USEtoryBoardSegue方法的iOS Swift问题

Question

我正在尝试将Objective-C方法转换为Swift.

Objective-C方法如下:

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{
    if ([[segue identifier] isEqualToString:@"showDetail"])
    {
        NSIndexPath *indexPath = [self.tableView indexPathForSelectedRow];
        NSString *string = [feeds[indexPath.row] objectForKey: @"link"];

        string = [string stringByReplacingOccurrencesOfString:@"\n" withString:@" "];

        [[segue destinationViewController] setUrl:string];
    }
}

到目前为止我遇到的Swift方法(有问题)如下:

override func prepareForSegue(segue: UIStoryboardSegue!, sender: AnyObject!)
{
    if (segue.identifier == "showDetail") {

        var indexPath:NSIndexPath = self.tableView.indexPathForSelectedRow()
        var string:NSString = self.feeds[indexPath.row] as String

        string = string.stringByReplacingOccurrencesOfString("\n", withString: "")

        segue.destinationViewController.setURL(string, forKey: nil)
    }
}

主要问题是最后一行:

segue.destinationViewController.setURL(string, forKey: nil)

我收到以下错误:

"无法将表达式的类型'Void'转换为'NSURL!'"

这是我试图调用的控制器:

import UIKit

class DetailViewController: UIViewController
{
    var url:NSString!
    var webView:UIWebView!

    override func viewDidLoad()
    {
        super.viewDidLoad()

        var url:NSURL = NSURL(string:self.url.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding))
        var request:NSURLRequest = NSURLRequest(URL: url)
        webView.loadRequest(request)
    }
}

我希望有人可以帮助我解决这个问题,我已经有一段时间在努力解决这个问题了.:/

谢谢!

Answer 1

在Objective C或Swift中访问destinationViewController时使用的正确模式是将其强制转换为正确的子类类型.这将允许您访问子类方法和属性.

所以你应该说

let myDestVC = segue.destinationViewController as MyViewControllerClass 

或者在目标C中

MyViewControllerClass *myDestVC =(MyViewControllerClass *)segue.destinationViewController;