我在 uitableview 的一个单元格内有两个图像，这些图像显示为来自外部服务器的图像，并且它们的每个标签都有一个该图像代表的项目 ID，如果我单击该图像，我需要将用户移动到新 View 显示该项目详细信息的 Controller ，我强加了一个问题，用户需要双击才能显示详细信息而不是单击，以下是我的代码:

func tableView(_ tableView: UITableView, didSelectRowAt indexPath: IndexPath) {
        let cell = self.tableView.cellForRow(at: indexPath as IndexPath) as! prodctCell
        Id1stMove =  cell.image1st.tag
        let tapGesture = UITapGestureRecognizer (target: self, action: #selector(ItemsController.imgTap))
        cell.image1st.addGestureRecognizer(tapGesture)
        cell.image1st.isUserInteractionEnabled = true

        let cell1 = self.tableView.cellForRow(at: indexPath as IndexPath) as! prodctCell
        Id2ndMove =  cell1.image2nd.tag
        let tapGesture1 = UITapGestureRecognizer (target: self, action: #selector(ItemsController.imgTap1))
        cell1.image2nd.addGestureRecognizer(tapGesture1)
    }


func imgTap()
    {
        let secondViewController = self.storyboard?.instantiateViewController(withIdentifier: "testViewController") as? testViewController
            let navController = UINavigationController(rootViewController: secondViewController!)
            navController.setViewControllers([secondViewController!], animated:true)
            self.revealViewController().setFront(navController, animated: true)
            revealViewController().pushFrontViewController(navController, animated: true)
        secondViewController?.movmentId = Id1stMove
        updateCount(itemId: Id1stMove)

    }

最佳答案

昨天我自己创建了示例并进行了尝试。我得到了解决方案。但是由于我有一些工作，我无法立即发布我的答案。

现在我会给你我的答案。我不希望我的以下答案有声誉。

当您第一次点击或点按图片时，它会导航。

didSelectRowAt方法中的imageView不需要添加TapGestureRecognizer，需要在cellForRowAt方法中为imageView添加TapGestureRecognizer。

import UIKit

class ViewController: UIViewController,UITableViewDelegate,UITableViewDataSource {
    let mobiles: [String] = ["iPhone", "Android"]
    let images: [String] = ["iPhone.png", "android.png"]
    @IBOutlet var tableViewImageTapping: UITableView!
    override func viewDidLoad() {
        super.viewDidLoad()
        // Do any additional setup after loading the view, typically from a nib.
    }

    override func didReceiveMemoryWarning() {
        super.didReceiveMemoryWarning()
        // Dispose of any resources that can be recreated.
    }

    // number of rows in table view
    func tableView(_ tableView: UITableView, numberOfRowsInSection section: Int) -> Int {
        return self.mobiles.count
    }

    // create a cell for each table view row
    func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell
    {
        var cell:UITableViewCell? = tableView.dequeueReusableCell(withIdentifier: "cell")
        if (cell == nil) {
            cell = UITableViewCell(style:UITableViewCellStyle.default, reuseIdentifier:"cell")
        }
        cell?.textLabel?.text = self.mobiles[indexPath.row]
        let strImageName = images[indexPath.row]
        cell?.imageView?.image = UIImage(named: strImageName)
        cell?.imageView?.tag = indexPath.row
        let tapGestureRecognizer = UITapGestureRecognizer(target: self, action: #selector(imageTapped(tapGestureRecognizer:)))
        tapGestureRecognizer.numberOfTapsRequired = 1
        cell?.imageView?.isUserInteractionEnabled = true
        cell?.imageView?.addGestureRecognizer(tapGestureRecognizer)
        return cell!
    }

    // method to run when table view cell is selected
    func tableView(_ tableView: UITableView, didSelectRowAt indexPath: IndexPath) {
        print("You tapped table view cell index is \(indexPath.row).")
    }

    // method to run when imageview is tapped
    func imageTapped(tapGestureRecognizer: UITapGestureRecognizer)
    {
        let imgView = tapGestureRecognizer.view as! UIImageView
        print("your taped image view tag is : \(imgView.tag)")
        if (imgView.tag == 0) //Give your image View tag
        {
            //navigate to next view
        }
        else{

        }
    }
} 

输出截图

打印结果为

当你第一次点击第一个图片时

然后在第一次点击第二张图片的时候

关于ios - 单击 UITableView 单元格内的图像，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/43188568/