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ios - 快速代码返回警报类型字符串不起作用

转载 作者:塔克拉玛干 更新时间:2023-11-02 22:58:55 26 4
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我发现无效返回警报但未找到字符串警报返回。尝试 Void 到 String

// return Void Alert
func returnAlert(title: String!, message: String! ,success: (() -> Void)? , cancel: (() -> Void)?) {
dispatch_async(dispatch_get_main_queue(), {
let alertController = UIAlertController(title:title,
message: "",
preferredStyle: UIAlertControllerStyle.Alert)

self.newQtyField = UITextField()
self.newQtyField.keyboardType = .NumberPad

func addTextField(textField: UITextField!){
// add the text field and make the result global
let row = self.array[Int(message)!]
textField.text = String(row.pbQty!)
self.newQtyField = textField
}


let cancelLocalized = NSLocalizedString("cancel", tableName: "activity", comment:"")
let okLocalized = NSLocalizedString("ok", tableName: "Localizable", comment:"")

let cancelAction: UIAlertAction = UIAlertAction(title: cancelLocalized,
style: .Cancel) {
action -> Void in cancel?()
}
let successAction: UIAlertAction = UIAlertAction(title: okLocalized,
style: .Default) {
action -> Void in success?()
}
alertController.addTextFieldWithConfigurationHandler(addTextField)
alertController.addAction(cancelAction)
alertController.addAction(successAction)


self.presentViewController(alertController, animated: true, completion: nil)
})
}

//底部调用乐趣 |我认为 success: { () to success: { textfield ,但无法正常工作......需要你的帮助

    returnAlert("title", message: "msg", success: { () -> Void in


})
{ () -> Void in
print("user canceled")
}

最佳答案

很难理解您在问题中所说的内容。

您应该将方法定义更改为:

func returnAlert(title: String!, message: String! ,success: ((text:String) -> Void)? , cancel: (() -> Void)?) {

并更改 successAction 以返回 UITextField 的值:

let successAction: UIAlertAction = UIAlertAction(title: okLocalized,
style: .Default) { action in
success?(text: self.newQtyField.text!)
}

关于ios - 快速代码返回警报类型字符串不起作用,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38517119/

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