Question

我一直在寻找，但找不到解决方案。有人可以让我知道我做错了什么。这是我的功能代码：在我的数据库中，我将小时= 5作为Int，分钟= 45作为Int 但是当我打印时，函数的值会打印0,0

var docRefF : DocumentReference!

func getTime()-> (Int, Int){
    var FHour = Int()
    var FMinute = Int()

    docRefF = Firestore.firestore().document("sampleTime/worktime")
    dataListener =  docRefF.addSnapshotListener { (docSnapshot, error) in
        guard let docSnapshot = docSnapshot, docSnapshot.exists else { return }

        let data = docSnapshot.data()
        let Hour:Int = data["Hour"]! as! Int
        let Minute: Int = data["Minute"]! as! Int

        FHour = Hour
        FMinute = Minute
    }

    return (FHour, FMinute)
}
let time = getTime()
print("\(time.0),\(time.1)" )
 //printed 0,0

Answer 1

显然是异步执行的问题。使用 * function with Closure * 而不是返回值的函数。

请参阅以下代码

var docRefF : DocumentReference!
func getTime(_ then:(_ first:Int, _ second:Int)->()){
    var FHour = Int()
    var FMinute = Int()


    docRefF = Firestore.firestore().document("sampleTime/worktime")
    dataListener =  docRefF.addSnapshotListener { (docSnapshot, error) in
     guard let docSnapshot = docSnapshot, docSnapshot.exists else { return }
   let data = docSnapshot.data()
    let Hour:Int = data["Hour"]! as! Int
    let Minute: Int = data["Minute"]! as! Int

        FHour = Hour
        FMinute = Minute

    //TODO:- Use other firebase related task same as your code, so that uncomment other lines same as your origional

    then(FHour, FMinute) // This is closure call back line

     }
   }

   //Calling your function
    getTime { (first, second) in

    print(first,second)

    }

Swift 4函数返回0表示多个返回值

1 个答案: