Ios 当我点击uitableviewcell中的uiimageview时，应用程序崩溃

我是iOS开发新手，使用xcode 8.2.1和swift 3。我在视图控制器中获取uitableview，在uitableviewcell中，当我点击it应用程序时获取uiimageview，当我获取按钮并执行按钮操作时，同样的问题也会发生

错误是：

sampleToRunBuild[3752:1620857]-[sampleToRunBuild.TapViewController TappedOnImage:：发送到实例0x141e07bc0的选择器无法识别 2017-04-13 18:18:00.531126样本构建[3752:1620857]* 由于未捕获异常而终止应用程序 “NSInvalidArgumentException”，原因： '-[sampleToRunBuild.TapViewController TappedOnImage:]：无法识别 选择器已发送到实例0x141e07bc0' *第一次抛出调用堆栈：0x18343d1b8 0x181e7455c 0x183444268 0x183441270 0x18333a80c 0x1898b3f80 0x1898b7688 0x18947e73c 0x18931d0f0 0x1898a7680 0x1898a71e0 0x1898a649c 0x18931b30c 0x1892ebda0 0x189ad575c 0x189acf130 0x1833eab5c 0x1833ea4a4 0x1833e80a4 0x1833162b8 0x184dca198 0x1893567fc 0x189351534 0x100065f30 0x1822f95b8 libc++abi.dylib:以未捕获终止 NSException类型的异常

我的代码是：

import UIKit

class TapViewController: UIViewController, UITableViewDelegate, UITableViewDataSource, UIGestureRecognizerDelegate {

    @IBOutlet weak var tableView: UITableView!

    override func viewDidLoad() {
        super.viewDidLoad()

        tableView.delegate = self
        tableView.dataSource = self
    }

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

    func tableView(_ tableView: UITableView, numberOfRowsInSection section: Int) -> Int {
        return 1
    }

    func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell {
        let cell = tableView.dequeueReusableCell(withIdentifier: "mycell")

        let img: UIImageView = cell?.viewWithTag(1) as! UIImageView
        let img2: UIImageView = cell?.viewWithTag(2) as! UIImageView

        img.tag = indexPath.row
        img2.tag = indexPath.row

        img.isUserInteractionEnabled = true
        img2.isUserInteractionEnabled = true

        let tapped:UITapGestureRecognizer = UITapGestureRecognizer(target: self, action: Selector(("TappedOnImage:")))
        tapped.numberOfTapsRequired = 1
        tapped.delegate = self
        img.addGestureRecognizer(tapped)


        let tapped1:UITapGestureRecognizer = UITapGestureRecognizer(target: self, action: Selector(("TappedOnImage:")))
        tapped1.numberOfTapsRequired = 1
        tapped1.delegate = self
        img.addGestureRecognizer(tapped1)

        return cell!
    }

    func TappedOnImage(sender:UITapGestureRecognizer){
        print("tap on imageview")
    }
}

提前感谢…

添加UITapGestureRecognitizer，如下面的代码所示

   let tapGestureRecognizer = UITapGestureRecognizer(target: self, action: #selector(imageTapped(tapGestureRecognizer:)))
    imageView.isUserInteractionEnabled = true
    imageView.addGestureRecognizer(tapGestureRecognizer)


func imageTapped(tapGestureRecognizer: UITapGestureRecognizer)
{
    let tappedImage = tapGestureRecognizer.view as! UIImageView

    // Your action
}

你需要使用

#selector(TapViewController.imageTapped(sender:))

---

如果您使用的是swift 3，请不要再使用选择器，而是使用选择器语法

e、 g

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代码无法工作的原因是您忘记将参数标签sender添加到选择器字符串。这是使用旧语法的缺点之一——它不会在编译时告诉您错误。使用新语法，如果您错误地编写了选择器，那么将出现编译器错误，您甚至不需要关心参数标签，只需要名称即可。

欢迎使用我的答案@KunalMalve

let tapped1:UITapGestureRecognizer = 
    UITapGestureRecognizer(target: self, action: #selector(TappedOnImage))