Benseler
Nov 22 at 17:44
add a comment |
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