jQuery.css cannot set repeating-linear-gradient

Question

I'm trying to make it by this code:

var $li = $('#marks').find('li[data-id=111222333]'),
color = 'rgb(' + $li.css('background-color').match(/\d+/g).map(function (i) {
    return (+i > 100 ? ((+i + 33) > 255 ? i : +i + 33) : +i + 99);
}).join(',') + ')',
gradient = 'repeating-linear-gradient(45deg, '+$li.css('background-color')+', '+$li.css('background-color')+' 3%, '+color+' 3%, '+color+' 6%);';
$li.css({'background-image': gradient});

But it doesn't works... I'm also trying this:

$li.css({'background': gradient});
$li.css('background', gradient);
$li.css('background-image', gradient);
$li[0].style.background = gradient;
$li[0].style.backgroundImage = gradient;

Butt result is same... Any ideas?!

UPD. Try here https://jsfiddle.net/kk77x3ru/

Does this make any difference $li.css({background: gradient});? Can you check the value in console? — rrk
– rrk, Commented Feb 4, 2016 at 7:33

Batu.Khan · Accepted Answer · 2016-02-04 08:06:19Z

1

It's because you have an unnecessary ; char in your gradient. Get rid of that

gradient = 'repeating-linear-gradient(45deg, '+$li.css('background-color')+', '+$li.css('background-color')+' 3%, '+color+' 3%, '+color+' 6%)';

UPDATED FIDDLE

answered Feb 4, 2016 at 8:06

Batu.Khan

3,0652 gold badges23 silver badges29 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Batu.Khan Over a year ago

@m0sk1t glad I could help!

Collectives™ on Stack Overflow

jQuery.css cannot set repeating-linear-gradient

1 Answer 1

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related