在项目中想要添加一个长按Cell弹出UIMenuController的功能,当我在tableview中添加时是可以弹出来的,然而当在自定义cell.m文件中添加UILongPressGestureRecognizer时,可以触发事件,但是MenuController死活不出来,捣鼓了半天无果,只好google搜索,最后发现忘记实现下面的第三个方法。下面把解决方案记录一下。
要实现长按弹出菜单栏需要做到以下三点:
1.在view(cell)或者viewController中调用-becomeFirstResponder方法;
2.你的view获得或者view controller需要实现 -canBecomeFirstResponder 方法,返回YES;
3.你的view获得或者view controller需要实现-canPerformAction:action withSender:sender 方法来隐藏或者现实响应的item;
1.添加longpress事件
[self addGestureRecognizer: [[UILongPressGestureRecognizer alloc]initWithTarget:self action:@selector(longTap:)]];
2.处理长按事件
-(void)longTap:(UILongPressGestureRecognizer *)longRecognizer { if (longRecognizer.state==UIGestureRecognizerStateBegan) { [self becomeFirstResponder]; UIMenuController *menu=[UIMenuController sharedMenuController]; UIMenuItem *copyItem = [[UIMenuItem alloc] initWithTitle:@"复制" action:@selector(copyItemClicked:)]; UIMenuItem *resendItem = [[UIMenuItem alloc] initWithTitle:@"转发" action:@selector(resendItemClicked:)]; [menu setMenuItems:[NSArray arrayWithObjects:copyItem,resendItem,nil]]; [menu setTargetRect:self.bounds inView:self]; [menu setMenuVisible:YES animated:YES]; } }
3.实现默认方法
#pragma mark 处理action事件 -(BOOL)canPerformAction:(SEL)action withSender:(id)sender{ if(action ==@selector(copyItemClicked:)){ return YES; }else if (action==@selector(resendItemClicked:)){ return YES; } return [super canPerformAction:action withSender:sender]; } #pragma mark 实现成为第一响应者方法 -(BOOL)canBecomeFirstResponder{ return YES; }
4.处理item点击事件
#pragma mark method -(void)resendItemClicked:(id)sender{ NSLog(@"转发"); //通知代理 } -(void)copyItemClicked:(id)sender{ NSLog(@"复制"); // 通知代理 }
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