所以我一直在努力创造一个"时期" class,具有以下属性:
class PTPeriod {
// MARK: Stored Properties
var start: Date
var end: Date
var place: PTPlace?
//MARK: DESIGNATED NITIALIZER
init(start: Date, end: Date, place: PTPlace? = nil) {
self.start = start
self.end = end
self.place = place
}
我想用一个接受如下字典的方便初始化它:
{
"close" : {
"day" : 1,
"time" : "0000"
},
"open" : {
"day" : 0,
"time" : "0900"
}
}
这是我的初始化程序,最初我将繁重的工作放入辅助方法中。但是,我得到了同样的错误,所以我删除了辅助方法,但错误仍在发生。不确定它认为我在呼唤自己。
更新:将日期格式移动到日期格式化程序扩展名中,但错误仍然存在!不确定为什么。新的init显示:
我的新方便初学者:
// MARK: Convenience init for initializing periods pulled from Google Place API
convenience init?(placeData: [String:Any], place: PTPlace? = nil) throws {
var start: Date? = nil
var end: Date? = nil
var formatter = DateFormatter()
do {
for (key, value) in placeData {
let period = value as! [String: Any]
if (key == "open") {
start = try formatter.upcoming_date(with: period)
} else if (key == "close") {
end = try formatter.upcoming_date(with: period)
} else {
break
}
}
if (start != nil && end != nil) { self.init(start: start!, end: end!, place: place) }
else { print("we f'd up") }
}
catch { print(error.localizedDescription) }
}
这是我的DateFormatter.upcoming_date方法:
func upcoming_date(with googlePlacePeriod: [String: Any]) throws -> Date? {
if let day = googlePlacePeriod["day"] as? Int {
if (day < 0) || (day > 6) { throw SerializationError.invalid("day", day) } // throw an error if day is not between 0-6
if let time = googlePlacePeriod["time"] as? String {
if time.characters.count != 4 { throw SerializationError.invalid("time", time) } // throw an error if time is not 4 char long
var upcoming_date: Date
let gregorian = Calendar(identifier: .gregorian)
let current_day_of_week = Date().getDayOfWeekInt()
let dayOfPeriod = day + 1
///TRUST THAT THIS WORKS... NO JODAS
var distance = dayOfPeriod - current_day_of_week // inverse = false
if (dayOfPeriod < current_day_of_week) {
switch distance {
case -1: distance = 6
case -2: distance = 5
case -3: distance = 4
case -4: distance = 3
case -5: distance = 2
case -6: distance = 1
default: break
}
}
/** time example: "1535" translates into 3:35PM
first two characters in string are "hour" (15)
last two characters in string are "minute(35)
**/
let hour = Int(time.substring(to: time.index(time.startIndex, offsetBy: 2)))!
let minute = Int(time.substring(from: time.index(time.endIndex, offsetBy: -2)))
upcoming_date = Date().fastForward(amount: distance, unit: "days", inverse: false)!
var components = gregorian.dateComponents([.year, .month, .day, .hour, .minute, .second], from: upcoming_date)
components.hour = hour
components.minute = minute
upcoming_date = gregorian.date(from: components)!
return upcoming_date
}
else { throw SerializationError.missing("time") }
}
else { throw SerializationError.missing("day") }
}
答案 0 :(得分:1)
你有一个可用的初始化程序,所以如果它失败了,你必须{ id: 'f212d399-f139-4d76-b892-7c2a83281f9b' }
让它知道它失败了:
return nil
另外,在if (start != nil && end != nil) {
self.init(start: start!, end: end!, place: place) }
} else {
print("we f'd up")
return nil
}
- do
中,您需要
重新catch
throw
块中的错误
catch
完全消除catch {
print(error.localizedDescription)
throw error
}
- do
个区块;或
只是catch
:
return nil
如果你这样做,你可能不会将这个可用的初始化程序定义为catch {
print(error.localizedDescription)
return nil
}
,因为你不再抛出它。
鉴于您不太可能对抛出的错误做任何有意义的事情,我只是让它成为一个不会抛出的可用的初始化器:
throws