如果以编程方式创建，则无法访问 UIImageView

Can#39;t access UIImageView if created programmatically(如果以编程方式创建，则无法访问 UIImageView)

问题描述

如果我通过 Storyboard 创建一个 UIImageView 并将其作为 @property 添加到 ViewController.h 中，我可以访问该 UIImageView 从 ViewController.m 中的任何位置通过 [self UIImageView] 或 self.UIImageView.

If I create a UIImageView via Storyboard and add it as a @property into a ViewController.h, I can access that UIImageView from anywhere in the ViewController.m via [self UIImageView] or self.UIImageView.

但是如果我从 viewDidLoad 以编程方式创建 UIImageView，如下所示，我似乎无法从 viewDidLoad 外部访问它.

But If I create the UIImageView programmatically from viewDidLoad as you see below, I seem to lose the ability to access it from outside viewDidLoad.

UIImageView *prevImgView = [[UIImageView alloc] init];
UIImageView *currImgView = [[UIImageView alloc] init];
UIImageView *nextImgView = [[UIImageView alloc] init];

NSArray *imageViews = [NSArray arrayWithObjects:prevImgView, currImgView, nextImgView, nil];

UIScrollView *scrollView = [[UIScrollView alloc] initWithFrame:self.view.bounds];
[self.view addSubview: scrollView];

CGRect cRect = scrollView.bounds;
UIImageView *cView;

for (int i=0; i<imageViews.count; i++) {
    cView = [imageViews objectAtIndex:i];
    cView.frame = cRect;
    [scrollView addSubview:cView];
    cRect.origin.x += cRect.size.width;
}

所以稍后如果我想修改从 viewDidLoad 创建的任何一个 UIImageView 中显示的图像，我似乎无法访问它.有谁知道我做错了什么?

So later on if I want to modify the image that is being displayed in any one of the UIImageView I've created from viewDidLoad, I can't seem to access it. Does anyone know what I'm doing wrong?

推荐答案

你需要在你的头文件中创建你的 UIImageView，然后它就可以在视图的其余部分中使用.不过，您需要将其添加到 viewDidLoad 中的视图中.

You would need to create your UIImageView in your header file and then it would be available throughout the rest of the view. You will need to add it to the view in viewDidLoad though.

         Header
         UIImageView *image;

         Main // ViewDidLoad
         image = [UIImageView alloc] .....
         [self.view addSubView image];


         Main Function .....
         [image setImage ....

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