从iOS桌面视图创建图像弹出

Question

我想创建一个类似于Twitter iOS应用程序的弹出图像，在该图像上单击表格视图中的图像，然后全屏显示图像，以便您可以更好地看到它。

我尝试以下：

  UITapGestureRecognizer *tapGesture = [[UITapGestureRecognizer alloc] initWithTarget:self action:@selector(handleTap:)]; //this is called from the configurecell function 

然后我调用该函数是：

-(void)handleTap:(UIGestureRecognizer *)sender 
{ 
CGPoint tapLocation = [sender locationInView:self.tableView]; 
NSIndexPath *tapIndexPath = [self.tableView indexPathForRowAtPoint:tapLocation]; 
Location *location = [self.fetchedResultsController objectAtIndexPath:tapIndexPath]; 
UIImage *overlayImage = [UIImage imageNamed:@"Default.PNG"]; //default.png is just the placeholder for now 
UIImageView *overlayImageView = [[UIImageView alloc] initWithImage:overlayImage]; 
[self.tableView addSubview:overlayImageView]; 
} 

这将导致图像以弹出，但只从该图像的横置电池上方。

编辑：

非常感谢您的投入。我发现我的问题是，uiimageview被固定到表视图的顶部，因为CGRectMake是0,0等，等（0,0，是左上角）。相反，我们需要将图像视图设置为当前可视区域，像这样

-(void)handleTap:(UIGestureRecognizer *)sender 
{ 
CGPoint tapLocation = [sender locationInView:self.tableView]; 
NSIndexPath *tapIndexPath = [self.tableView indexPathForRowAtPoint:tapLocation]; 
CGFloat distanceFromBottom = [self.tableView contentOffset].y; 
UIImageView *imview=[[UIImageView alloc] initWithFrame:CGRectMake(0, distanceFromBottom, self.view.frame.size.width, self.view.frame.size.height)]; 
imview.backgroundColor=[UIColor blackColor]; 
imview.image = @"Default.PNG"; 
imview.tag=12345; //give tag so we can find it and dismiss it 
imview.contentMode=UIViewContentModeScaleAspectFit; //make sure image isn't stretched 
[self.view addSubview:imview]; //add the subview 
self.tableView.scrollEnabled=NO; //make sure we can't scroll while image is popped out 

//now lets add a gesture recognizer to make sure we can dismiss the pop up uiimageview 
UITapGestureRecognizer *tapGesture = [[UITapGestureRecognizer alloc] initWithTarget:self action:@selector(dismissTap:)]; 
tapGesture.numberOfTapsRequired = 1; 
tapGesture.cancelsTouchesInView=YES; 
imview.userInteractionEnabled = YES; 
[imview addGestureRecognizer:tapGesture]; 
} 


-(void)dismissTap:(UIGestureRecognizer *)sender 
{ 
[[self.view viewWithTag:12345]removeFromSuperview]; 
self.tableView.scrollEnabled=YES; //re enable scrolling 
} 

Answer 1

做这样

UIAlertView *alert = [[UIAlertView alloc] initWithTitle:@"PopUP Title" 
               message:@"This is pop up window/ Alert" 
               delegate:nil 
             cancelButtonTitle:@"OK" 
             otherButtonTitles:nil]; 

UIImageView *tempImageView=[[UIImageView alloc]initWithFrame:CGRectMake(20,20,50,50)]; 
tempImageView.image=[UIImage imageNamed:@"abc.png"]; 

[alert addSubView:tempImageView] 

[alert show]; 

，或者你可以从here github

Answer 2

下载尝试使用：

-(void)handleTap:(UIGestureRecognizer *)sender 
{ 
CGPoint tapLocation = [sender locationInView:self.tableView]; 
NSIndexPath *tapIndexPath = [self.tableView indexPathForRowAtPoint:tapLocation]; 
Location *location = [self.fetchedResultsController objectAtIndexPath:tapIndexPath]; 
UIImage *overlayImage = [UIImage imageNamed:@"Default.PNG"]; //default.png is just the placeholder for now 
UIImageView *overlayImageView = [[UIImageView alloc] initWithImage:overlayImage]; 
overlayImageView.frame = self.view.frame; // or set it as tableview.frame...set frame as per requirement. 
[self.view addSubview:overlayImageView]; /You can also add a button to bring it back also. 
}