使用变量语法打开URL地址

有一个变量agencyWebsite和一个标签，应该在点击下面方法的时候打开一个网站。

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

在编译器的警报：

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

再点击网站应用就会崩溃。不知道应该怎么解决？

下面是设置label点击的代码：

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];

运行代码：

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