xcode 获取深层链接 url iOS 中的参数值
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Get value of parameters in deep link url iOS
提问by Linar sen
everyone. My question is: How can I get data from deep link URL? I have two apps and I want to send data from app1 to app2 using the deep link. I have a button on app1 to click and open app2 then app 2 will get data from app1 by deep link URL.
每个人。我的问题是：如何从深层链接 URL 获取数据？我有两个应用程序，我想使用深层链接将数据从 app1 发送到 app2。我在 app1 上有一个按钮可以单击并打开 app2，然后 app 2 将通过深层链接 URL 从 app1 获取数据。
Here is my code of button send in app1:
这是我在 app1 中发送按钮的代码：
@IBAction func btnSend_Clicked(_ sender: Any) {
let text = self.txtInput.text?.replacingOccurrences(of: " ", with: "%20")
UIApplication.shared.open(URL(string: "myapp://?code=\(text!)")!, options: [:], completionHandler: nil)
}
so, How can i get data from deeplink url (code parameter) in app2?
那么，如何从 app2 中的深层链接 url（代码参数）获取数据？
Really Thanks for your help !!!!
真的很感谢你的帮助！！！！
回答by Sour LeangChhean
You implement this code in Appdelegate:
您在 Appdelegate 中实现此代码：
func application(_ app: UIApplication, open url: URL, options: [UIApplicationOpenURLOptionsKey : Any] = [:]) -> Bool {
let urlComponents = URLComponents(url: url, resolvingAgainstBaseURL: false)
let items = (urlComponents?.queryItems)! as [NSURLQueryItem]
if (url.scheme == "myapp") {
var vcTitle = ""
if let _ = items.first, let propertyName = items.first?.name, let propertyValue = items.first?.value {
vcTitle = url.query!//"propertyName"
}
}
return false
}
