Question

我有这个功能，根本没有被调用。我没有事件准备印制segue ...

override func prepareForSegue(segue: UIStoryboardSegue, sender: AnyObject?) {
        println("PREPARING FOR SEGUE");
        if (segue.identifier == "ToChatRoom") {
            var chatView:ChatRoomViewController = segue.destinationViewController as ChatRoomViewController;
            var index = coralReefTableView.indexPathForSelectedRow()!.row;
            var id = roomIDArray.objectAtIndex(index);
            println("ID IS : \(id)");
            chatView.selectedRoomID = id as Int;
        }
    }

我用这些代码行调用segue ......

    func tableView(tableView: UITableView, didSelectRowAtIndexPath indexPath: NSIndexPath) {
        var cell = tableView.cellForRowAtIndexPath(indexPath);
        //performSegueWithIdentifier("ToChatRoom", sender: self);
        let nextController:AnyObject! = self.storyboard?.instantiateViewControllerWithIdentifier("chatRoom");
        self.showViewController(nextController as UIViewController, sender: nextController);

当我取消注释performSegueWithIdentifier时，我收到此错误：exc_breakpoint（code = exc_i386_bpt subcode = 0x0）。我想知道这会是什么原因？

Answer 1

您的ChatRoomViewController是否在导航控制器中？如果是，则segue.destinationViewController不会指向ChatRoom控制器？只是一个猜测。在这种情况下，您需要以下内容：

if segue.identifier == "ToChatRoom" {
   let navigationController = segue.destinationViewController as UINavigationController
   let chatView = navigationController.viewControllers[0] as ChatRoomViewController
   var index = ...

当performSegueWithIdentifier行被注释掉时，从不调用prepareForSegue。从故事板中明确实例化的下一行绕过了它的需要。

swift prepareForSegue not working / exc_breakpoint（code = exc_i386_bpt subcode = 0x0）

1 个答案: