Question

我有一个文本字段，可以在图表中显示十个字符串。如果textfield one有文本，那么它将显示在textfield 2上，如果textfield 2有文本，那么它将转到textfield 3，依此类推。但由于某种原因，它总是在textfield 1中显示字符串。我附加了我的代码，这可能是一个愚蠢的错误。

@IBOutlet weak var name: UITextField!
let pone = UserDefaults.standard
let ptwo = UserDefaults.standard
let pthree = UserDefaults.standard
let pfour = UserDefaults.standard
let pfive = UserDefaults.standard
let psix = UserDefaults.standard
let pseven = UserDefaults.standard
let peight = UserDefaults.standard
let pnine = UserDefaults.standard
let pten = UserDefaults.standard
let pelleven = UserDefaults.standard
let ptwelve = UserDefaults.standard
let pthirteen = UserDefaults.standard
let pfourteen = UserDefaults.standard
let pfifteen = UserDefaults.standard

if pone.string(forKey: "number1") != nil{
        if !(pone.string(forKey: "number1")?.isEmpty)! {
            pone.set(name.text, forKey: "number1")
        }
    }else
    if ptwo.string(forKey: "number2") != nil{
        if !(ptwo.string(forKey: "number2")?.isEmpty)! {
            ptwo.set(name.text, forKey: "number2")
        }
    }else
    if pthree.string(forKey: "number3") != nil{
        if !(pthree.string(forKey: "number3")?.isEmpty)! {
            pthree.set(name.text, forKey: "number3")
        }
    }else
    if pfour.string(forKey: "number4") != nil{
        if !(pfour.string(forKey: "number4")?.isEmpty)! {
            pfour.set(name.text, forKey: "number4")
        }
    }else
    if pfive.string(forKey: "number5") != nil{
        if !(pfive.string(forKey: "number5")?.isEmpty)! {
            pfive.set(name.text, forKey: "number5")
        }
    }else
    if psix.string(forKey: "number6") != nil{
        if !(psix.string(forKey: "number6")?.isEmpty)! {
            psix.set(name.text, forKey: "number6")
        }
    }else
    if pseven.string(forKey: "number7") != nil{
        if !(pseven.string(forKey: "number7")?.isEmpty)! {
            pseven.set(name.text, forKey: "number7")
        }
    }else
    if peight.string(forKey: "number8") != nil{
        if !(peight.string(forKey: "number8")?.isEmpty)! {
            peight.set(name.text, forKey: "number8")
        }
    }else
    if pnine.string(forKey: "number9") != nil{
        if !(pnine.string(forKey: "number9")?.isEmpty)! {
            pnine.set(name.text, forKey: "number9")
        }
    }else
    if pten.string(forKey: "number10") != nil{
        if !(pten.string(forKey: "number10")?.isEmpty)! {
            pten.set(name.text, forKey: "number10")
        }
    }else
    if pelleven.string(forKey: "number11") != nil{
        if !(pelleven.string(forKey: "number11")?.isEmpty)! {
            pelleven.set(name.text, forKey: "number11")
        }
    }else
    if ptwelve.string(forKey: "number12") != nil{
        if !(ptwelve.string(forKey: "number12")?.isEmpty)! {
            ptwelve .set(name.text, forKey: "number12")
        }
    }else
    if pthirteen.string(forKey: "number13") != nil{
        if !(pthirteen.string(forKey: "number13")?.isEmpty)! {
            pthirteen.set(name.text, forKey: "number13")
        }
    }else
    if pfourteen.string(forKey: "number14") != nil{
        if !(pfourteen.string(forKey: "number14")?.isEmpty)! {
            pfourteen.set(name.text, forKey: "number14")
        }
    }else
    if pfifteen.string(forKey: "number15") != nil{
        if !(pfifteen.string(forKey: "number15")?.isEmpty)! {
        pfifteen.set(name.text, forKey: "number15")
        }
    }

Answer 1

这是我提出的替代所有这些：

let keys = (0...15).map{ "number\($0)" }
for key in keys {
    guard let existingValue = UserDefaults.standard.string(forKey: key),
        existingValue.isEmpty else {
        continue
    }

    UserDefaults.standard.set(name.text, forKey: key)
}

价值永远是零

1 个答案: