UIView处理手势，与内部UIButton，应该赶上触及

Question

我有一个UIView包含一个UIButton。
了UIView捕获使用下面的方法触摸事件：

[self.view addGestureRecognizer:[[[UITapGestureRecognizer alloc] initWithTarget:self action:@selector(open:)] autorelease]]; 

在某些情况下，我不想做任何事情视图时感动：

- (void) open:(UITapGestureRecognizer *)recognizer 
{ 
    if (self.someCondition == YES) return; 
    // Do very interesting stuff 
} 

的UIButton的链接到方法是这样的：

[self.deleteButton addTarget:self action:@selector(deleteTheWorld:) forControlEvents:UIControlEventTouchUpInside]; 

问题是，当someCondition = YES时，UIButton不响应触摸事件。我该如何回应？

注意：当someCondition == YES时，我只显示UIButton。

Answer 1

首先尝试使用tapRecognizer.cancelsTouchesInView = NO;如果这是不行的，我建议使用UIGestureRecognizerDelegate方法来防止触摸你的看法是这样的：

- (BOOL)gestureRecognizer:(UIGestureRecognizer *)gestureRecognizer shouldReceiveTouch:(UITouch *)touch { 
     if ([touch.view isKindOfClass:[UIButton class]]) { 
       return YES; // button is touched then, yes accept the touch 
     } 
     else if (self.someContiditon == YES) { 
      return NO; // we don't want to receive touch on the view because of the condition 
     } 
     else { 
      return YES; // tap detected on view but no condition is required 
     } 
    } 

Answer 2

我认为你最好的选择是管理点击你的open选择器中的按钮。

只是把像

CGPoint location = [recognizer locationInView:self.view]; 
if(self.someCondition == YES) 
    if(recognizer.state == UIGestureRecognizerStateRecognized && 
     CGRectContainsPoint(self.deleteButton.frame, location)) 
     [self deleteTheWorld]; 
    else 
     return; 

代替

- (void) open:(UITapGestureRecognizer *)recognizer 
{ 
    if (self.someCondition == YES) return; 
    // Do very interesting stuff 
} 

，当然你也不需要注册按钮的目标行动，那么！