Swift turn a country code into a emoji flag via unicode
SwiftUnicodeSwift Problem Overview
I'm looking for a quick way to turn something like:
let germany = "DE"
into
let flag = "\u{1f1e9}\u{1f1ea}"
ie, what's the mapping of D to 1f1e9 and E to 1f1ea
I was looking at .utf8 for the string, but this returns an integer.
FWIW my general goal is to be able to take an arbitrary country code and get the corresponding emoji flag.
EDIT: I'm also fine with just holding a table that does this mapping if its available somewhere. I googled around but didn't find it.
Swift Solutions
Solution 1 - Swift
Here's a general formula for turning a two-letter country code into its emoji flag:
func flag(country:String) -> String {
let base = 127397
var usv = String.UnicodeScalarView()
for i in country.utf16 {
usv.append(UnicodeScalar(base + Int(i)))
}
return String(usv)
}
let s = flag("DE")
EDIT Ooops, no need to pass through the nested String.UnicodeScalarView struct. It turns out that String has an append method for precisely this purpose. So:
func flag(country:String) -> String {
let base : UInt32 = 127397
var s = ""
for v in country.unicodeScalars {
s.append(UnicodeScalar(base + v.value))
}
return s
}
EDIT Oooops again, in Swift 3 they took away the ability to append a UnicodeScalar to a String, and they made the UnicodeScalar initializer failable (Xcode 8 seed 6), so now it looks like this:
func flag(country:String) -> String {
let base : UInt32 = 127397
var s = ""
for v in country.unicodeScalars {
s.unicodeScalars.append(UnicodeScalar(base + v.value)!)
}
return String(s)
}
Solution 2 - Swift
If anyone looking for solution in ObjectiveC here is convenient category:
@interface NSLocale (RREmoji)
+ (NSString *)emojiFlagForISOCountryCode:(NSString *)countryCode;
@end
@implementation NSLocale (RREmoji)
+ (NSString *)emojiFlagForISOCountryCode:(NSString *)countryCode {
NSAssert(countryCode.length == 2, @"Expecting ISO country code");
int base = 127462 -65;
wchar_t bytes[2] = {
base +[countryCode characterAtIndex:0],
base +[countryCode characterAtIndex:1]
};
return [[NSString alloc] initWithBytes:bytes
length:countryCode.length *sizeof(wchar_t)
encoding:NSUTF32LittleEndianStringEncoding];
}
@end
test:
for ( NSString *countryCode in [NSLocale ISOCountryCodes] ) {
NSLog(@"%@ - %@", [NSLocale emojiFlagForISOCountryCode:countryCode], countryCode);
}
output: - AD - AE - AF - AG - AI ...
Solution 3 - Swift
Two optimizations of matt's answer.
- No need to pass through the nested String in Swift 4
- To avoid pass lower case string, I added uppercased()
Here is the code.
func flag(from country:String) -> String {
let base : UInt32 = 127397
var s = ""
for v in country.uppercased().unicodeScalars {
s.unicodeScalars.append(UnicodeScalar(base + v.value)!)
