Allow useRefValue on some tokens, but not on others

[mihkeleidast]

I'm trying to work around this Figma Tokens issue: tokens-studio/figma-plugin#706

Our shadow tokens are defined as follows:

{
  "shadow": {
    "core": {
      "2": {
        "value": {
          "x": "0",
          "y": "1",
          "blur": "2",
          "spread": "0",
          "color": "{color.core.black.8}",
          "type": "dropShadow"
        }
      },
      "4": {
        "value": {
          "x": "0",
          "y": "2",
          "blur": "4",
          "spread": "0",
          "color": "{color.core.black.8}",
          "type": "dropShadow"
        }
      },
      "8": {
        "value": {
          "x": "0",
          "y": "4",
          "blur": "8",
          "spread": "0",
          "color": "{color.core.black.8}",
          "type": "dropShadow"
        }
      }
    },
    "elevation": {
      "small": {
        "value": ["{shadow.core.4}", "{shadow.core.2}"]
      },
      "medium": {
        "value": ["{shadow.core.8}", "{shadow.core.2}"]
      }
    }
  }
}

Our elevation shadows stack the different core shadows and we'd like to not have to duplicate those definitions. We would like to use useRefValue, so the token relationships would be visible in Figma Tokens as well. This is impossible currently due to how our tokens are defined and the FT issue I linked above.

Is there an easy way of adding some sort of filtering to useRefValue, so that the specific shadow.elevation.small.value would not use references, but everything else still would?

[jorenbroekema]

I see, yeah I guess we can make a little something to work around that problem, I propose a boolean metadata flag that you can put in your token (as a sibling to the value prop) to flag it was "ignore this one for useRefValue function".

#18

I think that should allow you to work around the problem then?

Let me know!

[mihkeleidast]

Ooh that seems nice! This is definitely a better workaround than not using references at all, but it does lose the internal color reference currently. If this is more difficult to solve, I understand, does not matter too much.