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;
cell中关键代码展示:
1.添加longpress事件
1 | [ self addGestureRecognizer: [[UILongPressGestureRecognizer alloc]initWithTarget: self action: @selector (longTap:)]];
|
2.处理长按事件
1 2 3 4 5 6 7 8 9 10 11 12 13 | -( 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
1 2 3 4 5 6 7 8 9 10 11 12 13 | #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
1 2 3 4 5 6 7 8 9 | #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
}
|
喜欢我们的内容,关注官方微信公众账号:乐Coding。
|