繁体   English   中英

如何释放带有参数的UIButton的父UIView?

[英]How do I release a parent UIView of/with a UIButton with parameters?

我想知道是否有一种释放UIView的方法,该方法通过subViews UIButton在特定功能之外不存在。 由于无法通过IBAction的按钮提供指向UIView的指针的参数,因此我陷入了困境。

- (IBAction)statsTemp1:(id)sender{


CGRect newSize = CGRectMake(0,13,322,434);
 UIView *overl = [[UIView alloc] initWithFrame:newSize];
 [self.view addSubview:overl];
 [overl setBackgroundColor:[UIColor grayColor]];
 [overl setAlpha:0.85];
 [self doTheGraphIn:overl];
 CGRect btnFrame = CGRectMake(150, 400, 30, 30);
 UIButton *closeStatButton = [UIButton buttonWithType:UIButtonTypeRoundedRect];
 closeStatButton.frame = btnFrame;
 [overl addSubview:closeStatButton];
 [closeStatButton addTarget:self action:@selector(closeStat:) forControlEvents:UIControlEventTouchUpInside];}-(IBAction)closeStat:(id)sender{
 [self release];
    }
    -(IBAction)closeStat:(id)sender{
 [overl release];
    }

只是为了方便:我想用按钮释放(UIView *)覆盖,并且不能事先创建。

感谢您的帮助,我非常感激:)

听起来您想从self.view中删除over1。 (这意味着释放)

尝试一下...(从内存中,请检查)

- (IBAction)statsTemp1:(id)sender{

// Create UIView
     CGRect newSize = CGRectMake(0,13,322,434);
     UIView *overl = [[UIView alloc] initWithFrame:newSize];
     [overl setBackgroundColor:[UIColor grayColor]];
     [overl setAlpha:0.85];
     [over1 setTag:99];

// Add UIButton to overl
       CGRect btnFrame = CGRectMake(150, 400, 30, 30);
       UIButton *closeStatButton = [UIButton buttonWithType:UIButtonTypeRoundedRect];
       closeStatButton.frame = btnFrame;
       [overl addSubview:closeStatButton];
       [closeStatButton addTarget:self action:@selector(closeStat:)   forControlEvents:UIControlEventTouchUpInside];

// Add overl to view
     [self.view addSubview:overl];

// over1 can now be released because self.view has a pointer to it
       [overl release];

// Do God knows what
     [self doTheGraphIn:overl];

 }

// Remove overl from self.view
-(IBAction)closeStat:(id)sender {
     UIView *overl = [self.view viewForTag:99];
     [over1 removeFromSuperView];
  }

// The UIButton is autorelease, so you should be good.

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM