Almost iSO8601 format

user_9400730 0
Viewed 54

I need to parse the following date in a JSON payload and it's almost in ISO8601 format.

"2020-06-05 14:52:54 UTC"

To conform to ISO8601 it needs to be altered slightly.

"2020-06-05T14:52:54Z"

It's super annoying because I now have to make a customer date decoding strategy.

static func make() -> JSONDecoder {
    let decoder = JSONDecoder()
    decoder.keyDecodingStrategy = .convertFromSnakeCase
//        decoder.dateDecodingStrategy = .iso8601
    decoder.dateDecodingStrategy = .custom({ decoder in
        let container = try decoder.singleValueContainer()
        let dateStr = try container.decode(String.self)
        guard let date = formatter.date(from: dateStr) else {
            preconditionFailure("Unexpected date format.")
        }
        return date
    })
    return decoder
}

I don't have control of the data source. Is there anything I can do to avoid a custom decoding strategy in this case?

swift json ios
3 Answers

Since you seem to just want to use a DateFormatter to parse the date string, use the formatted strategy.

let formatter = DateFormatter()

formatter.dateFormat = "yyyy-MM-dd HH:mm:ss zzzz" 
// or 
// formatter.dateFormat = "yyyy-MM-dd HH:mm:ss 'UTC'"
// formatter.timeZone = .init(identifier: "UTC")

formatter.locale = Locale(identifier: "en_US_POSIX")
decoder.dateDecodingStrategy = .formatted(formatter)
user_51335850

An alternative to Sweeper's answer is to write an extension of DateFormatter

extension DateFormatter {
    static let almostISO8601Formatter: DateFormatter = {
        let formatter = DateFormatter()
        formatter.calendar = Calendar(identifier: .iso8601)
        formatter.locale = Locale(identifier: "en_US_POSIX")
        formatter.dateFormat = "yyyy-MM-dd HH:mm:ss Z"
        return formatter
    }()
}

Then your make() function simply becomes

static func make() -> JSONDecoder {
    let decoder = JSONDecoder()
    decoder.keyDecodingStrategy = .convertFromSnakeCase
    decoder.dateDecodingStrategy = .formatted(.almostISO8601Formatter)
    return decoder
}

Consider also to put make() (with a more meaningful name) in an extension of JSONDecoder

user_50440420

You could map your input into iso8601, to use an existing decoder:

func iso8601ify(_ str: String) -> String {
    str.split(separator: " ")
        .prefix(2)
        .joined(separator: "T")
        .appending("Z")
}

This example, there's no error handling, of course

user_47736100
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