map.getstyle() can't get custom layer

[aji-geo]

map.getstyle() method can get many layer in style but no custom layers, like deck.gl layer. how can I get all the layers include deck.gl layer?

[kkaefer]

Custom layers are explicitly excluded from getStyle in

mapbox-gl-js/src/style/style.js

Lines 322 to 331 in 254674e

	_serializeLayers(ids: Array<string>): Array<Object> {
	const serializedLayers = [];
	for (const id of ids) {
	const layer = this._layers[id];
	if (layer.type !== 'custom') {
	serializedLayers.push(layer.serialize());
	}
	}
	return serializedLayers;
	}

. getStyle is used for serializing a style to a JSON data structure, and custom layers contain functions that are cannot be represented in JSON.

To obtain an individual custom layer, use getLayer().

[mayteio]

Can we make this a feature request then? We have a scenario where we're adding n custom layers, and need to track what has been added without knowing their ids ahead of time. I thought prefixing and filtering map.getStyle().layers for this prefix would work, then I found this issue.

[mayteio]

Seems it just searches the _layers object?

mapbox-gl-js/src/style/style.js

Lines 750 to 758 in 254674e

	/**
	* Return the style layer object with the given `id`.
	*
	* @param {string} id - id of the desired layer
	* @returns {?Object} a layer, if one with the given `id` exists
	*/
	getLayer(id: string): Object {
	return this._layers[id];
	}

Something like this could work:

/**
 * Return the style layer object with the given `id`.
 *
 * @param {string} id - id of the desired layer
 * @returns {?Object} a layer, if one with the given `id` exists
 */
getLayer(id: string): Object {
  return this._layers[id];
}

/**
 * Return style layers whose `id` matches a search string
 *
 * @param {string} search - search string to filter layer ids
 * @returns {?Object[]} an array of layers, if any that include the given `search` in their `id` exists
 */
getLayers(search: string): Object[] {
  return Object.entries(this._layers)
    .filter(([id]) => id.includes(search))
    .map(([,layer]) => layer)
}

// or if you want to keep it OOP for referential equality
getLayers(search: string): Object[] {
  const results = [];
  const ids = Object.entries(this._layers)
    .filter(([id]) => id.includes(search))
    .foreach(([id]) => results.push(this._layers[id]))
  return results;
}

Would be happy to make a PR if you'll accept it!

[mayteio]

map.getstyle() method can get many layer in style but no custom layers, like deck.gl layer. how can I get all the layers include deck.gl layer?

@AJIKO You can access it currently with map.style._layers[customLayerId] - though I'm sure there's a reason _layers is private. Are you able to elaborate @kkaefer?

[kkaefer]

@mayteio same explanation as above: the idea of getStyle is to return a valid Style JSON file that conforms to the style specification. Custom layers inherently can't be part of a style document because they contain functions and always require the use of runtime styling. Therefore, you'll have to use individual getLayer calls. I recommend against using map.style._layers[customLayerId] in production code because this API is private and might change in future versions of Mapbox GL.

[mayteio]

OK great, thanks for the explanation.

Unfortunately we have a situation with deck.gl where we have an array of layers that we add to Mapbox via @deck.gl/mapbox. If an item is removed from that array, we have no way of knowing what it’s ID was. Thus we prefix all layers with “custom-“, find them in mapbox, find the difference (I.e those who are in mapbox and not in the deck.gl array) and then remove them. The only way for us is to use map._layers as these custom layers are (rightly) not in the style JSON.

[marcofugaro]

So how do I get a list of all the layers I added?

getLayers even exists in mapbox-gl-native

[mayteio]

When you call map.addLayer, prefix your layer id with custom-. Then you can get a list of all added layers with map._layers.filter(l => l.id.includes('custom-')). You can then loop through these and remove them one by one.