如何在 UiButton 操作中传递任何类型的参数? [英] how can i pass any kind of parameter in UiButton action?

问题描述

我想通过按钮动作传递一个 url，因为我有 10 个按钮，它们是通过代码动态创建的，点击后为每个相应的按钮分配一个特定的 url.这是我的代码

I want to pass a url through button action, because i have 10 buttons which are created through code dynamically and on clicking then there a specific url are assign for each corresponding button. Here is my code

NSString *linkUrl = [NSString stringWithFormat:@"%@",[AllrecordG objectForKey:@"Link"]];
[leftLabelG addTarget:self action:@selector(clickMe:linkUrl:) forControlEvents:UIControlEventTouchUpInside];

-(void)clickMe:(UIButton *)sender{
}

但在这里我收到警告为未使用的变量 linkUrl".

But here I get warning as "unused variable linkUrl".

我研究了不同的文章，但有人告诉我不可能传递参数链接.任何人都可以告诉我如何为每个按钮操作传递 url 以及如何检索这些值.在 clickMe 定义中.

I have studied different articles but somebody told that it is not possible to pass argument linke that. Anybody can tell me that how can i pass url for each button action and also how I retrieve those values. in the clickMe definition.

提前致谢.

推荐答案

像这样子类 UIButton :

Subclass UIButton like that :

MyButton.h

@interface MyButton : UIButton
{
     NSString *_linkUrl;
}

@property (nonatomic, retain) NSString *linkUrl

MyButton.m

@implementation MyButton
@synthesize linkUrl = _linkUrl

传递linkUrl:

NSString *linkUrl = [NSString stringWithFormat:@"%@",[AllrecordG objectForKey:@"Link"]];
[leftLabelG addTarget:self action:@selector(clickMe:) forControlEvents:UIControlEventTouchUpInside];
[leftLabelG setLinkUrl:linkUrl];

现在您可以像这样在操作中获取linkUrl :

Now you can get the linkUrl in your action like that :

-(void)clickMe:(MyButton *)sender{
      NSString *url = sender.linkUrl;
}

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