快速将参数传递给#selector方法[重复]

Question

我想在单击图像时将多个参数传递给函数。 这是我的代码

var param1 = 120
var param2 = "hello"
var param3 =  "world"

let image: UIImage = UIImage(named: "imageName")!
 bgImage = UIImageView(image: image)

let singleTap = UITapGestureRecognizer(target: self, action:#selector(WelcomeViewController.tapDetected(_:secondParam:thirdParam:)))
singleTap.numberOfTapsRequired = 1
bgImage!.userInteractionEnabled = true
bgImage!.addGestureRecognizer(singleTap)

调用函数

func tapDetected(firstParam: Int, secondParam: String, thirdParam: String) {
print("Single Tap on imageview")
            print(firstParam)  //print 120
            print(secondParam) // print hello
            print(thirdParam) /print world
        }

如何传递参数以便获得正确的值？

Answer 1

你不能。来自文档：

手势识别器有一个或多个与之关联的目标-动作对。
...
调用的操作方法必须符合以下签名之一：

- (void)handleGesture;
- (void)handleGesture:(UIGestureRecognizer *)gestureRecognizer;

您可以使用实例变量来传递参数。

Answer 2

在 swift2 中，您不能在操作中传递参数。您可以只使用类属性：

var param1 = 120
var param2 = "hello"
var param3 = "world"
let image: UIImage = UIImage(named: "imageName")!
var bgImage: UIImageView?

override func viewDidLoad() {
    super.viewDidLoad()

    bgImage = UIImageView(image: image)
    let singleTap = UITapGestureRecognizer(target: self, action: #selector(WelcomeViewController.tapDetected))
    singleTap.numberOfTapsRequired = 1
    bgImage!.userInteractionEnabled = true
    bgImage!.addGestureRecognizer(singleTap)
}

调用函数

func tapDetected() {
    print("Single Tap on imageview")
    print(param1) // print 120
    print(param2) // print hello
    print(param3) // print world
}