我知道当错误是''等于'未在此范围内声明时“意味着对象未被正确创建,但是当我创建通知中心用于iOS的widget,使用WeeLoader模板和THEOS进行编译,我得到这个错误:'UITapGestureRecognizer'没有在这个范围内声明。(WeeLoader)'UITapGestureRecognizer'未在此范围内声明
这是我.mm文件:
- (void)loadFullView {
UITapGestureRecognizer *Tap = [[UITapGestureRecognizer alloc] initWithTarget:self action:@selector(handleSingleTap:)];
UIImage *bg = [[UIImage imageWithContentsOfFile:@"/System/Library/WeeAppPlugins/WeeAppTest.bundle/WeeAppBackground.png"] stretchableImageWithLeftCapWidth:5 topCapHeight:71];
UIImageView *bgView = [[UIImageView alloc] initWithImage:bg];
bgView.frame = CGRectMake(0, 0, 316, 71);
bgView.userInteractionEnabled = YES;
[bgView addGestureRecognizer:Tap];
[_view addSubview:bgView];
[bgView release];
[Tap release];
UILabel *lbl = [[UILabel alloc] initWithFrame:CGRectMake(0, 0, 316, 71)];
lbl.backgroundColor = [UIColor clearColor];
lbl.textColor = [UIColor whiteColor];
lbl.text = @"Hello world";
lbl.textAlignment = UITextAlignmentCenter;
[_view addSubview:lbl];
[lbl release];
}
-(void) handleTapGesture:(UIGestureRecognizer *) sender {
}
我在做什么错?我确信我正确地声明了一切,顺便说一下,代码在UITapGestureRecognizer中工作。
谢谢。
谢谢你的回应,我曾试图在init创建'UITapGestureRecognizer',viewDidLoad中,并且还试图在界面中创建它。仍然是同样的错误:(也是方法名称是一个错字写在这个问题,对不起。 – 2013-03-04 01:14:19
@RandomAwesomeGuy你可以发布更多的代码吗? 这对我有用http://pastebin.com/VuH4Gk8G – badger0053 2013-03-04 01:30:27
这里是我的代码http://pastebin.com/fG9EdTsT – 2013-03-04 02:23:59