UILable链接打开safari浏览器

有自定义UITableViewCell 其中包含两枚UILabel。表格单元显示information/text。一些单元设置：

cell.myTextlabel.text = @"http://www.google.de"

我想在点击这些text链接时，safari浏览器可以打开网页。应该怎么实现？

设置userInteractionEnabled 为 YES 。添加一个姿势识别器：

myLabel.userInteractionEnabled = YES;
UITapGestureRecognizer *gestureRec = [[UITapGestureRecognizer alloc] initWithTarget:self action:@selector(openUrl:)];
gestureRec.numberOfTouchesRequired = 1;
gestureRec.numberOfTapsRequired = 1;
[myLabel addGestureRecognizer:gestureRec];
[gestureRec release];

然后实习操作方法：

- (void)openUrl:(id)sender
{
    UIGestureRecognizer *rec = (UIGestureRecognizer *)sender;

    id hitLabel = [self.view hitTest:[rec locationInView:self.view] withEvent:UIEventTypeTouches];

    if ([hitLabel isKindOfClass:[UILabel class]]) {
        [[UIApplication sharedApplication] openURL:[NSURL URLWithString:((UILabel *)hitLabel).text]];
    }
}

如果你用UITextView代替UILabel，它会自动检测链接。

设置视图的dataDetectorTypes到UIDataDetectorTypeLink