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ios - Swift 函数异步

转载 作者:行者123 更新时间:2023-11-30 11:38:39 25 4
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func passcodeViewController(_ passcodeViewController: TOPasscodeViewController, isCorrectCode code: String) -> Bool {

let userDefault = UserDefaults.standard
let tokenPinCode = userDefault.string(forKey: "tokenPinCode")
let mailData = self.emailField.text
let dataStruct = mailData!+"|"+tokenPinCode!
print("1")
self.checkToken(code: dataStruct) { (response) in
if(response[0] == "OK"){
print("2")
self.alertPasswordChange(text: "Podaj nowe hasło", code: dataStruct)
}else{
self.standardAlert(title: "Znaleziono błędy", message: "Podany kod jest błedny", ok: "Rozumiem")
self.werifyButton.isEnabled = true
}
}
print("3")
return false
}

函数返回:打印 -> 1 -> 3 -> 2

如何实现效果:打印 -> 1 -> 2 -> 3

最佳答案

使您的函数无效并传递可以处理 bool 值的完成处理程序。

func passcodeViewController(_ controller: Controller, code: String, @escaping handler: (Bool) -> ()) {
// Your logic
asyncRequest(...) {
response in
let result = ... // find whether code ok
handler(result)
}
}

你可以这样调用它:

passcodeViewController(controller, code: "$&36_$") {
(isOk: Bool) in
print(3)
print("code is ok: \(isOk)")
}

关于ios - Swift 函数异步,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/49462799/

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