Question

我正在关注iOS的示例： https://github.com/mapbox/ios-sdk-examples/blob/master/Examples/Swift/FeatureSelectionExample.swift 基本上它是这样的： 1）用户点击地图。 2）获取所选状态的名称。 3）突出显示所选状态。这是iOS中的Swift代码：

@objc @IBAction func handleMapTap(sender: UITapGestureRecognizer) {
    // Get the CGPoint where the user tapped.
    let spot = sender.location(in: mapView)

    // Access the features at that point within the state layer.
    let features = mapView.visibleFeatures(at: spot, styleLayerIdentifiers: Set([layerIdentifier]))

    // Get the name of the selected state.
    if let feature = features.first, let state = feature.attribute(forKey: "name") as? String {
        changeOpacity(name: state)
    } else {
        changeOpacity(name: "")
    }
}

func changeOpacity(name: String) {
    let layer = mapView.style?.layer(withIdentifier: layerIdentifier) as! MGLFillStyleLayer

    // Check if a state was selected, then change the opacity of the states that were not selected.
    if name.count > 0 {
        layer.fillOpacity = NSExpression(format: "TERNARY(name = %@, 1, 0)", name)
    } else {
        // Reset the opacity for all states if the user did not tap on a state.
        layer.fillOpacity = NSExpression(forConstantValue: 1)
    }
}

我需要在Android中做到这一点。我发现的最接近的示例是这样的： https://github.com/mapbox/mapbox-android-demo/blob/master/MapboxAndroidDemo/src/main/java/com/mapbox/mapboxandroiddemo/examples/query/SelectBuildingActivity.java

我可以使用以下代码获取属性并从那里获取状态名称：

 List<Feature> features = mapboxMap.queryRenderedFeatures(finalPoint, "building");

但是如何在Android中做到这一点？

layer.fillOpacity = NSExpression(format: "TERNARY(name = %@, 1, 0)", name)

Android版Mapbox：如何在图层中选择要素？

0 个答案: