[英]Almost iSO8601 format
I need to parse the following date in a JSON payload and it's almost in ISO8601 format.我需要在 JSON 有效载荷中解析以下日期,它几乎是 ISO8601 格式。
"2020-06-05 14:52:54 UTC"
To conform to ISO8601 it needs to be altered slightly.为了符合 ISO8601,它需要稍作改动。
"2020-06-05T14:52:54Z"
It's super annoying because I now have to make a customer date decoding strategy.这非常烦人,因为我现在必须制定客户日期解码策略。
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.我无法控制数据源。 Is there anything I can do to avoid a custom decoding strategy in this case?
在这种情况下,我能做些什么来避免自定义解码策略吗?
Since you seem to just want to use a DateFormatter
to parse the date string, use the formatted
strategy.由于您似乎只想使用
DateFormatter
来解析日期字符串,因此请使用formatted
策略。
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)
An alternative to Sweeper's answer is to write an extension of DateFormatter
Sweeper 答案的替代方法是编写
DateFormatter
的扩展
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然后你的
make()
function 就变成了
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
还考虑将
make()
(具有更有意义的名称)放在JSONDecoder
的扩展中
You could map your input into iso8601, to use an existing decoder:您可以将map 输入到 iso8601,以使用现有的解码器:
func iso8601ify(_ str: String) -> String {
str.split(separator: " ")
.prefix(2)
.joined(separator: "T")
.appending("Z")
}
This example, there's no error handling, of course这个例子,当然没有错误处理
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