與變量語法

Question

打開URL我有一個變量agencyWebsite，當用這種方法點擊應該打開網站標籤：

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

我得到的編譯器警告說：

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

當鏈接被點擊時，應用程序崩潰。有什麼建議麼？

編輯：這是我在做什麼，以使標籤可點擊

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爲UILabel*類型的，你需要訪問它的而不是將對象本身傳遞給URLWithString:。

- (void)website1LblTapped { 

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

調用self.agencyWebsite將返回你UILabel*對象，而self.agencyWebsite.text將返回一個包含從標籤中的文本NSString*對象。