我需要在我的应用程序中创建 firebase 深层链接,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 来获取我将分享给其他应用程序的 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,'如果事件名称由“'”字符组成(Jeremie 的事件),URLQueryItem 不会正确编码此字符(在我的 android 应用程序中,系统函数将其转换为“'”--> %27)
我如何对其进行编码?
PS:我尝试使用 addingPercentEncoding(withAllowedCharacters: .urlFragmentAllowed)
没有成功......
感谢您的帮助!
最佳答案
我找到了一个比其他解决方案更简单的解决方案。 (至少对我来说是这样)
我像往常一样构造我的 URLComponents,但在从中获取 url 之前,我获取了 urlComponents.string 并将特殊代码替换为符号。我通过 addingPercentEncoding 运行该字符串,然后从该字符串创建一个 url。
let stringWithoutBS = urlComponents.string!.replacingOccurrences(of: "%27", with: "'")
//ignore the force unwrap
guard let encodedURL = stringWithoutBS.addingPercentEncoding(withAllowedCharacters: CharacterSet.urlQueryAllowed) else {
print("unhelpful message")
return nil
}
var correctURL = URL(string: encodedURL)
关于swift - 特殊字符在 URLQueryItems 中没有被正确编码,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/41273994/