2015-09-05 44 views
1

一个WKInterfaceLabel的Text属性我想键入InterfaceController if语句基本上去喜欢 -如何访问Apple关注应用

if label.text == "Hello" { 
//execute some code 
} 

WKInterfaceLabel似乎只是有setText()财产。那么如何访问WKInterfaceLabel的文本属性?提前致谢!

+0

回答每个问题的第一步:谷歌你遇到问题的类。这将永远给你一个链接从苹果与所有可用的功能,变量和类常量。 [WKInterfaceLabel类参考](https://developer.apple.com/library/prerelease/ios/documentation/WatchKit/Reference/WKInterfaceLabel_class/) –

回答

2

简答:你不能。

较长的答案,将字符串存储在一个单独的变量中,并将其用于if语句。

class InterfaceController: WKInterfaceController { 

    @IBOutlet var myLabel: WKInterfaceLabel! 
    var myString : String = "" 

    override func awakeWithContext(context: AnyObject?) { 
     super.awakeWithContext(context) 

     myString = "hello" 
     myLabel.setText(myString) 

     if myString == "hello" { 

      //so awesome stuff here 

     } 
     // Configure interface objects here. 
    } 

    override func willActivate() { 
     // This method is called when watch view controller is about to be visible to user 
     super.willActivate() 
    } 

    override func didDeactivate() { 
     // This method is called when watch view controller is no longer visible 
     super.didDeactivate() 
    } 

} 
+0

再次感谢! –