我需要在我的应用程序中创建Firebase深度链接,以便使用UIActivityViewController在不同平台上共享它们。 为了创建这些链接,我使用以下代码和URLQueryItem来获取我将与其他应用程序共享的URL:
func createURLWithComponents(EventToUrl: Event) -> URL? {
var urlComponents = URLComponents()
urlComponents.scheme = "https"
urlComponents.host = "www.example.com"
urlComponents.path = "/"
// add params
let id = URLQueryItem(name: "id", value: EventToUrl.eventID)
print(EventToUrl.name)
let name = URLQueryItem(name: "name", value: EventToUrl.name)
print(name.description)
print(name.value)
let photo = URLQueryItem(name: "photo", value: EventToUrl.photoUrl)
let created = URLQueryItem(name: "created", value: EventToUrl.creationDate )
let participants = URLQueryItem(name: "participants", value: String(EventToUrl.Participants.count) )
let place = URLQueryItem(name: "place", value: EventToUrl.Places[0].name)
let time = URLQueryItem(name: "time", value: EventToUrl.time[0].TimestampString)
urlComponents.queryItems = [id, name, photo, created, participants , place , time]
let dynamicLinkDomain : String = "***.app.goo.gl"
var deepLink = URLComponents()
deepLink.scheme = "https"
deepLink.host = dynamicLinkDomain
deepLink.path = "/"
let link = URLQueryItem(name: "link", value: urlComponents.url?.absoluteString)
let apn = URLQueryItem(name: "apn", value: "***" )
let ibi = URLQueryItem(name: "ibi", value: "***" )
let d = URLQueryItem(name: "d", value: "1")
deepLink.queryItems = [link, apn, ibi, d]
print(deepLink.url?.absoluteString)
print(deepLink.url)
return deepLink.url
}
这是我发送给他的链接:
https://***.app.goo.gl/?link=https://www.example.com/?id%3D-KZYfNRCXdSqUG72FmEQ%26name%3Djeremie's%2520Event%26photo%3D%26created%3D%26participants%3D1%26place%3DLa%2520Piazza%26time%3D1482362284.50304&apn=***&ibi=***&d=1
https://.app.goo.gl/?link=https://www.example.com/?id%3D-KZYfNRCXdSqUG72FmEQ%26name%3Djeremie's%2520Event%26photo%3D%26created%3D%26participants%3D1%26place%3DLa%2520Piazza%26time%3D1482362284.50304&apn=&ibi=***&d=1
如您所见,如果活动名称包含“ ' ”字符(Jeremie's event),URLQueryItem将无法正确编码此字符(在我的Android应用程序中,系统功能会将其转换为“'” -->%27)。我该如何对其进行编码?
PS:我尝试使用 addingPercentEncoding(withAllowedCharacters: .urlFragmentAllowed)
,但没有成功...
谢谢您的帮助!
+
字符。你要么使用.urlQueryAllowed
并删除一些字符,要么自己手动构建。请参见 https://dev59.com/oJTfa4cB1Zd3GeqPXeqS#35912606(适用于Swift 2)或https://dev59.com/yZnga4cB1Zd3GeqPa5mP#38798445(适用于Swift 3)。 - Rob