等待调度信号时,将控制器视为弹出窗口

func tableView(_ tableView: UITableView, didSelectRowAt indexPath: IndexPath) {
    self.showPopUp() 
    let semaphore = DispatchSemaphore(value: 0)

    self.api.requestMedicationsByReagent(method: 1, ean: "", hash: medHash!, gen: generic) { output in
        semaphore.signal()
        self.objects = output
    }
     // Thread will wait here until async task closure is complete
     let _ = semaphore.wait(timeout: DispatchTime.distantFuture)
}

showPopUp的作用是:

func showPopUp() {
    let popOverVC = UIStoryboard(name: "Main", bundle: nil).instantiateViewController(withIdentifier: "loadingPopUp") as! PopUpViewController

    self.addChildViewController(popOverVC)
    popOverVC.view.frame = self.view.frame
    self.view.addSubview(popOverVC.view)
    popOverVC.didMove(toParentViewController: self)
}