Scripting a slider for color control between multiple layers

[Brendon Murphy]

If you’re wanting each color control to change based on the slider, the expression on each color control should be like this:

slider = Math.round(effect(“Slider Control”)(“Slider”));

colorValue1 = [1,1,0,1];//put your desired color values here

colorValue2 = [1,0,0,1];//put your desired color values here

if(slider==1){colorValue1};

if(slider==2){colorValue2};

And so on. I’d also put stipulations for if the slider is less than 1 or greater than your highest assigned value so there are no errors.

[Michael Szalapski]

Although, if you’re using the current versions, you might be better-suited using a drop-down menu instead of a slider like this.

[Daniel Gillies]

Not sure if this is especially helpful, but Ray Dynamic Color is an elegant solution for these types of problems.

[Walter Soyka]

I strongly agree with the suggestion to use a dropdown — they’re much easier for your template’s users to understand and use.

Here’s some documentation on how to set them up:

https://helpx.adobe.com/after-effects/using/create_dropdowns_using_expressions.html

[Dan Ebberts]

I think you just need to a default case so the expression has something to do if the slider goes outside the 1-13 range. So you could just add this to the end of your expression so the color goes to black if the slider goes out of range:

else
[0,0,0,1]

[Roland R. kahlenberg]

You should use curly brackets to encapsulate your “if … else” conditionals.
Something like the following –

slider = Math.floor(thisComp.layer(“<wbr>Color”).effect(“Slider Control”)(“Slider”).value);
if (slider == 1) {
effect(“Color Control”)(“Color”)
} else if (slider == 2) {
effect(“Color Control 2”)(“Color”)
} else if (slider == 3) {
…
}

And as others have suggested, a Dropdown Menu is preferred for a more elegant UI/UX. You’ll want to be use descriptive a name for each Color Theme/Selection to make it onto the “This Week’s Awesome Dropdown Menu List”. 🙂

And for completeness, if you ever need to use a Slider Control for discreet selection/action, you’ll want to apply the Math.floor Expression directly onto the Slider Control instead of applying it onto the effect being controlled by the Slider Control.

[Lee Pines]

Thanks for the response. The problem isn’t the slider, it’s connecting it to another layer. BUT, how would you make the dropdown?

[Lee Pines]

Thanks! The problem isn’t with this part of the code, it’s with the first line. I can’t seem to make it connect to another layer properly. The code works fine without thisComp.layer(“Color”) so I’m wondering what’s wrong with this part of the code

[Michael Szalapski]

Walter’s reply should cover you on how to make the dropdown with this link.