}
return s
}
Solution 4 - Swift
I know this is not exactly what was asked, but maybe it will help someone:
let flags: [String: String] = [
"AD": "🇦🇩", "AE": "🇦🇪", "AF": "🇦🇫", "AG": "🇦🇬", "AI": "🇦🇮", "AL": "🇦🇱", "AM": "🇦🇲", "AO": "🇦🇴", "AQ": "🇦🇶", "AR": "🇦🇷", "AS": "🇦🇸", "AT": "🇦🇹", "AU": "🇦🇺", "AW": "🇦🇼", "AX": "🇦🇽", "AZ": "🇦🇿", "BA": "🇧🇦", "BB": "🇧🇧", "BD": "🇧🇩", "BE": "🇧🇪", "BF": "🇧🇫", "BG": "🇧🇬", "BH": "🇧🇭", "BI": "🇧🇮", "BJ": "🇧🇯", "BL": "🇧🇱", "BM": "🇧🇲", "BN": "🇧🇳", "BO": "🇧🇴", "BQ": "🇧🇶", "BR": "🇧🇷", "BS": "🇧🇸", "BT": "🇧🇹", "BV": "🇧🇻", "BW": "🇧🇼", "BY": "🇧🇾", "BZ": "🇧🇿", "CA": "🇨🇦", "CC": "🇨🇨", "CD": "🇨🇩", "CF": "🇨🇫", "CG": "🇨🇬", "CH": "🇨🇭", "CI": "🇨🇮", "CK": "🇨🇰", "CL": "🇨🇱", "CM": "🇨🇲", "CN": "🇨🇳", "CO": "🇨🇴", "CR": "🇨🇷", "CU": "🇨🇺", "CV": "🇨🇻", "CW": "🇨🇼", "CX": "🇨🇽", "CY": "🇨🇾", "CZ": "🇨🇿", "DE": "🇩🇪", "DJ": "🇩🇯", "DK": "🇩🇰", "DM": "🇩🇲", "DO": "🇩🇴", "DZ": "🇩🇿", "EC": "🇪🇨", "EE": "🇪🇪", "EG": "🇪🇬", "EH": "🇪🇭", "ER": "🇪🇷", "ES": "🇪🇸", "ET": "🇪🇹", "FI": "🇫🇮", "FJ": "🇫🇯", "FK": "🇫🇰", "FM": "🇫🇲", "FO": "🇫🇴", "FR": "🇫🇷", "GA": "🇬🇦", "GB": "🇬🇧", "GD": "🇬🇩", "GE": "🇬🇪", "GF": "🇬🇫", "GG": "🇬🇬", "GH": "🇬🇭", "GI": "🇬🇮", "GL": "🇬🇱", "GM": "🇬🇲", "GN": "🇬🇳", "GP": "🇬🇵", "GQ": "🇬🇶", "GR": "🇬🇷", "GS": "🇬🇸", "GT": "🇬🇹", "GU": "🇬🇺", "GW": "🇬🇼", "GY": "🇬🇾", "HK": "🇭🇰", "HM": "🇭🇲", "HN": "🇭🇳", "HR": "🇭🇷", "HT": "🇭🇹", "HU": "🇭🇺", "ID": "🇮🇩", "IE": "🇮🇪", "IL": "🇮🇱", "IM": "🇮🇲", "IN": "🇮🇳", "IO": "🇮🇴", "IQ": "🇮🇶", "IR": "🇮🇷", "IS": "🇮🇸", "IT": "🇮🇹", "JE": "🇯🇪", "JM": "🇯🇲", "JO": "🇯🇴", "JP": "🇯🇵", "KE": "🇰🇪", "KG": "🇰🇬", "KH": "🇰🇭", "KI": "🇰🇮", "KM": "🇰🇲", "KN": "🇰🇳", "KP": "🇰🇵", "KR": "🇰🇷", "KW": "🇰🇼", "KY": "🇰🇾", "KZ": "🇰🇿", "LA": "🇱🇦", "LB": "🇱🇧", "LC": "🇱🇨", "LI": "🇱🇮", "LK": "🇱🇰", "LR": "🇱🇷", "LS": "🇱🇸", "LT": "🇱🇹", "LU": "🇱🇺", "LV": "🇱🇻", "LY": "🇱🇾", "MA": "🇲🇦", "MC": "🇲🇨", "MD": "🇲🇩", "ME": "🇲🇪", "MF": "🇲🇫", "MG": "🇲🇬", "MH": "🇲🇭", "MK": "🇲🇰", "ML": "🇲🇱", "MM": "🇲🇲", "MN": "🇲🇳", "MO": "🇲🇴", "MP": "🇲🇵", "MQ": "🇲🇶", "MR": "🇲🇷", "MS": "🇲🇸", "MT": "🇲🇹", "MU": "🇲🇺", "MV": "🇲🇻", "MW": "🇲🇼", "MX": "🇲🇽", "MY": "🇲🇾", "MZ": "🇲🇿", "NA": "🇳🇦", "NC": "🇳🇨", "NE": "🇳🇪", "NF": "🇳🇫", "NG": "🇳🇬", "NI": "🇳🇮", "NL": "🇳🇱", "NO": "🇳🇴", "NP": "🇳🇵", "NR": "🇳🇷", "NU": "🇳🇺", "NZ": "🇳🇿", "OM": "🇴🇲", "PA": "🇵🇦", "PE": "🇵🇪", "PF": "🇵🇫", "PG": "🇵🇬", "PH": "🇵🇭", "PK": "🇵🇰", "PL": "🇵🇱", "PM": "🇵🇲", "PN": "🇵🇳", "PR": "🇵🇷", "PS": "🇵🇸", "PT": "🇵🇹", "PW": "🇵🇼", "PY": "🇵🇾", "QA": "🇶🇦", "RE": "🇷🇪", "RO": "🇷🇴", "RS": "🇷🇸", "RU": "🇷🇺", "RW": "🇷🇼", "SA": "🇸🇦", "SB": "🇸🇧", "SC": "🇸🇨", "SD": "🇸🇩", "SE": "🇸🇪", "SG": "🇸🇬", "SH": "🇸🇭", "SI": "🇸🇮", "SJ": "🇸🇯", "SK": "🇸🇰", "SL": "🇸🇱", "SM": "🇸🇲", "SN": "🇸🇳", "SO": "🇸🇴", "SR": "🇸🇷", "SS": "🇸🇸", "ST": "🇸🇹", "SV": "🇸🇻", "SX": "🇸🇽", "SY": "🇸🇾", "SZ": "🇸🇿", "TC": "🇹🇨", "TD": "🇹🇩", "TF": "🇹🇫", "TG": "🇹🇬", "TH": "🇹🇭", "TJ": "🇹🇯", "TK": "🇹🇰", "TL": "🇹🇱", "TM": "🇹🇲", "TN": "🇹🇳", "TO": "🇹🇴", "TR": "🇹🇷", "TT": "🇹🇹", "TV": "🇹🇻", "TW": "🇹🇼", "TZ": "🇹🇿", "UA": "🇺🇦", "UG": "🇺🇬", "UM": "🇺🇲", "US": "🇺🇸", "UY": "🇺🇾", "UZ": "🇺🇿", "VA": "🇻🇦", "VC": "🇻🇨", "VE": "🇻🇪", "VG": "🇻🇬", "VI": "🇻🇮", "VN": "🇻🇳", "VU": "🇻🇺", "WF": "🇼🇫", "WS": "🇼🇸", "YE": "🇾🇪", "YT": "🇾🇹", "ZA": "🇿🇦", "ZM": "🇿🇲", "ZW": "🇿🇼"
]
Solution 5 - Swift
Another function for turning a two-letter country code into its emoji flag using Swift 5.
internal func getFlag(from countryCode: String) -> String {
countryCode
.unicodeScalars
.map({ 127397 + $0.value })
.compactMap(UnicodeScalar.init)
.map(String.init)
.joined()
}
Solution 6 - Swift
To give more insight into matt answer
Swift 2 version
public static func flag(countryCode: String) -> Character {
let base = UnicodeScalar("🇦").value - UnicodeScalar("A").value
let string = countryCode.uppercaseString.unicodeScalars.reduce("") {
var string = $0
string.append(UnicodeScalar(base + $1.value))
return string
}
return Character(string)
}
Swift 3 version, taken from https://github.com/onmyway133/Smile/blob/master/Sources/Smile.swift#L52
public func emoji(countryCode: String) -> Character {
let base = UnicodeScalar("🇦").value - UnicodeScalar("A").value
var string = ""
countryCode.uppercased().unicodeScalars.forEach {
if let scala = UnicodeScalar(base + $0.value) {
string.append(String(describing: scala))
}
}
return Character(string)
}
Solution 7 - Swift
For a more functional approach, using no mutable variables, use this:
private func flag(country: String) -> String {
let base: UInt32 = 127397
return country.unicodeScalars
.flatMap({ UnicodeScalar(base + $0.value) })
|> String.UnicodeScalarView.init
|> String.init
}
Where the |> operator is the function application operator, working like a "pipe" for a more natural reading order: We take the scalars, map them into new scalars, turn that into a view, and that into a string.
It's defined like so:
infix operator |> : MultiplicationPrecedence
func |> <T, U>(left: T, right: (T) -> U) -> U {
return right(left)
}
Without custom operators, we can still do without mutable state, like so:
private func flag(country: String) -> String {
let base: UInt32 = 127397
return String(String.UnicodeScalarView(
country.unicodeScalars.flatMap({ UnicodeScalar(base + $0.value) })
))
}
But IMHO, this reads a little "backwards", since the natural flow of operations read neither out-in, nor in-out, but a little bit of both.