Question

我正在使用Xcode做一个项目。我有四个按钮，其中两个是在视图之间传递数据，并希望其他两个按钮只是跳转到另一个视图而不传递任何数据。前两个表现非常好，将数据传递给另一个视图。但当我点击其余两个时，应用程序崩溃说：

＆＃34; 2017-10-30 23:08:11.970500-0400 xxx [2557：65602]无法施放价值类型＆＃39; xxx.goalInfoViewController＆＃39; （0x107f01580）来＆＃39; xxx.IOViewController＆＃39; （0x107f019a0）＆＃34;

以下是代码：

@IBAction func monthlyIncomeAction(_ sender: Any) {
    numberForSegue = 1
    performSegue(withIdentifier: "outcomeSegue", sender: self)
}
@IBAction func MonthlyOutcomeAction(_ sender: Any) {
    numberForSegue = 2
    performSegue(withIdentifier: "incomeSegue", sender: self)
}
override func prepare(for segue: UIStoryboardSegue, sender: (Any)?) {
    let ioController = segue.destination as! IOViewController
    ioController.segueIndex = numberForSegue
}

Answer 1

您应该确保只有两个在视图之间传递数据的segue才能在prepare方法中运行代码。让我们尝试下面的prepare方法代码。

override func prepare(for segue: UIStoryboardSegue, sender: (Any)?) {
    if segue.identifier == "outcomeSegue" || segue.identifier == "incomeSegue" {
      return;
    }

    let ioController = segue.destination as! IOViewController
    ioController.segueIndex = numberForSegue
}

或者

override func prepare(for segue: UIStoryboardSegue, sender: (Any)?) {
    if segue.destination is IOViewController {
        let ioController = segue.destination as! IOViewController
        ioController.segueIndex = numberForSegue
    }
}

segue如何仅适用于某些指定的按钮

1 个答案: