Swift (Xcode) - 从 Table Cell -> Action Button -> Segue 将值传递给下一个控制器

Question

我有 table 视图并且在每个单元格中都有一个按钮可以打开一个操作 sheet。一旦该操作 sheet 打开，就会有一个操作 "Report Progress"。当我按下那个动作按钮时，我想打开另一个具有 segue 标识符 "ShowProgressReport" 的视图控制器。在那个新的视图控制器中，我有一个属性 "ProjectName" 默认情况下是空的。我希望属性从以前的视图控制器中获取值。但是我无法在 "prepareForSegue" 方法中获取索引值。这是我编码的内容：

TableView

func tableView(tableView: UITableView, cellForRowAtIndexPath indexPath: NSIndexPath) -> ProjectTableViewCell {

        let cell = tableView.dequeueReusableCellWithIdentifier("ProjectViewCell", forIndexPath:indexPath) as! ProjectTableViewCell

        // Remove indenting of cell
        cell.separatorInset = UIEdgeInsetsZero
        cell.layoutMargins = UIEdgeInsetsZero

        // Set project name
        cell.ProjectName.text = self.ProjectsArray[indexPath.row] as? String

        // Set action button
        cell.ActionButton.tag = indexPath.row
        cell.ActionButton.addTarget(self, action: #selector(ProjectsController.projectActions(_:)), forControlEvents: .TouchUpInside)

        return cell

    }

按钮动作Sheet

@IBAction func projectActions(sender: UIButton) {
        let index = sender.tag

        let optionMenu = UIAlertController(title: nil, message: self.ProjectsArray[index] as? String, preferredStyle: .ActionSheet)

        // Report Progress
        let reportProgressAction = UIAlertAction(title: "Report Progress", style: .Default, handler: {
            (alert: UIAlertAction!) -> Void in
            self.performSegueWithIdentifier("ShowReportProgress", sender: self)
        })

        // Cancel
        let cancelAction = UIAlertAction(title: "Cancel", style: .Cancel, handler: {
            (alert: UIAlertAction!) -> Void in

        })

        optionMenu.addAction(reportProgressAction)
        optionMenu.addAction(cancelAction)

        self.presentViewController(optionMenu, animated: true, completion: nil)
    }

执行 Segue

override func prepareForSegue(segue: UIStoryboardSegue, sender: AnyObject?) {

        let index = sender?.tag

        if segue.identifier == "ShowReportProgress"
        {
            let upcoming: ReportProgressController = segue.destinationViewController as! ReportProgressController

            let ProjectName = self.ProjectsArray[index!] as? String --> Here it says: fatal error: unexpectedly found nil while unwrapping an Optional value
            upcoming.ProjectName = ProjectName!
        }

    }

有什么帮助吗？

Answer 1

在这一行

self.performSegueWithIdentifier("ShowReportProgress", sender: self)

您正在传递代表视图控制器的 self。您可能是想通过按钮。

self.performSegueWithIdentifier("ShowReportProgress", sender: sender)

您应该始终使用可选绑定来避免崩溃和解包保证存在的值。

if let button = sender where segue.identifier == "ShowReportProgress" {
    let upcoming: ReportProgressController = segue.destinationViewController as! ReportProgressController

    let ProjectName = self.ProjectsArray[button.tag] as! String
    upcoming.ProjectName = ProjectName
}

Answer 2

替换这个

self.performSegueWithIdentifier("ShowReportProgress", sender: self)

与

self.performSegueWithIdentifier("ShowReportProgress", sender: sender)

您在 performSegueWithIdentifier 中传递控制器引用而不是按钮引用。

Swift (Xcode) - 从 Table Cell -> Action Button -> Segue 将值传递给下一个控制器

Swift (Xcode) - Pass Value to Next Controller From Table Cell -> Action Button -> Segue

xcode

ios

uistoryboardsegue

swift