Transform GEOSwift.JSON into a Swift struct

Question

Transform GEOSwift.JSON into a Swift struct

In my GEOSwift implementation, I have a feature structured like this:

{
  "type" : "Feature",
  "geometry" : {
    "type" : "Point",
    "coordinates" : [
      -xx.xxxxxxxxxxxxxxx,
      xx.xxxxxxxxxxxxxxx
    ]
  },
  "properties" : {
    "mapLayer" : "MyMapLayer",
    "data" : {
      "id" : 42,
      "sizeClass" : "Large",
    // and so on...
    },
    "featureType" : "MyFeatureType"
  }
}

My goal is to extract the data section and map it to a struct as follows:

struct MyStruct: Decodable {
    var id: Int
    var sizeClass: String?
    // and more properties...
}

This snippet of code above retrieves the data part, however, it is of type GEOSwift.JSON, and converting it into string to decode using JSONDecoder class is unclear.

if case let .object(data) = feature.properties?["data"] {
  // process data: GEOSwift.JSON to convert it into MyStruct
}

Below is the GEOSwift.JSON enum definition:

import Foundation

public enum JSON: Hashable, Sendable {
    case string(String)
    case number(Double)
    case boolean(Bool)
    case array([JSON])
    case object([String: JSON])
    case null

    /// Recursively unwraps and returns the associated value
    public var untypedValue: Any {
        switch self {
        case let .string(string):
            return string
        case let .number(number):
            return number
        case let .boolean(boolean):
            return boolean
        case let .array(array):
            return array.map { $0.untypedValue }
        case let .object(object):
            return object.mapValues { $0.untypedValue }
        case .null:
            return NSNull()
        }
    }

   // Additional extensions for JSON initialization
}

json swift geojson geoswift

Answer 1

Answer №1

To convert feature.properties["data"] back to JSON and then decode it into MyStruct, you need to first encode it as GEOSwift.JSON does not conform to Decodable directly. However, it does conform to Encodable, so the process involves encoding it and then decoding it back.

let myStructData = feature.properties?["data"] ?? nil
let jsonData = try! JSONEncoder().encode(myStructData)
let decoder = JSONDecoder()
myStruct = try decoder.decode(MyStruct.self, from: jsonData)

Answer 2

To convert feature.properties["data"] back to JSON and then decode it into MyStruct, you need to first encode it as GEOSwift.JSON does not conform to Decodable directly. However, it does conform to Encodable, so the process involves encoding it and then decoding it back.

let myStructData = feature.properties?["data"] ?? nil
let jsonData = try! JSONEncoder().encode(myStructData)
let decoder = JSONDecoder()
myStruct = try decoder.decode(MyStruct.self, from: jsonData)

Answer 3

Answer №2

In addition to the answer provided earlier, it is worth noting that the GEOSwift.JSON enum is encodable, allowing for direct conversion to JSON data using JSONEncoder. This eliminates the need for an intermediate step of converting GEOSwift.JSON to a dictionary and then to Data. The code snippet would look something like this:

import Foundation
import GEOSwift

struct MyStruct: Decodable {
    let id: Int
    let sizeClass: String
    // and so forth...
}

if let feature = mySomething as? GeoJSON.Feature {
    // Extract the properties as GEOSwift.JSON
    if let data = feature.properties?["data"] {
        do {
            // Re-encode GEOSwift.JSON back to JSON Data
            let jsonData = try JSONEncoder().encode(data)

            // Decode JSON Data into your struct
            let decoder = JSONDecoder()
            let myStruct = try decoder.decode(MyStruct.self, from: jsonData)

            // Utilize `myStruct`
            print(myStruct)
        } catch {
            print("Error: \(error)")
        }
    }
} else {
    // Handle other scenarios here
    print("GeoJSON object is not a Feature")
}

This method should work effectively as long as the structure of the GEOSwift.JSON object aligns with your MyStruct. If there is a mismatch in structure, errors will occur during the decoding process.

Alternatively, another option mentioned in the comments is to use MKGeoJSONDecoder which can be a suitable alternative to GEOSwift when working specifically with map-related data using Apple's MapKit framework.
This approach offers a convenient way to parse GeoJSON data and convert it into objects suitable for display on a map.

A basic example of how you could utilize MKGeoJSONDecoder for parsing GeoJSON data is outlined below:

import MapKit

// Assuming you have GeoJSON data stored in a variable named `geoJsonData` of type Data
let geoJsonObject: [MKGeoJSONObject]
do {
    geoJsonObject = try MKGeoJSONDecoder().decode(geoJsonData)
} catch {
    print("Failed to decode GeoJSON: \(error)")
    return
}

// You can then handle the geoJsonObject based on its type
for object in geoJsonObject {
    switch object {
    case let feature as MKGeoJSONFeature:
        // Process feature
        print("Feature: \(feature)")
    case let geometry as MKShape & MKGeoJSONObject:
        // Handle geometry
        print("Geometry: \(geometry)")
    default:
        print("Unexpected type")
    }
}

This snippet parses GeoJSON data and displays each feature or geometry present in the data. MKGeoJSONFeature objects contain details about the feature's properties which can be accessed via the properties property of type Data. Should your features include properties represented as JSON objects, the JSONDecoder can assist in decoding these properties into Swift types defined by you.

It is important to note that MKGeoJSONDecoder is tailored for use with the MapKit framework and intended for map-related data operations. For applications not involving maps or requiring processing of non-map-centric GeoJSON data, an alternative such as GEOSwift or a general-purpose GeoJSON library might be more suitable.

Answer 4

In addition to the answer provided earlier, it is worth noting that the GEOSwift.JSON enum is encodable, allowing for direct conversion to JSON data using JSONEncoder. This eliminates the need for an intermediate step of converting GEOSwift.JSON to a dictionary and then to Data. The code snippet would look something like this:

import Foundation
import GEOSwift

struct MyStruct: Decodable {
    let id: Int
    let sizeClass: String
    // and so forth...
}

if let feature = mySomething as? GeoJSON.Feature {
    // Extract the properties as GEOSwift.JSON
    if let data = feature.properties?["data"] {
        do {
            // Re-encode GEOSwift.JSON back to JSON Data
            let jsonData = try JSONEncoder().encode(data)

            // Decode JSON Data into your struct
            let decoder = JSONDecoder()
            let myStruct = try decoder.decode(MyStruct.self, from: jsonData)

            // Utilize `myStruct`
            print(myStruct)
        } catch {
            print("Error: \(error)")
        }
    }
} else {
    // Handle other scenarios here
    print("GeoJSON object is not a Feature")
}

This method should work effectively as long as the structure of the GEOSwift.JSON object aligns with your MyStruct. If there is a mismatch in structure, errors will occur during the decoding process.

Alternatively, another option mentioned in the comments is to use MKGeoJSONDecoder which can be a suitable alternative to GEOSwift when working specifically with map-related data using Apple's MapKit framework.
This approach offers a convenient way to parse GeoJSON data and convert it into objects suitable for display on a map.

A basic example of how you could utilize MKGeoJSONDecoder for parsing GeoJSON data is outlined below:

import MapKit

// Assuming you have GeoJSON data stored in a variable named `geoJsonData` of type Data
let geoJsonObject: [MKGeoJSONObject]
do {
    geoJsonObject = try MKGeoJSONDecoder().decode(geoJsonData)
} catch {
    print("Failed to decode GeoJSON: \(error)")
    return
}

// You can then handle the geoJsonObject based on its type
for object in geoJsonObject {
    switch object {
    case let feature as MKGeoJSONFeature:
        // Process feature
        print("Feature: \(feature)")
    case let geometry as MKShape & MKGeoJSONObject:
        // Handle geometry
        print("Geometry: \(geometry)")
    default:
        print("Unexpected type")
    }
}

This snippet parses GeoJSON data and displays each feature or geometry present in the data. MKGeoJSONFeature objects contain details about the feature's properties which can be accessed via the properties property of type Data. Should your features include properties represented as JSON objects, the JSONDecoder can assist in decoding these properties into Swift types defined by you.

It is important to note that MKGeoJSONDecoder is tailored for use with the MapKit framework and intended for map-related data operations. For applications not involving maps or requiring processing of non-map-centric GeoJSON data, an alternative such as GEOSwift or a general-purpose GeoJSON library might be more suitable.

Transform GEOSwift.JSON into a Swift struct

Answer №1

Answer №2

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