Swift addsubview并将其删除

Question

我想添加子视图并一键删除。这是我的代码：

/ 添加子视图 /

var testView: UIView = UIView(frame: CGRectMake(0, 0, 320, 568))
testView.backgroundColor = UIColor.blueColor()
testView.alpha = 0.5
testView.tag = 100
super.view.userInteractionEnabled = false
self.view.userInteractionEnabled = true
self.view.addSubview(testView)

/ 删除子视图 /

override func touchesBegan(touches: NSSet, withEvent event: UIEvent) {
    let touch = touches.anyObject() as UITouch
    let point = touch.locationInView(self.view)

    if(testView.tag==100){
        println("Tag 100")
        testView.removeFromSuperview()
    }
    else{
        println("tag not found")
    }

}

但是删除它不起作用有人可以帮助我吗？谢谢！

Answer 1

感谢帮助。这是解决方案：我创建了子视图，并添加了一个手势来删除它

@IBAction func infoView(sender: UIButton) {
    var testView: UIView = UIView(frame: CGRectMake(0, 0, 320, 568))
    testView.backgroundColor = UIColor.blueColor()
    testView.alpha = 0.5
    testView.tag = 100
    testView.userInteractionEnabled = true
    self.view.addSubview(testView)

    let aSelector : Selector = "removeSubview"
    let tapGesture = UITapGestureRecognizer(target:self, action: aSelector)
    testView.addGestureRecognizer(tapGesture)
}

func removeSubview(){
    println("Start remove sibview")
    if let viewWithTag = self.view.viewWithTag(100) {
        viewWithTag.removeFromSuperview()
    }else{
        println("No!")
    }
}

更新：

迅捷3+

@IBAction func infoView(sender: UIButton) {
    let testView: UIView = UIView(frame: CGRect(x: 0, y: 0, width: 320, height: 568))
    testView.backgroundColor = .blue
    testView.alpha = 0.5
    testView.tag = 100
    testView.isUserInteractionEnabled = true
    self.view.addSubview(testView)

    let aSelector : Selector = #selector(GasMapViewController.removeSubview)
    let tapGesture = UITapGestureRecognizer(target:self, action: aSelector)
    testView.addGestureRecognizer(tapGesture)
}

func removeSubview(){
    print("Start remove sibview")
    if let viewWithTag = self.view.viewWithTag(100) {
        viewWithTag.removeFromSuperview()
    }else{
        print("No!")
    }
}