恢复 segued iOS UIViewControllers 的更好方法答案

【问题标题】：A better way to restore segued iOS UIViewControllers恢复 segued iOS UIViewControllers 的更好方法
【发布时间】：2015-12-02 11:30:04
【问题描述】：

我有一个显示选项列表的 UITableViewController。点击每个选项会使用 segue 导航到另一个视图控制器

需要以下代码来恢复应用的状态。但是有更清洁的方法吗？我在列表中包含了处理两个选项的代码，但图像有 10 个选项！

class SettingsViewController:UITableViewController {

var isRestoration = false
var aboutVC:AboutViewController?
var feedbackVC:FeedbackViewController?

override func viewWillAppear(animated: Bool) {
    super.viewWillAppear(animated)
    if isRestoration {
        isRestoration = false
        restoreChildIfPresent()
    } else {
        // nullify so when child is closed, we don't reopen
        // on next restoration
        forgetChildren()
    }
}

override func restoreChildIfPresent() {
    if let aboutVC = aboutVC {
        self.navigationController?.pushViewController(aboutVC, animated:false)
    } else if let feedbackVC = feedbackVC {
        self.navigationController?.pushViewController(feedbackVC, animated:false)
    }
}

func forgetChildren() {
    aboutVC = nil
    charterVC = nil
    feedbackVC = nil
}

override func prepareForSegue(segue:UIStoryboardSegue, sender:AnyObject?) {
    aboutVC = nil
    feedbackVC = nil
    if segue.identifier == "aboutSegue" {
        aboutVC = segue.destinationViewController as? AboutViewController    
    } else if segue.identifier == "feedbackSegue" {
        feedbackVC = segue.destinationViewController as? FeedbackViewController
    }
}

    override func decodeRestorableStateWithCoder(coder:NSCoder) {
    isRestoration = true
    aboutVC = coder.decodeObjectForKey("aboutVC") as? AboutUsViewController
    feedbackVC = coder.decodeObjectForKey("feedbackVC") as? FeedbackViewController
    super.decodeRestorableStateWithCoder(coder)
}

override func encodeRestorableStateWithCoder(coder:NSCoder) {
    coder.encodeObject(aboutVC, forKey:"aboutVC")
    coder.encodeObject(feedbackVC, forKey:"feedbackVC")
    super.encodeRestorableStateWithCoder(coder)
}

我很惊讶我需要 encodeObject() 子 VC，因为它们没有嵌入（对于那些嵌入 UITabbarController 的子 VC 确实需要 encodeObject() 子）。

我的主要“不喜欢”是 restoreChildIfPresent()，随着更多 ViewControllers 的添加和对 forgetChildren() 的丑陋需求的增加，它会增长

【问题讨论】：

标签： segue state-restoration

【解决方案1】：

通过使用对 UIViewController 的单个引用，我已将问题中的代码简化为更易于管理的代码。但是结构是一样的。可以添加进一步的 Swifty-ness，但我对替代方法更感兴趣，或者确认这种方法是推荐的做法。

class SettingsViewController:UITableViewController {

var isRestoration = false
var childVC:UIViewController?

override func viewWillAppear(animated: Bool) {
    super.viewWillAppear(animated)
    if isRestoration {
        isRestoration = false
        restoreChildIfPresent()
    } else {
        // nullify so when child is closed, we don't reopen
        // on next restoration
        forgetChild()
    }
}

override func restoreChildIfPresent() {
    if let childVC = childVC {
        self.navigationController?.pushViewController(childVC, animated:false)
    }
}

func forgetChild() {
    childVC = nil
}

override func prepareForSegue(segue:UIStoryboardSegue, sender:AnyObject?) {
    childVC = nil
    if segue.identifier == "aboutSegue" {
        childVC = segue.destinationViewController    
    } else if segue.identifier == "feedbackSegue" {
        childVC = segue.destinationViewController
    }
}

override func decodeRestorableStateWithCoder(coder:NSCoder) {
    isRestoration = true
    childVC = coder.decodeObjectForKey("childVC")
    super.decodeRestorableStateWithCoder(coder)
}

override func encodeRestorableStateWithCoder(coder:NSCoder) {
    coder.encodeObject(childVC, forKey:"childVC")
    super.encodeRestorableStateWithCoder(coder)
}

【讨论】：