ios - 在 objective-c 中创建 url 方法的最佳方法

Question

我的应用中有两个仅包含 URL 的标签:

-(void)openURLA{
    NSString *url = @"http://urla.com";
    [[UIApplication sharedApplication] openURL: [NSURL URLWithString:url]];
}
-(void)openURLB{
    NSString *url = @"http://urlb.com";
    [[UIApplication sharedApplication] openURL: [NSURL URLWithString:url]];
}

并且此代码位于现有方法中:

UITapGestureRecognizer *gra = [[UITapGestureRecognizer alloc] initWithTarget:self action:@selector(openURLA)];
UITapGestureRecognizer *grb = [[UITapGestureRecognizer alloc] initWithTarget:self action:@selector(openURLB)];

当用户点击这些标签之一时，openURL 方法运行良好，URL 在 Safari 中打开。

我想知道如何只创建一种方法来打开 URL 并传递包含 label1.text 或 label2.text 值的参数？

我不是 100% 确定从哪里开始做这件事，所以我会感谢一些帮助。

Answer 1

已编辑:

遵循整个代码:

UILabel  * label1 = [[UILabel alloc] initWithFrame:CGRectMake(40, 70, 300, 50)];
    label1.backgroundColor = [UIColor redColor];
    label1.userInteractionEnabled = YES;
    label1.textColor=[UIColor whiteColor];
    label1.text = @"http://urla.com";
    [self.view addSubview:label1];

    UILabel  * label2 = [[UILabel alloc] initWithFrame:CGRectMake(40, 130, 300, 50)];
    label2.backgroundColor = [UIColor redColor];
    label2.userInteractionEnabled = YES;
    label2.textColor=[UIColor whiteColor];
    label2.text =  @"http://urlb.com";
    [self.view addSubview:label2];


    UITapGestureRecognizer *gsture1 = [[UITapGestureRecognizer alloc] initWithTarget:self action:@selector(openURLS:)];
    [label1 addGestureRecognizer:gsture1];

    UITapGestureRecognizer *gesture2 = [[UITapGestureRecognizer alloc] initWithTarget:self action:@selector(openURLS:)];
    [label2 addGestureRecognizer:gesture2];

并调用UITapGestureRecognizer的方法

- (void)openURLS:(UITapGestureRecognizer*)gesture
{
    UILabel *lblUrl=(UILabel *)[gesture view];
    NSLog(@"%@", lblUrl.text); // here you get your selected label text.
    [[UIApplication sharedApplication] openURL: [NSURL URLWithString:lblUrl.text]];
}