Question

当我尝试对另一个视图控制器执行segue时出现此错误。我不知道为什么会收到此错误？

线程1：EXC_BAD_ACCESS

override func prepare(for segue: UIStoryboardSegue, sender: Any?) {
        if segue.identifier == "NormalPushupSegue" {
            let normalVC = segue.destination as! PopupViewController
            normalVC.formType = "Normal"
            performSegue(withIdentifier: "NormalPushupSegue", sender: self)
        }
        if segue.identifier == "DiamondPushupSegue" {
            let diamondVC = segue.destination as! PopupViewController
            diamondVC.formType = "Diamond"
            performSegue(withIdentifier: "DiamondPushupSegue", sender: self)
        }
        if segue.identifier == "WidePushupSegue" {
            let wideVC = segue.destination as! PopupViewController
            wideVC.formType = "Wide"
            performSegue(withIdentifier: "WidePushupSegue", sender: self)
        }
        if segue.identifier == "DeclinePushupSegue" {
            let declineVC = segue.destination as! PopupViewController
            declineVC.formType = "Decline"
            performSegue(withIdentifier: "DeclinePushupSegue", sender: self)
        }
    }

Answer 1

您不能在performSegue中调用prepare(for segue，因为它将被递归调用，因此总是会导致错误

全部删除

performSegue(withIdentifier: "NormalPushupSegue", sender: self)

Answer 2

首先最好安全地解开视图控制器。像这样：

if let myViewController = segue.destination as? MyViewController { 
    // Set up the VC, add some values and options
}

第二个-您不需要调用 performSegue ，执行segue已经在调用视图控制器。只需删除 performSegue

第三-您可以简化并应用这样的逻辑：

enum AppSegueName: String {
case Normal = "NormalPushupSegue"
case Diamond = "DiamondPushupSegue"
case Wide = "WidePushupSegue"
case Decline = "DeclinePushupSegue"
}

extension AppSegueName: CaseIterable {}

在准备函数中，使用switch \ case语句并将AppSegueName原始值与segue.identifier进行比较

像这样：

override func prepare(for segue: UIStoryboardSegue, sender: Any?) {
    AppSegueName.allCases.forEach {
        if $0.rawValue == segue.identifier {

            if let myViewController = segue.destination as? MyViewController { 
            myViewController.formType = $0 // formType is of type AppSegueName
            }
        }
    }
}

我正在尝试Segue并传递字符串，但是出现错误

2 个答案: