How to use a dictionary in `codable` response? - ios

Question

I have the following response from a JSON query. How do I represent a dictionary as codable? I have shortened the JSON response to save space.
{
"result":[
{
"delivery_address": "",
"made_sla": "true",
"watch_list": "",
"upon_reject": "cancel",
"location": {
"link": "https://foo.com/api/now/table/cmn_location/753ac76f4fd9320066bfb63ca310c79b",
"value": "753ac76f4fd9320066bfb63ca310c79b"
}
}
]
}
struct ResultList : Codable {
let result: [Result]
}
struct Result : Codable {
let delivery_address: String
let made_sla: String
let watch_list: String
let upon_reject: String
let location: Location
}
struct Location: Codable {
let link: String?
let value: String?
}
let decoder = JSONDecoder()
do {
let todo = try decoder.decode(ResultList.self, from: responseData)
print("todo \(todo)")
} catch {
print("error trying to convert data to JSON")
print(error)
}
I'm getting the following error:
"Expected to decode Dictionary<String, Any> but found a string/data instead.", underlyingError: nil))

Answer 1

Based on all the comments, I believe the JSON you're actually decoding looks more like this:
{
"result": [{
"delivery_address": "",
"made_sla": "true",
"watch_list": "",
"upon_reject": "cancel",
"location": {
"link": "https://foo.com/api/now/table/cmn_location/753ac76f4fd9320066bfb63ca310c79b",
"value": "753ac76f4fd9320066bfb63ca310c79b"
}
},
{
"delivery_address": "",
"made_sla": "true",
"watch_list": "",
"upon_reject": "cancel",
"location": ""
}
]
}
So there are some records that have a location, and some that encode location as an empty string. Basically this is really messed up JSON on a whole lot of levels and whatever code generates it should be fixed. The bad news I'm sure that won't happen. The good news is we can fix it locally at least.
We're going to have to decode this by hand, so we might as well clean up all the rest of the mess while we're in here. The first thing is that the names don't match Swift naming conventions. We can fix that using CodingKeys. We can also provide real types (Bool and a Rejection enum) for the things that are currently mis-typed strings.
enum Rejection: String, Codable {
case cancel
}
struct Result : Codable {
let deliveryAddress: String
let madeSLA: Bool
let watchList: String
let uponReject: Rejection
let location: Location?
private enum CodingKeys: String, CodingKey {
case deliveryAddress = "delivery_address"
case madeSLA = "made_sla"
case watchList = "watch_list"
case uponReject = "upon_reject"
case location
}
}
Now we just need to be able to decode it. Notice that I made Location optional. Clearly it sometimes doesn't exist, so you either need a default value or it needs to be optional. I chose the latter. Decoding all of this is very straight forward:
init(from decoder: Decoder) throws {
let container = try decoder.container(keyedBy: CodingKeys.self)
deliveryAddress = try container.decode(String.self, forKey: .deliveryAddress)
madeSLA = try container.decode(String.self, forKey: .madeSLA) == "true"
watchList = try container.decode(String.self, forKey: .watchList)
uponReject = try container.decode(Rejection.self, forKey: .uponReject)
location = try? container.decode(Location.self, forKey: .location)
}
The last line is your actual question. It just says if we can't decode it as a Location, set it to nil. We could be stricter here and try first to decode it as a Location, and then as a String, and then check if the String is empty, but it feels reasonable to use nil here for any decoding failure.

Answer 2

Assuming your JSON is (please note the missing closing brace)
{
"result": [
{
"delivery_address": "",
"made_sla": "true",
"watch_list": "",
"upon_reject": "cancel",
"location": {
"link": "https://blah/blah/foo",
"value": "fsfdfr32r34rwfwffas"
}
}
]
}
You can decode these structs
struct Root : Decodable {
let result : [Result]
struct Result : Decodable {
let location: Location
}
}
struct Location: Decodable {
let link: String
let value: String
}
with
JSONDecoder().decode(Root.self, from: data)