Utilizing AngularJS to Implement Gradient Effects Using the ng-style Directive

Having trouble setting gradients using the angular js ng-style directive. I can successfully set normal colors using the code below:

<span ng-style="{'background-color': slidercolor}">works</span>

However, when trying to set a gradient it does not work.

Here is the jsfiddle link for reference:

<span ng-style="{'background-color':'-webkit-gradient(linear, left top, right top, color-stop(0%, 'slidercolor'), color-stop(100%, rgba(125,185,232,0)))'}">didnt work</span>

http://jsfiddle.net/sT8ar/1/

Furthermore, please advise on which values should be enclosed in quotes and which should not.

javascripthtmlcssangularjs

Answer №1

You have two issues to address:

  1. -webkit-gradient() returns <image>, not <color>; it should be utilized as background-image.
  2. When concatenating in Angular expressions, use a + (plus sign).

The correct syntax is demonstrated here (example):

<span ng-style="{'background-image':'-webkit-gradient(linear, left top, right top, color-stop(0%,'+ slidercolor+'), color-stop(100%,rgba(125,185,232,0)))'}">
  Now it works properly.
</span>

Answer №2

I trust this will be beneficial for you.

regulator

private setStyles() {
  const rgbaGradientStart = hexToRgba(this.hexGradientStart, 1);
  const rgbaGradientEnd = hexToRgba(this.hexGradientEnd, 0.1);
  const gradientParams = `left, ${rgbaGradientStart} 0%, ${rgbaGradientEnd} 50%`;

  this.gradientStyles = {
    'background-image': `-webkit-linear-gradient(${gradientParams})`,
    // 'background-image': `-moz-linear-gradient(${gradientParams})`,
    // 'background-image': `-o-linear-gradient(${gradientParams})`,
    // 'background-image': `-ms-linear-gradient(${gradientParams})`,
    // 'background-image': `linear-gradient(${gradientParams})`,
  }
}

model

<div
  class="super-gradient"
  [ngStyle]="gradientStyles"
></div>

P.S.: you may also find interest in this
https://github.com/angular/angular/issues/11362#issuecomment-244816438

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