在swift中使用url启动应用程序，然后在启动应用程序之前调用API

// 定义URL Scheme
let url = URL(string: "myapp://open")!

// 尝试打开URL
if UIApplication.shared.canOpenURL(url) {
    if let openURL = UIApplication.shared.open(url, options: [:], completionHandler: nil) {
        print("Opened URL: \(openURL)")
    }
} else {
    print("Cannot open URL")
}

// 定义Universal Link
let url = URL(string: "https://example.com/open")!

// 尝试打开URL
if let openURL = UIApplication.shared.open(url, options: [:], completionHandler: nil) {
    print("Opened URL: \(openURL)")
} else {
    print("Cannot open URL")
}

// 定义API URL
let apiURL = URL(string: "https://api.example.com/data")!

// 创建URLSession
let session = URLSession.shared

// 创建URLRequest
var request = URLRequest(url: apiURL)
request.httpMethod = "GET"

// 发送请求
let task = session.dataTask(with: request) { data, response, error in
    if let error = error {
        print("Error: \(error)")
        return
    }
    
    if let data = data {
        do {
            let json = try JSONSerialization.jsonObject(with: data, options: []) as? [String: Any]
            print("JSON: \(json ?? [:])")
        } catch {
            print("JSON parsing error: \(error)")
        }
    }
}

task.resume()