与变量语法

Question

打开URL我有一个变量agencyWebsite，当用这种方法点击应该打开网站标签：

- (void)website1LblTapped { 
    NSURL *url = [NSURL URLWithString:self.agencyWebsite]; 
    [[UIApplication sharedApplication] openURL:url]; 
} 

我得到的编译器警告说：

Incompatible pointer types sending UILabel* to parameter of type NSString* 

当链接被点击时，应用程序崩溃。有什么建议么？

编辑：这是我在做什么，以使标签可点击

UITapGestureRecognizer* website1LblGesture = [[UITapGestureRecognizer alloc] initWithTarget:self action:@selector(website1LblTapped)]; 
    // if labelView is not set userInteractionEnabled, you must do so 
    [self.agencyWebsite setUserInteractionEnabled:YES]; 
    [self.agencyWebsite addGestureRecognizer:website1LblGesture]; 

是我用得到它的工作

NSURL *url = [NSURL URLWithString:[NSString stringWithFormat:@"http://%@", self.agencyWebsite.text]]; 

Answer 1

如果agencyWebsite为UILabel*类型的，你需要访问它的而不是将对象本身传递给URLWithString:。

- (void)website1LblTapped { 

    NSURL *url = [NSURL URLWithString:self.agencyWebsite.text]; 
    [[UIApplication sharedApplication] openURL:url]; 
} 

调用self.agencyWebsite将返回你UILabel*对象，而self.agencyWebsite.text将返回一个包含从标签中的文本NSString*对象。