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swift - 使用未解析的标识符 'tableView'

转载 作者:行者123 更新时间:2023-11-28 06:35:37 25 4
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I had an a problem with this code, this identifier didn't run with me

import UIKit

class RestaurantDetailViewController: UIViewController {


@IBOutlet var restaurantImageView: UIImageView!
var restaurantImage = ""

override func viewDidLoad () {
super.viewDidLoad ()

restaurantImageView.image = UIImage(named: restaurantImage)
}

override func prepareForSegue(segue: UIStoryboardSegue, sender: AnyObject?) {
if segue.identifier == "showRestaurantDetail" {
if let indexPath = tableView.indexPathSelectedRow {
let destinationController = segue.destinationViewController as! RestaurantDetailViewController
destinationController.restaurantImage = restaurantImageView[indexPath.row]
}
}
}
}

错误是:使用了未解析的标识符“tableView”

寻找解决方案。谢谢,

最佳答案

prepareForSegue 方法显然在错误的类中(复制粘贴错误?)。

它属于 Controller 。

关于swift - 使用未解析的标识符 'tableView',我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/39182209/

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