在 UILabel 中处理触摸事件并将其连接到 IBAction

好的，我用 Interface Builder 创建了一个 UILabel，它显示一些“点击开始”的默认文本。

当用户点击 UILabel时，我希望它触发一个 IBAction 方法: -(IBAction)next;，它更新标签上的文字，以表示一些新的内容。
如果这允许我简单地将一个连接从我的方法拖动到我的标签，然后在内部选择触摸，就像使用一个按钮一样，那将是非常方便的。可惜没有雪茄。

所以不管怎样，我想我的问题是，我是否必须继承 UILabel才能让它工作？或者有什么方法可以在标签上拖动一个按钮，但使其0% 不透明。还是我错过了一个更简单的解决方案？

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小开

UILabel inherits from UIView which inherits from UIResponder. All UIresponder objects can handle touch events. So in your class file which knows about your view (which contains the UIlabel) implement: 

-(void)touchesBegan:(NSSet*)touches withEvent:(UIEvent*)event;

In interface builder set the UILabel's tag value. when touches occur in your touchesBegan method, check the tag value of the view to which the tag belongs:

UITouch *touch = [touches anyObject];


if(touch.view.tag == MY_TAG_VAL)
label.text = @"new text";

You connect your code in your class file with the UILabel object in interface builder by declaring your UILabel instance variable with the IBOutlet prefix:

IBOutlet UILabel *label;

Then in interface builder you can connect them up.

小开

You can use a UIButton, make it transparent, i.e. custom type without an image, and add a UILabel on it (centered). Then wire up the normal button events.

小开
最佳答案

Check it out:

UILabel *label = ...
label.userInteractionEnabled = YES;
UITapGestureRecognizer *tapGesture =
[[UITapGestureRecognizer alloc] initWithTarget:self
action:@selector(labelTap)];
[label addGestureRecognizer:tapGesture];

The trick is to enable user interaction.

小开

Swift 3

You have an IBOutlet

@IBOutlet var label: UILabel!

In which you enable user interaction and add a gesture recognizer

label.isUserInteractionEnabled = true
let tapGesture = UITapGestureRecognizer(target: self, action: #selector(userDidTapLabel(tapGestureRecognizer:)))
label.addGestureRecognizer(tapGesture)

And finally, handle the tap

func userDidTapLabel(tapGestureRecognizer: UITapGestureRecognizer) {
// Your code goes here
}

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